#set document(title: "9.3 Independent Samples for Two Means", author: "Kathryn Kozak") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 9.3#h(0.6em)Independent Samples for Two Means This section will look at how to analyze when two samples are collected that are independent. As with all other hypothesis tests and confidence intervals, the process is the same though the formulas and assumptions are different. The only difference with the independent t-test, as opposed to the other tests that have been done, is that there are actually two different formulas to use depending on if a particular assumption is met or not. === Hypothesis Test for Independent t-Test (2-Sample t-Test) + State the random variables and the parameters in words. #linebreak() #math.equation(block: false, alt: "x sub 1")[$x_(1)$] = random variable 1 #linebreak() #math.equation(block: false, alt: "x sub 2")[$x_(2)$] = random variable 2 #linebreak() #math.equation(block: false, alt: "μ sub 1")[$μ_(1)$] = mean of random variable 1 #linebreak() #math.equation(block: false, alt: "μ sub 2")[$μ_(2)$]= mean of random variable 2 + State the null and alternative hypotheses and the level of significance The normal hypotheses would be #linebreak() #math.equation(block: false, alt: "H sub o : μ sub 1 equals μ sub 2 or H sub o : μ sub 1 minus μ sub 2 equals 0; H sub A : μ sub 1 less than μ sub 2 H sub A : μ sub 1 minus μ sub 2 less than 0; H sub A : μ sub 1 greater than μ sub 2 H sub A : μ sub 1 minus μ sub 2 greater than 0; H sub A : μ sub 1 not equal to μ sub 2 H sub A : μ sub 1 minus μ sub 2 not equal to 0")[$H_(o) : μ_(1) = μ_(2) " or " #h(1em) H_(o) : μ_(1) − μ_(2) = 0 \ H_(A) : μ_(1) < μ_(2) #h(1em) #h(1em) #h(0.222em) H_(A) : μ_(1) − μ_(2) < 0 \ H_(A) : μ_(1) > μ_(2) #h(1em) #h(1em) #h(0.222em) H_(A) : μ_(1) − μ_(2) > 0 \ H_(A) : μ_(1) ≠ μ_(2) #h(1em) #h(1em) #h(0.222em) H_(A) : μ_(1) − μ_(2) ≠ 0$] #linebreak() Also, state your #math.equation(block: false, alt: "α")[$α$] level here. + State and check the assumptions for the hypothesis test + A random sample of size #math.equation(block: false, alt: "n sub 1")[$n_(1)$] is taken from population 1. A random sample of size #math.equation(block: false, alt: "n sub 2")[$n_(2)$] is taken from population 2. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] The samples do not need to be the same size, but the test is more robust if they are. ] + The two samples are independent. + Population 1 is normally distributed. Population 2 is normally distributed. Just as before, the t-test is fairly robust to the assumption if the sample size is large. This means that if this assumption isn’t met, but your sample sizes are quite large (over 30), then the results of the t-test are valid. + The population variances are unknown and not assumed to be equal. The old assumption is that the variances are equal. However, this assumption is no longer an assumption that most statisticians use. This is because it isn’t really realistic to assume that the variances are equal. So we will just assume the assumption of the variances being unknown and not assumed to be equal is true, and it will not be checked. + Find the sample statistic, test statistic, and p-value #linebreak() Sample Statistic: #linebreak() Calculate #math.equation(block: false, alt: "x ― sub 1 , x ― sub 2 , s sub 1 , s sub 2 , n sub 1 , n sub 2")[$limits(x)^(―)_(1) , limits(x)^(―)_(2) , s_(1) , s_(2) , n_(1) , n_(2)$] #linebreak() Test Statistic: #linebreak() Since the assumption that #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$] isn’t being satisfied, then #linebreak() #math.equation(block: false, alt: "t equals the fraction open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus open parenthesis μ sub 1 minus μ sub 2 close parenthesis over the square root of n sub 1 s 1 2 plus n sub 2 s 2 2")[$t = frac(( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − ( μ_(1) − μ_(2) ), sqrt(limits(n_(1))^(s_(1)^(2)) + limits(n_(2))^(s_(2)^(2))))$] #linebreak() Usually #math.equation(block: false, alt: "μ sub 1 minus μ sub 2 equals 0")[$μ_(1) − μ_(2) = 0$], since #math.equation(block: false, alt: "H sub o : μ sub 1 minus μ sub 2 equals 0")[$H_(o) : μ_(1) − μ_(2) = 0$] #linebreak() Degrees of freedom: (the Welch–Satterthwaite equation) #linebreak() #math.equation(block: false, alt: "d f equals the fraction open parenthesis A plus B close parenthesis squared over the fraction A squared over n sub 1 minus 1 plus the fraction B squared over n sub 2 minus 1")[$d f = frac(( A + B )^(2), frac(A^(2), n_(1) − 1) + frac(B^(2), n_(2) − 1))$] #linebreak() where #math.equation(block: false, alt: "A equals the fraction s 1 2 over n sub 1 and B equals the fraction s 2 2 over n sub 2")[$A = frac(s_(1)^(2), n_(1)) " and " B = frac(s_(2)^(2), n_(2))$] #linebreak() p-value: #linebreak() Using the TI-83/84: tcdf(lower limit, upper limit, #emph[df]) Using R: pt(t, #emph[df]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] If #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 less than 0")[$H_(A) : μ_(1) − μ_(2) < 0$], then lower limit is #math.equation(block: false, alt: "minus 1 E 99")[$− 1 E 99$] and upper limit is your test statistic. If #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 greater than 0")[$H_(A) : μ_(1) − μ_(2) > 0$], then lower limit is your test statistic and the upper limit is #math.equation(block: false, alt: "1 E 99")[$1 E 99$]. If #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 not equal to 0")[$H_(A) : μ_(1) − μ_(2) ≠ 0$], then find the p-value for #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 less than 0")[$H_(A) : μ_(1) − μ_(2) < 0$], and multiply by 2. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] If #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 less than 0")[$H_(A) : μ_(1) − μ_(2) < 0$], then use pt(t, #emph[df]). If #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 greater than 0")[$H_(A) : μ_(1) − μ_(2) > 0$], then use 1 - pt(t, #emph[df]). If #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 not equal to 0")[$H_(A) : μ_(1) − μ_(2) ≠ 0$], then find the p-value for #math.equation(block: false, alt: "H sub A : μ sub 1 minus μ sub 2 less than 0")[$H_(A) : μ_(1) − μ_(2) < 0$], and multiply by 2. ] + Conclusion #linebreak() This is where you write reject #math.equation(block: false, alt: "H sub o")[$H_(o)$] or fail to reject #math.equation(block: false, alt: "H sub o")[$H_(o)$]. The rule is: if the p-value \< #math.equation(block: false, alt: "α")[$α$], then reject #math.equation(block: false, alt: "H sub o")[$H_(o)$]. If the p-value #math.equation(block: false, alt: "greater than or equal to α")[$≥ α$], then fail to reject #math.equation(block: false, alt: "H sub o")[$H_(o)$] + Interpretation This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show #math.equation(block: false, alt: "H sub A")[$H_(A)$] is true, or you do not have enough evidence to show #math.equation(block: false, alt: "H sub A")[$H_(A)$] is true. === Confidence Interval for the Difference in Means from Two Independent Samples (2 Samp T-Int) The confidence interval for the difference in means has the same random variables and means and the same assumptions as the hypothesis test for independent samples. If you have already completed the hypothesis test, then you do not need to state them again. If you haven’t completed the hypothesis test, then state the random variables and means and state and check the assumptions before completing the confidence interval step. + Find the sample statistic and confidence interval #linebreak() Sample Statistic: #linebreak() Calculate Confidence Interval: #math.equation(block: false, alt: "x ― sub 1 , x ― sub 2 , s sub 1 , s sub 2 , n sub 1 , n sub 2")[$limits(x)^(―)_(1) , limits(x)^(―)_(2) , s_(1) , s_(2) , n_(1) , n_(2)$] #linebreak() The confidence interval estimate of the difference is #math.equation(block: false, alt: "μ sub 1 minus μ sub 2")[$μ_(1) − μ_(2)$] #linebreak() Since the assumption that #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$] isn’t being satisfied, then #linebreak() #math.equation(block: false, alt: "open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus E less than μ sub 1 minus μ sub 2 less than open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis plus E")[$( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − E < μ_(1) − μ_(2) < ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) + E$] #linebreak() where #math.equation(block: false, alt: "E equals t sub c the square root of the fraction s 1 2 over n sub 1 plus the fraction s 2 2 over n sub 2")[$E = t_(c) sqrt(frac(s_(1)^(2), n_(1)) + frac(s_(2)^(2), n_(2)))$] #linebreak() where #math.equation(block: false, alt: "t sub c")[$t_(c)$] is the critical value with degrees of freedom: #linebreak() Degrees of freedom: (the Welch–Satterthwaite equation) #linebreak() #math.equation(block: false, alt: "d f equals the fraction open parenthesis A plus B close parenthesis squared over the fraction A squared over n sub 1 minus 1 plus the fraction B squared over n sub 2 minus 1")[$d f = frac(( A + B )^(2), frac(A^(2), n_(1) − 1) + frac(B^(2), n_(2) − 1))$] #linebreak() where #math.equation(block: false, alt: "A equals the fraction s 1 2 over n sub 1 and B equals the fraction s 2 2 over n sub 2")[$A = frac(s_(1)^(2), n_(1)) " and " B = frac(s_(2)^(2), n_(2))$] + Statistical Interpretation: In general this looks like, “there is a C% chance that #math.equation(block: false, alt: "open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus E less than μ sub 1 minus μ sub 2 less than open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis plus E")[$( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − E < μ_(1) − μ_(2) < ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) + E$] contains the true mean difference.” + Real World Interpretation: This is where you state what interval contains the true difference in means, though often you state how much more (or less) the first mean is from the second mean. The critical value is a value from the Student’s t-distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the difference in sample means, and the interval has a probability of containing the true difference in means, then you can think of this as the statement #math.equation(block: false, alt: "P open parenthesis open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus E less than μ sub 1 minus μ sub 2 less than open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis plus E close parenthesis equals C")[$P ( ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − E < μ_(1) − μ_(2) < ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) + E ) = C$]. To find the critical value you use table A.2 in the Appendix. ==== How to check the assumptions of two sample t-test and confidence interval: In order for the t-test or confidence interval to be valid, the assumptions of the test must be true. So whenever you run a t-test or confidence interval, you must make sure the assumptions are true. So you need to check them. Here is how you do this: + For the random sample assumption, describe how you took the two samples. Make sure your sampling technique is random for both samples. + For the independent assumption, describe how they are independent samples. + For the assumption about each population being normally distributed, remember the process of assessing normality from chapter 6. Make sure you assess each sample separately. + You do not need to check the equal variance assumption since it is not being assumed. #examplebox("Example 1")[hypothesis test for two means][ The cholesterol level of patients who had heart attacks was measured two days after the heart attack. The researchers want to see if patients who have heart attacks have higher cholesterol levels over healthy people, so they also measured the cholesterol level of healthy adults who show no signs of heart disease. The data is in#emph[Table #math.equation(block: false, alt: "1")[$1$]] ("Cholesterol levels after," 2013). Do the data show that people who have had heart attacks have higher cholesterol levels over patients that have not had heart attacks? Test at the 1% level. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The 2-SampTTest on the cholesterol data] Heart-attack patients load into L1 (n = 28) and healthy adults into L2 (n = 30). The test opens right-tailed at α = 0.01 with the variances not pooled, matching the calculator setup. - 2-SampTTest: t = 6.145, p-value ≈ 1.9×10⁻⁷ - 2-SampTInt at 99% (Welch df): ≈ (33.96, 87.63) ] #figure(table( columns: 2, align: left, inset: 6pt, table.header([Cholesterol Level of Heart Attack Patients], [Cholesterol Level of Healthy Individual]), [270], [196], [236], [232], [210], [200], [142], [242], [280], [206], [272], [178], [160], [184], [220], [198], [226], [160], [242], [182], [186], [182], [266], [198], [206], [182], [318], [238], [294], [198], [282], [188], [234], [166], [224], [204], [276], [182], [282], [178], [360], [212], [310], [164], [280], [230], [278], [186], [288], [162], [288], [182], [244], [218], [236], [170], [], [200], [], [176], )) + State the random variables and the parameters in words. + State the null and alternative hypotheses and the level of significance. + State and check the assumptions for the hypothesis test. + Find the sample statistic, test statistic, and p-value. + Conclusion + Interpretation #solutionbox[ 1. #math.equation(block: false, alt: "x sub 1")[$x_(1)$] = Cholesterol level of patients who had a heart attack #math.equation(block: false, alt: "x sub 2")[$x_(2)$] = Cholesterol level of healthy individuals #math.equation(block: false, alt: "μ sub 1")[$μ_(1)$] = mean cholesterol level of patients who had a heart attack #math.equation(block: false, alt: "μ sub 2")[$μ_(2)$] = mean cholesterol level of healthy individuals 2. The normal hypotheses would be #math.equation(block: true, alt: "H sub o : μ sub 1 equals μ sub 2 or H sub o : μ sub 1 minus μ sub 2 equals 0; H sub A : μ sub 1 greater than μ sub 2 H sub A : μ sub 1 minus μ sub 2 greater than 0; α equals 0.01")[$H_(o) : μ_(1) = μ_(2) " or " #h(1em) H_(o) : μ_(1) − μ_(2) = 0 \ H_(A) : μ_(1) > μ_(2) #h(1em) #h(1em) #h(0.222em) H_(A) : μ_(1) − μ_(2) > 0 \ α = 0.01$] 3. + A random sample of 28 cholesterol levels of patients who had a heart attack is taken. A random sample of 30 cholesterol levels of healthy individuals is taken. The problem does not state if either sample was randomly selected. So this assumption may not be valid. + The two samples are independent. This is because either they were dealing with patients who had heart attacks or healthy individuals. + Population of all cholesterol levels of patients who had a heart attack is normally distributed. Population of all cholesterol levels of healthy individuals is normally distributed. Patients who had heart attacks: #figure(figph[Histogram of cholesterol levels for heart attack patients, with most observations between 200 and 300 and a peak frequency of 12 in the 250–300 bin.], alt: "Histogram of cholesterol levels for heart attack patients, with most observations between 200 and 300 and a peak frequency of 12 in the 250–300 bin.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Histogram of Cholesterol Levels of Patients who had Heart Attacks]) This looks somewhat bell shaped. #figure(figph[A horizontal boxplot showing cholesterol levels for 28 heart attack patients, with values ranging from 142 to about 360 and a median near 268.], alt: "A horizontal boxplot showing cholesterol levels for 28 heart attack patients, with values ranging from 142 to about 360 and a median near 268.", caption: [Figure #math.equation(block: false, alt: "2")[$2$]: Modified Box Plot of Cholesterol Levels of Patients who had Heart Attacks]) There are no outliers #figure(figph[A normal Q-Q plot of the 28 heart-attack cholesterol measurements, with ordered sample values from 142 to about 360 plotted against theoretical normal quantiles.], alt: "A normal Q-Q plot of the 28 heart-attack cholesterol measurements, with ordered sample values from 142 to about 360 plotted against theoretical normal quantiles.", caption: [Figure #math.equation(block: false, alt: "3")[$3$]: Normal Quantile Plot of Cholesterol Levels of Patients who had Heart Attacks]) This looks somewhat linear. So, the population of all cholesterol levels of patients who had heart attacks is probably somewhat normally distributed. Healthy individuals: #figure(figph[Histogram of cholesterol levels for healthy adults, with values ranging from 160 to 242.], alt: "Histogram of cholesterol levels for healthy adults, with values ranging from 160 to 242.", caption: [Figure #math.equation(block: false, alt: "4")[$4$]: Histogram of Cholesterol Levels of Healthy Individuals]) This does not look bell shaped. #figure(figph[A horizontal boxplot of cholesterol levels for healthy adults, with values ranging from 160 to 242 and a median near 187.], alt: "A horizontal boxplot of cholesterol levels for healthy adults, with values ranging from 160 to 242 and a median near 187.", caption: [Figure #math.equation(block: false, alt: "5")[$5$]: Modified Box Plot of Cholesterol Levels of Healthy Individuals]) There are no outliers. #figure(figph[A normal Q-Q plot of the 30 healthy-adult cholesterol measurements, with sample values from 160 to 242 plotted against theoretical normal quantiles.], alt: "A normal Q-Q plot of the 30 healthy-adult cholesterol measurements, with sample values from 160 to 242 plotted against theoretical normal quantiles.", caption: [Figure #math.equation(block: false, alt: "6")[$6$]: Normal Quantile Plot of Cholesterol Levels of Healthy Individuals]) This doesn't look linear. So, the population of all cholesterol levels of healthy individuals is probably not normally distributed. This assumption is not valid for the second sample. Since the sample is fairly large, and the t-test is robust, it may not be an issue. However, just realize that the conclusions of the test may not be valid. 4. Sample Statistic: #math.equation(block: true, alt: "x ― sub 1 approximately equals 252.32 , x ― sub 2 approximately equals 193.13 , s sub 1 approximately equals 47.0642 , s sub 2 approximately equals 22.3000 , n sub 1 equals 28 , n sub 2 equals 30")[$limits(x)^(―)_(1) ≈ 252.32 , limits(x)^(―)_(2) ≈ 193.13 , s_(1) ≈ 47.0642 , s_(2) ≈ 22.3000 , n_(1) = 28 , n_(2) = 30$] Test Statistic: #math.equation(block: true, alt: "t equals the fraction open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus open parenthesis μ sub 1 minus μ sub 2 close parenthesis over the square root of the fraction s 1 2 over n sub 1 plus the fraction s 2 2 over n sub 2")[$t = frac(( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − ( μ_(1) − μ_(2) ), sqrt(frac(s_(1)^(2), n_(1)) + frac(s_(2)^(2), n_(2))))$] #math.equation(block: true, alt: "equals the fraction open parenthesis 252.32 minus 193.13 close parenthesis minus 0 over the square root of the fraction 47.0642 squared over 28 plus the fraction 22.3000 squared over 30")[$= frac(( 252.32 − 193.13 ) − 0, sqrt(frac(47.0642^(2), 28) + frac(22.3000^(2), 30)))$] #math.equation(block: true, alt: "approximately equals 6.051")[$≈ 6.051$] Degrees of freedom: (the Welch-Satterthwaite equation) #math.equation(block: true, alt: "A equals the fraction s 1 2 over n sub 1 equals the fraction 47.0642 squared over 28 approximately equals 79.1085")[$A = frac(s_(1)^(2), n_(1)) = frac(47.0642^(2), 28) ≈ 79.1085$] #math.equation(block: true, alt: "B equals the fraction s 2 2 over n sub 2 equals the fraction 22.3000 squared over 30 approximately equals 16.5763")[$B = frac(s_(2)^(2), n_(2)) = frac(22.3000^(2), 30) ≈ 16.5763$] #math.equation(block: true, alt: "d f equals the fraction open parenthesis A plus B close parenthesis squared over the fraction A squared over n sub 1 minus 1 plus the fraction B squared over n sub 2 minus 1 equals the fraction open parenthesis 79.1085 plus 16.5763 close parenthesis squared over the fraction 79.1085 squared over 28 minus 1 plus the fraction 16.5763 squared over 30 minus 1 approximately equals 37.9493")[$d f = frac(( A + B )^(2), frac(A^(2), n_(1) − 1) + frac(B^(2), n_(2) − 1)) = frac(( 79.1085 + 16.5763 )^(2), frac(79.1085^(2), 28 − 1) + frac(16.5763^(2), 30 − 1)) ≈ 37.9493$] p-value: Using TI-83/84: #math.equation(block: false, alt: "tcdf open parenthesis 6.051 , 1 E 99 , 37.9493 close parenthesis approximately equals 2.44 times 10 to the power minus 7")[$op("tcdf") ( 6.051 , 1 E 99 , 37.9493 ) ≈ 2.44 × 10^(− 7)$] Using R: #math.equation(block: false, alt: "1 minus p t open parenthesis 6.051 , 37.9493 close parenthesis approximately equals 2.44 times 10 to the power minus 7")[$1 − upright(p) upright(t) ( 6.051 , 37.9493 ) ≈ 2.44 × 10^(− 7)$] pt(6.051, 37.9493) Using Technology: Using the TI-83/84: #figure(figph[TI-83/84 calculator 2-SampTTest input screen showing List1 L1, List2 L2, frequencies of 1, the alternative hypothesis μ1\>μ2 selected, and pooled set to No.], alt: "TI-83/84 calculator 2-SampTTest input screen showing List1 L1, List2 L2, frequencies of 1, the alternative hypothesis μ1>μ2 selected, and pooled set to No.", caption: [Figure #math.equation(block: false, alt: "7")[$7$]: Setup for 2-SampTTest on TI-83/84 Calculator]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] The Pooled question on the calculator is for whether you are assuming the variances are equal. Since this assumption is not being made, then the answer to this question is no. Pooled means that you assume the variances are equal and can pool the sample variances together. ] #figure(figph[Side-by-side TI-84 calculator screens display 2-SampTTest results for unequal variances, including t statistics, p-values, degrees of freedom, sample means, standard deviations, and sample sizes.], alt: "Side-by-side TI-84 calculator screens display 2-SampTTest results for unequal variances, including t statistics, p-values, degrees of freedom, sample means, standard deviations, and sample sizes.", caption: [Figure #math.equation(block: false, alt: "8")[$8$]: Results for 2-SampTTest on TI-83/84 Calculator]) Using R: command in general: t.test(variable1, variable2, alternative = "less" or "greater") For this example, the R command is: t.test(heartattack, healthy, alternative="greater") healthy \<- c(196, 232, 200, 242, 206, 178, 184, 198, 160, 182, 182, 198, 182, 238, 198, 188, 166, 204, 182, 178, 212, 164, 230, 186, 162, 182, 218, 170, 200, 176) heartattack \<- c(270, 236, 210, 142, 280, 272, 160, 220, 226, 242, 186, 266, 206, 318, 294, 282, 234, 224, 276, 282, 360, 310, 280, 278, 288, 288, 244, 236) t.test(heartattack, healthy, alternative="greater") Welch Two Sample t-test data: heartattack and healthy t = 6.1452, df = 37.675, p-value = 1.86e-07 alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: 44.1124 Inf sample estimates: mean of x mean of y 253.9286 193.1333 The test statistic is t = 6.1452. The p-value is #math.equation(block: false, alt: "1.86 times 10 to the power minus 7")[$1.86 × 10^(− 7)$] 5. Reject #math.equation(block: false, alt: "H sub o")[$H_(o)$] since the p-value \< #math.equation(block: false, alt: "α")[$α$]. 6. This is enough evidence to show that patients who have had heart attacks have higher cholesterol level on average from healthy individuals. (Though do realize that some of assumptions are not valid, so this interpretation may be invalid.) ] ] #examplebox("Example 2")[confidence interval for #math.equation(block: false, alt: "μ sub 1 minus μ sub 2")[$μ_(1) − μ_(2)$]][ The cholesterol level of patients who had heart attacks was measured two days after the heart attack. The researchers want to see if patients who have heart attacks have higher cholesterol levels over healthy people, so they also measured the cholesterol level of healthy adults who show no signs of heart disease. The data is in Example #math.equation(block: false, alt: "1")[$1$] ("Cholesterol levels after," 2013). Find a 99% confidence interval for the mean difference in cholesterol levels between heart attack patients and healthy individuals. + State the random variables and the parameters in words. + State and check the assumptions for the hypothesis test. + Find the sample statistic and confidence interval. + Statistical Interpretation + Real World Interpretation #solutionbox[ 1. These were stated in Example #math.equation(block: false, alt: "1")[$1$], but are reproduced here for reference. #math.equation(block: false, alt: "x sub 1")[$x_(1)$] = Cholesterol level of patients who had a heart attack #math.equation(block: false, alt: "x sub 2")[$x_(2)$] = Cholesterol level of healthy individuals #math.equation(block: false, alt: "μ sub 1")[$μ_(1)$] = mean cholesterol level of patients who had a heart attack #math.equation(block: false, alt: "μ sub 2")[$μ_(2)$] = mean cholesterol level of healthy individuals 2. The assumptions were stated and checked in Example #math.equation(block: false, alt: "1")[$1$]. 3. Sample Statistic: #math.equation(block: true, alt: "x ― sub 1 approximately equals 252.32 , x sub 2 ― approximately equals 193.13 , s sub 1 approximately equals 47.0642 , s sub 2 approximately equals 22.3000 , n sub 1 equals 28 , n sub 2 equals 30")[$limits(x)^(―)_(1) ≈ 252.32 , limits(x_(2))^(―) ≈ 193.13 , s_(1) ≈ 47.0642 , s_(2) ≈ 22.3000 , n_(1) = 28 , n_(2) = 30$] Test Statistic: Degrees of freedom: (the Welch–Satterthwaite equation) #math.equation(block: true, alt: "A equals the fraction s 1 2 over n sub 1 equals the fraction 47.0642 squared over 28 approximately equals 79.1085")[$A = frac(s_(1)^(2), n_(1)) = frac(47.0642^(2), 28) ≈ 79.1085$] #math.equation(block: true, alt: "B equals the fraction s 2 2 over n sub 2 equals the fraction 22.3000 squared over 30 approximately equals 16.5763")[$B = frac(s_(2)^(2), n_(2)) = frac(22.3000^(2), 30) ≈ 16.5763$] #math.equation(block: true, alt: "d f equals the fraction open parenthesis A plus B close parenthesis squared over the fraction A squared over n sub 1 minus 1 plus the fraction B squared over n sub 2 minus 1 equals the fraction open parenthesis 79.1085 plus 16.5763 close parenthesis squared over the fraction 79.1085 squared over 28 minus 1 plus the fraction 16.5763 squared over 30 minus 1 approximately equals 37.9493")[$d f = frac(( A + B )^(2), frac(A^(2), n_(1) − 1) + frac(B^(2), n_(2) − 1)) = frac(( 79.1085 + 16.5763 )^(2), frac(79.1085^(2), 28 − 1) + frac(16.5763^(2), 30 − 1)) ≈ 37.9493$] Since this df is not in the table, round to the nearest whole number. #math.equation(block: true, alt: "t sub c equals 2.712")[$t_(c) = 2.712$] #math.equation(block: true, alt: "E equals t sub c the square root of the fraction s 1 2 over n sub 1 plus the fraction s 2 2 over n sub 2 equals 2.712 the square root of the fraction 47.0642 squared over 28 plus the fraction 22.3000 squared over 30 approximately equals 26.53")[$E = t_(c) sqrt(frac(s_(1)^(2), n_(1)) + frac(s_(2)^(2), n_(2))) = 2.712 sqrt(frac(47.0642^(2), 28) + frac(22.3000^(2), 30)) ≈ 26.53$] #math.equation(block: true, alt: "open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus E less than μ sub 1 minus μ sub 2 less than open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis plus E")[$( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − E < μ_(1) − μ_(2) < ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) + E$] #math.equation(block: true, alt: "open parenthesis 252.32 minus 193.13 close parenthesis minus 26.53 less than μ sub 1 minus μ sub 2 less than open parenthesis 252.32 minus 193.13 close parenthesis plus 26.53")[$( 252.32 − 193.13 ) − 26.53 < μ_(1) − μ_(2) < ( 252.32 − 193.13 ) + 26.53$] #math.equation(block: true, alt: "32.66 m g / d L less than μ sub 1 minus μ sub 2 less than 85.72 m g / d L")[$32.66 upright(m) upright(g) / upright(d) upright(L) < μ_(1) − μ_(2) < 85.72 upright(m) upright(g) / upright(d) upright(L)$] Using Technology: #linebreak() Using TI-83/84: #figure(figph[TI-83/84 calculator 2-SampTInt setup screen showing lists L1 and L2, frequencies 1, confidence level .99, and pooled setting No.], alt: "TI-83/84 calculator 2-SampTInt setup screen showing lists L1 and L2, frequencies 1, confidence level .99, and pooled setting No.", caption: [Figure #math.equation(block: false, alt: "9")[$9$]: Setup for 2-SampTInt on TI-83/84 Calculator]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] The Pooled question on the calculator is for whether you are assuming the variances are equal. Since this assumption is not being made, then the answer to this question is no. Pooled means that you assume the variances are equal and can pool the sample variances together. ] #figure(figph[TI-84 calculator screen showing two-sample t-interval results for heart attack patients and healthy individuals, including the confidence interval, sample means, standard deviations, and sample sizes.], alt: "TI-84 calculator screen showing two-sample t-interval results for heart attack patients and healthy individuals, including the confidence interval, sample means, standard deviations, and sample sizes.", caption: [Figure #math.equation(block: false, alt: "10")[$10$]: Results for 2-SampTInt on TI-83/84 Calculator]) Using R: the commands is t.test(variable1, variable2, conf.level=C), where C is in decimal form. For this example, the command is t.test(heartattack, healthy, conf.level=.99) Output: Welch Two Sample t-test data: heartattack and healthy t = 6.1452, df = 37.675, p-value = 3.721e-07 alternative hypothesis: true difference in means is not equal to 0 99 percent confidence interval: 33.95750 87.63298 sample estimates: mean of x mean of y 253.9286 193.1333 The confidence interval is #math.equation(block: false, alt: "33.96 less than μ sub 1 minus μ sub 2 less than 87.63")[$33.96 < μ_(1) − μ_(2) < 87.63$] 4. There is a 99% chance that #math.equation(block: false, alt: "33.96 less than μ sub 1 minus μ sub 2 less than 87.63")[$33.96 < μ_(1) − μ_(2) < 87.63$] contains the true difference in means. 5. The mean cholesterol level for patients who had heart attacks is anywhere from 32.66 mg/dL to 85,72 mg/dL more than the mean cholesterol level for healthy patients. (Though do realize that many of assumptions are not valid, so this interpretation may be invalid.) ] ] If you do assume that the variances are equal, that is #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$], then the test statistic is: #math.equation(block: true, alt: "t equals the fraction open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus open parenthesis μ sub 1 minus μ sub 2 close parenthesis over s sub p the square root of the fraction 1 over n sub 1 plus the fraction 1 over n sub 2")[$t = frac(( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − ( μ_(1) − μ_(2) ), s_(p) sqrt(frac(1, n_(1)) + frac(1, n_(2))))$] where #math.equation(block: false, alt: "s sub p equals the square root of the fraction open parenthesis n sub 1 minus 1 close parenthesis s 1 2 plus open parenthesis n sub 2 minus 1 close parenthesis s 2 2 over open parenthesis n sub 1 minus 1 close parenthesis plus open parenthesis n sub 2 minus 1 close parenthesis")[$s_(p) = sqrt(frac(( n_(1) − 1 ) s_(1)^(2) + ( n_(2) − 1 ) s_(2)^(2), ( n_(1) − 1 ) + ( n_(2) − 1 )))$] #math.equation(block: false, alt: "s sub p")[$s_(p)$] = pooled standard deviation The Degrees of Freedom is: #emph[df] = #math.equation(block: false, alt: "n sub 1 plus n sub 2 minus 2")[$n_(1) + n_(2) − 2$] The confidence interval if you do assume that #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$] has been met, is #math.equation(block: true, alt: "open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus E less than μ sub 1 minus μ sub 2 less than open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis plus E")[$( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − E < μ_(1) − μ_(2) < ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) + E$] where #math.equation(block: false, alt: "E equals t sub c s sub p the square root of the fraction 1 over n sub 1 plus the fraction 1 over n sub 2")[$E = t_(c) s_(p) sqrt(frac(1, n_(1)) + frac(1, n_(2)))$] and #math.equation(block: false, alt: "s sub p equals the square root of the fraction open parenthesis n sub 1 minus 1 close parenthesis s 1 2 plus open parenthesis n sub 2 minus 1 close parenthesis s 2 2 over open parenthesis n sub 1 minus 1 close parenthesis plus open parenthesis n sub 2 minus 1 close parenthesis")[$s_(p) = sqrt(frac(( n_(1) − 1 ) s_(1)^(2) + ( n_(2) − 1 ) s_(2)^(2), ( n_(1) − 1 ) + ( n_(2) − 1 )))$] Degrees of Freedom: #emph[df] = #math.equation(block: false, alt: "n sub 1 plus n sub 2 minus 2")[$n_(1) + n_(2) − 2$] #math.equation(block: false, alt: "t sub c")[$t_(c)$] is the critical value where #emph[C] = 1 - #math.equation(block: false, alt: "α")[$α$] To show that the variances are equal, just show that the ratio of your sample variances is not unusual (probability is greater than 0.05). In other words, make sure the following is true. #math.equation(block: false, alt: "P open parenthesis F greater than s 1 2 / s 2 2 close parenthesis greater than or equal to 0.05 open parenthesis or P open parenthesis F greater than s 2 2 / s 1 2 close parenthesis greater than or equal to 0.05")[$P \( F > s_(1)^(2) / s_(2)^(2) \) ≥ 0.05 \( " or " P \( F > s_(2)^(2) / s_(1)^(2) \) ≥ 0.05$] so that the larger variance is in the numerator). This probability is from an F-distribution. To find the probability on the TI-83/84 calculator use #math.equation(block: false, alt: "Fcdf open parenthesis s 1 2 / s 2 2 , 1 E 99 , n sub 1 minus 1 , n sub 2 minus 1 close parenthesis")[$op("Fcdf") ( s_(1)^(2) / s_(2)^(2) , 1 E 99 , n_(1) − 1 , n_(2) − 1 )$]. To find the probability on R, use #math.equation(block: false, alt: "1 minus pf open parenthesis s 1 2 / s 2 2 , n sub 1 minus 1 , n sub 2 minus 1 close parenthesis")[$1 − op("pf") ( s_(1)^(2) / s_(2)^(2) , n_(1) − 1 , n_(2) − 1 )$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] The F-distribution is very sensitive to the normal distribution. A better test for equal variances is Levene's test, though it is more complicated. It is best to do Levene’s test when using statistical software (such as SPSS or Minitab) to perform the two-sample independent t-test. ] #examplebox("Example 3")[hypothesis test for two means][ The amount of sodium in beef hotdogs was measured. In addition, the amount of sodium in poultry hotdogs was also measured ("SOCR 012708 id," 2013). The data is in Example #math.equation(block: false, alt: "2")[$2$]. Is there enough evidence to show that beef has less sodium on average than poultry hotdogs? Use a 5% level of significance. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The 2-SampTTest on the hotdog data] Beef sodium levels load into L1 and poultry into L2. The calculator pooled the variances here (its F-test justified it); left-tailed at α = 0.05. - 2-SampTTest (pooled): t = −1.678, p-value = 0.0507 - 2-SampTInt at 95%: (−108.2, 10.114) ] #figure(table( columns: 2, align: left, inset: 6pt, table.header([Sodium in Beef Hotdogs], [Sodium in Poultry Hotdogs]), [495], [430], [477], [375], [425], [396], [322], [383], [482], [387], [587], [542], [370], [359], [322], [357], [479], [528], [375], [513], [330], [426], [300], [513], [386], [358], [401], [581], [645], [588], [440], [522], [317], [545], [319], [430], [298], [375], [253], [396], )) + State the random variables and the parameters in words. + State the null and alternative hypotheses and the level of significance. + State and check the assumptions for the hypothesis test. + Find the sample statistic, test statistic, and p-value. + Conclusion + Interpretation #solutionbox[ 1. #math.equation(block: false, alt: "x sub 1")[$x_(1)$] = sodium level in beef hotdogs #math.equation(block: false, alt: "x sub 2")[$x_(2)$] = sodium level in poultry hotdogs #math.equation(block: false, alt: "μ sub 1")[$μ_(1)$] = mean sodium level in beef hotdogs #math.equation(block: false, alt: "μ sub 2")[$μ_(2)$] = mean sodium level in poultry hotdogs 2. The normal hypotheses would be #math.equation(block: true, alt: "H sub o : μ sub 1 equals μ sub 2 or H sub o : μ sub 1 minus μ sub 2 equals 0; H sub A : μ sub 1 less than μ sub 2 H sub A : μ sub 1 minus μ sub 2 less than 0; α equals 0.05")[$H_(o) : μ_(1) = μ_(2) " or " #h(1em) H_(o) : μ_(1) − μ_(2) = 0 \ H_(A) : μ_(1) < μ_(2) #h(1em) #h(1em) #h(0.222em) H_(A) : μ_(1) − μ_(2) < 0 \ α = 0.05$] 3. + A random sample of 20 sodium levels in beef hotdogs is taken. A random sample of 20 sodium levels in poultry hotdogs. The problem does not state if either sample was randomly selected. So this assumption may not be valid. + The two samples are independent since these are different types of hotdogs. + Population of all sodium levels in beef hotdogs is normally distributed. Population of all sodium levels in poultry hotdogs is normally distributed. Beef Hotdogs: #figure(figph[Histogram of sodium levels for 20 beef hotdogs, grouped into 100-milligram bins from 200 to 700.], alt: "Histogram of sodium levels for 20 beef hotdogs, grouped into 100-milligram bins from 200 to 700.", caption: [Figure #math.equation(block: false, alt: "11")[$11$]: Histogram of Sodium Levels in Beef Hotdogs]) This looks somewhat bell shaped. #figure(figph[A horizontal boxplot of sodium levels in beef hotdogs, with values ranging from about 253 to 644 milligrams and a median near 381 milligrams.], alt: "A horizontal boxplot of sodium levels in beef hotdogs, with values ranging from about 253 to 644 milligrams and a median near 381 milligrams.", caption: [Figure #math.equation(block: false, alt: "12")[$12$]: Modified Box Plot of Sodium Levels in Beef Hotdogs]) There are no outliers. #figure(figph[A normal Q-Q plot of the 20 beef-hotdog sodium levels, with most observations following an increasing pattern and two high values near 587 and 644 milligrams.], alt: "A normal Q-Q plot of the 20 beef-hotdog sodium levels, with most observations following an increasing pattern and two high values near 587 and 644 milligrams.", caption: [Figure #math.equation(block: false, alt: "13")[$13$]: Normal Quantile Plot of Sodium Levels in Beef Hotdogs]) This looks somewhat linear. So, the population of all sodium levels in beef hotdogs may be normally distributed. Poultry Hotdogs: #figure(figph[Histogram of sodium levels for 20 poultry hotdogs, grouped into 100-milligram bins from 300 to 600, with the most hotdogs in the 300 to 400 milligram bin.], alt: "Histogram of sodium levels for 20 poultry hotdogs, grouped into 100-milligram bins from 300 to 600, with the most hotdogs in the 300 to 400 milligram bin.", caption: [Figure #math.equation(block: false, alt: "14")[$14$]: Histogram of Sodium Levels in Poultry Hotdogs]) This does not look bell shaped. #figure(figph[A horizontal boxplot showing the distribution of sodium levels in 20 poultry hotdogs, ranging from 357 to about 588 milligrams and a median near 428 milligrams.], alt: "A horizontal boxplot showing the distribution of sodium levels in 20 poultry hotdogs, ranging from 357 to about 588 milligrams and a median near 428 milligrams.", caption: [Figure #math.equation(block: false, alt: "15")[$15$]: Modified Box Plot of Sodium Levels in Poultry Hotdogs]) There are no outliers. #figure(figph[A normal Q-Q plot of sodium levels in poultry hotdogs, with ordered sample values compared against theoretical normal quantiles.], alt: "A normal Q-Q plot of sodium levels in poultry hotdogs, with ordered sample values compared against theoretical normal quantiles.", caption: [Figure #math.equation(block: false, alt: "16")[$16$]: Normal Quantile Plot of Sodium Levels in Poultry Hotdogs]) This does not look linear. So, the population of all sodium levels in poultry hotdogs is probably not normally distributed. This assumption is not valid. Since the samples are fairly large, and the t-test is robust, it may not be a large issue. However, just realize that the conclusions of the test may not be valid. d. The population variances are equal, i.e. #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$]. #math.equation(block: true, alt: "s sub 1 approximately equals 102.4347; s sub 2 approximately equals 81.1786; the fraction s 1 2 over s 2 2 equals the fraction 102.4347 squared over 81.1786 squared approximately equals 1.592")[$s_(1) ≈ 102.4347 \ s_(2) ≈ 81.1786 \ frac(s_(1)^(2), s_(2)^(2)) = frac(102.4347^(2), 81.1786^(2)) ≈ 1.592$] Using TI-83/84: Fcdf #math.equation(block: false, alt: "open parenthesis 1.592 , 1 E 99 , 19 , 19 close parenthesis approximately equals 0.1597 greater than or equal to 0.05")[$( 1.592 , 1 E 99 , 19 , 19 ) ≈ 0.1597 ≥ 0.05$] Using R: 1 - pf #math.equation(block: false, alt: "open parenthesis 1.592 , 19 , 19 close parenthesis approximately equals 0.1597 greater than or equal to 0.05")[$( 1.592 , 19 , 19 ) ≈ 0.1597 ≥ 0.05$] So you can say that these variances are equal. 4. Find the sample statistic, test statistic, and p-value Sample Statistic: #math.equation(block: true, alt: "x ― sub 1 equals 401.15 , x ― sub 2 equals 450.2 , s sub 1 approximately equals 102.4347 , s sub 2 approximately equals 81.1786 , n sub 1 equals 20 , n sub 2 equals 20")[$limits(x)^(―)_(1) = 401.15 , limits(x)^(―)_(2) = 450.2 , s_(1) ≈ 102.4347 , s_(2) ≈ 81.1786 , n_(1) = 20 , n_(2) = 20$] Test Statistic: The assumption #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$] has been met, so #math.equation(block: true, alt: "s sub p equals the square root of the fraction open parenthesis n sub 1 minus 1 close parenthesis s 1 2 plus open parenthesis n sub 2 minus 1 close parenthesis s 2 2 over open parenthesis n sub 1 minus 1 close parenthesis plus open parenthesis n sub 2 minus 1 close parenthesis")[$s_(p) = sqrt(frac(( n_(1) − 1 ) s_(1)^(2) + ( n_(2) − 1 ) s_(2)^(2), ( n_(1) − 1 ) + ( n_(2) − 1 )))$] #math.equation(block: true, alt: "equals the square root of the fraction 102.4347 squared * 19 plus 81.1786 squared * 19 over open parenthesis 20 minus 1 close parenthesis plus open parenthesis 20 minus 1 close parenthesis")[$= sqrt(frac(102.4347^(2) * 19 + 81.1786^(2) * 19, ( 20 − 1 ) + ( 20 − 1 )))$] #math.equation(block: true, alt: "approximately equals 92.4198")[$≈ 92.4198$] Though you should try to do the calculations in the problem so you don’t create round off error. #math.equation(block: true, alt: "t equals the fraction open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus open parenthesis μ sub 1 minus μ sub 2 close parenthesis over s sub P the square root of the fraction 1 over n sub 1 plus the fraction 1 over n sub 2")[$t = frac(( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − ( μ_(1) − μ_(2) ), s_(P) sqrt(frac(1, n_(1)) + frac(1, n_(2))))$] #math.equation(block: true, alt: "equals the fraction open parenthesis 401.15 minus 450.2 close parenthesis minus 0 over 92.4198 the square root of the fraction 1 over 20 plus the fraction 1 over 20")[$= frac(( 401.15 − 450.2 ) − 0, 92.4198 sqrt(frac(1, 20) + frac(1, 20)))$] #math.equation(block: true, alt: "approximately equals minus 1.678")[$≈ − 1.678$] #emph[df] = 20 + 20 - 2 = 38 p-value: Using TI-83/84: tcdf #math.equation(block: false, alt: "open parenthesis minus 1 E 99 , minus 1.678 , 38 close parenthesis approximately equals 0.0508")[$( − 1 E 99 , − 1.678 , 38 ) ≈ 0.0508$] Using R: pt #math.equation(block: false, alt: "open parenthesis minus 1.678 , 38 close parenthesis approximately equals 0.0508")[$( − 1.678 , 38 ) ≈ 0.0508$] Using technology to find the t and p-value: Using TI-83/84: #figure(figph[TI-83/84 calculator 2-SampTTest setup screen with Data input, List1 L1, List2 L2, frequencies 1, the alternative hypothesis μ1\<μ2 selected, and pooled set to Yes.], alt: "TI-83/84 calculator 2-SampTTest setup screen with Data input, List1 L1, List2 L2, frequencies 1, the alternative hypothesis μ1<μ2 selected, and pooled set to Yes.", caption: [Figure #math.equation(block: false, alt: "17")[$17$]: Setup for 2-SampTTest on TI-83/84 Calculator]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] The Pooled question on the calculator is for whether you are using the pooled standard deviation or not. In this example, the pooled standard deviation was used since you are assuming the variances are equal. That is why the answer to the question is Yes. ] #figure(figph[TI-84 calculator screen showing 2-SampTTest results for sodium in beef versus poultry hotdogs, including t=-1.678317463, p=0.0507460087, df=38, and sample means 401.15 and 450.2.], alt: "TI-84 calculator screen showing 2-SampTTest results for sodium in beef versus poultry hotdogs, including t=-1.678317463, p=0.0507460087, df=38, and sample means 401.15 and 450.2.", caption: [Figure #math.equation(block: false, alt: "18")[$18$]: Results for 2-SampTTest on TI-83/84 Calculator]) Using R: the command is t.test(variable1, variable2, alternative="less" or "greater") For this example, the command is t.test(beef, poultry, alternative="less", equalvar=TRUE) beef \<- c(495, 477, 425, 322, 482, 587, 370, 322, 479, 375, 330, 300, 386, 401, 645, 440, 317, 319, 298, 253) poultry \<- c(430, 375, 396, 383, 387, 542, 359, 357, 528, 513, 426, 513, 358, 581, 588, 522, 545, 430, 375, 396) t.test(beef, poultry, alternative="less", equalvar=TRUE) Welch Two Sample t-test data: beef and poultry t = -1.6783, df = 36.115, p-value = 0.05096 alternative hypothesis: true difference in means is less than 0 95 percent confidence interval: -Inf 0.2875363 sample estimates: mean of x mean of y 401.15 450.20 The #emph[t] = -1.6783 and the p-value = 0.05096. 5. Fail to reject #math.equation(block: false, alt: "H sub o")[$H_(o)$] since the p-value \> #math.equation(block: false, alt: "α")[$α$]. 6. This is not enough evidence to show that beef hotdogs have less sodium than poultry hotdogs. (Though do realize that many of assumptions are not valid, so this interpretation may be invalid.) ] ] #examplebox("Example 4")[confidence interval for #math.equation(block: false, alt: "μ sub 1 minus μ sub 2")[$μ_(1) − μ_(2)$]][ The amount of sodium in beef hotdogs was measured. In addition, the amount of sodium in poultry hotdogs was also measured ("SOCR 012708 id," 2013). The data is in Example #math.equation(block: false, alt: "2")[$2$]. Find a 95% confidence interval for the mean difference in sodium levels between beef and poultry hotdogs. + State the random variables and the parameters in words. + State and check the assumptions for the hypothesis test. + Find the sample statistic and confidence interval. + Statistical Interpretation + Real World Interpretation #solutionbox[ 1. These were stated in Example #math.equation(block: false, alt: "1")[$1$], but are reproduced here for reference. #math.equation(block: false, alt: "x sub 1")[$x_(1)$] = sodium level in beef hotdogs #math.equation(block: false, alt: "x sub 2")[$x_(2)$] = sodium level in poultry hotdogs #math.equation(block: false, alt: "μ sub 1")[$μ_(1)$] = mean sodium level in beef hotdogs #math.equation(block: false, alt: "μ sub 2")[$μ_(2)$] = mean sodium level in poultry hotdogs 2. The assumptions were stated and checked in Example #math.equation(block: false, alt: "3")[$3$]. 3. Sample Statistic: #math.equation(block: true, alt: "x ― sub 1 equals 401.15 , x ― sub 2 equals 450.2 , s sub 1 approximately equals 102.4347 , s sub 2 approximately equals 81.1786 , n sub 1 equals 20 , n sub 2 equals 20")[$limits(x)^(―)_(1) = 401.15 , limits(x)^(―)_(2) = 450.2 , s_(1) ≈ 102.4347 , s_(2) ≈ 81.1786 , n_(1) = 20 , n_(2) = 20$] Confidence Interval: The confidence interval estimate of the difference #math.equation(block: false, alt: "μ sub 1 minus μ sub 2")[$μ_(1) − μ_(2)$] is The assumption #math.equation(block: false, alt: "σ 1 2 equals σ 2 2")[$σ_(1)^(2) = σ_(2)^(2)$] has been met, so #math.equation(block: true, alt: "s sub p equals the square root of the fraction open parenthesis n sub 1 minus 1 close parenthesis s 1 2 plus open parenthesis n sub 2 minus 1 close parenthesis s 2 2 over open parenthesis n sub 1 minus 1 close parenthesis plus open parenthesis n sub 2 minus 1 close parenthesis")[$s_(p) = sqrt(frac(( n_(1) − 1 ) s_(1)^(2) + ( n_(2) − 1 ) s_(2)^(2), ( n_(1) − 1 ) + ( n_(2) − 1 )))$] #math.equation(block: true, alt: "equals the square root of the fraction 102.4347 squared * 19 plus 81.1786 squared * 19 over open parenthesis 20 minus 1 close parenthesis plus open parenthesis 20 minus 1 close parenthesis")[$= sqrt(frac(102.4347^(2) * 19 + 81.1786^(2) * 19, ( 20 − 1 ) + ( 20 − 1 )))$] #math.equation(block: true, alt: "approximately equals 92.4198")[$≈ 92.4198$] Though you should try to do the calculations in the formula for E so you don’t create round off error. #emph[df] = #math.equation(block: false, alt: "equals n sub 1 plus n sub 2 minus 2 equals 20 plus 20 minus 2 equals 38")[$= n_(1) + n_(2) − 2 = 20 + 20 − 2 = 38$] #math.equation(block: true, alt: "t sub c equals 2.024")[$t_(c) = 2.024$] #math.equation(block: true, alt: "E equals t sub c s sub p the square root of the fraction 1 over n sub 1 plus the fraction 1 over n sub 2")[$E = t_(c) s_(p) sqrt(frac(1, n_(1)) + frac(1, n_(2)))$] #math.equation(block: true, alt: "equals 2.024 open parenthesis 92.4198 close parenthesis the square root of the fraction 1 over 20 plus the fraction 1 over 20")[$= 2.024 ( 92.4198 ) sqrt(frac(1, 20) + frac(1, 20))$] #math.equation(block: true, alt: "approximately equals 59.15")[$≈ 59.15$] #math.equation(block: true, alt: "open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis minus E less than μ sub 1 minus μ sub 2 less than open parenthesis x ― sub 1 minus x ― sub 2 close parenthesis plus E")[$( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) − E < μ_(1) − μ_(2) < ( limits(x)^(―)_(1) − limits(x)^(―)_(2) ) + E$] #math.equation(block: true, alt: "open parenthesis 401.15 minus 450.2 close parenthesis minus 59.15 less than μ sub 1 minus μ sub 2 less than open parenthesis 401.15 minus 450.2 close parenthesis plus 59.15")[$( 401.15 − 450.2 ) − 59.15 < μ_(1) − μ_(2) < ( 401.15 − 450.2 ) + 59.15$] #math.equation(block: true, alt: "minus 108.20 g less than μ sub 1 minus μ sub 2 less than 10.10 g")[$− 108.20 g < μ_(1) − μ_(2) < 10.10 g$] Using technology: Using the TI-83/84: #figure(figph[TI-83/84 2-SampTInt setup screen showing Data input with List1 L1, List2 L2, frequencies of 1, confidence level 0.95, and pooled set to Yes.], alt: "TI-83/84 2-SampTInt setup screen showing Data input with List1 L1, List2 L2, frequencies of 1, confidence level 0.95, and pooled set to Yes.", caption: [Figure #math.equation(block: false, alt: "19")[$19$]: Setup for 2-SampTInt on Ti-83/84 Calculator]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] The Pooled question on the calculator is for whether you are using the pooled standard deviation or not. In this example, the pooled standard deviation was used since you are assuming the variances are equal. That is why the answer to the question is Yes. ] #figure(figph[TI-84 calculator screen showing 2-SampTInt output for beef versus poultry hotdog sodium means, including a confidence interval of (-108.2, 10.114).], alt: "TI-84 calculator screen showing 2-SampTInt output for beef versus poultry hotdog sodium means, including a confidence interval of (-108.2, 10.114).", caption: [Figure #math.equation(block: false, alt: "20")[$20$]: Results for 2-SampTInt on TI-83/84 Calculator]) Using R: the command is t.test(variable1, variable2, equalvar=TRUE, conf.level=C), where C is in decimal form. For this example, the command is t.test(beef, poultry, conf.level=.95, equalvar=TRUE) Welch Two Sample t-test data: beef and poultry t = -1.6783, df = 36.115, p-value = 0.1019 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -108.31592 10.21592 sample estimates: mean of x mean of y 401.15 450.20 The confidence interval is #math.equation(block: false, alt: "minus 108.32 less than μ sub 1 minus μ sub 2 less than 10.22")[$− 108.32 < μ_(1) − μ_(2) < 10.22$]. 4. There is a 95% chance that #math.equation(block: false, alt: "minus 108.20 g less than μ sub 1 minus μ sub 2 less than 10.10 g")[$− 108.20 g < μ_(1) − μ_(2) < 10.10 g$] contains the true difference in means. 5. The mean sodium level of beef hotdogs is anywhere from 108.20 g less than the mean sodium level of poultry hotdogs to 10.10 g more. (The negative sign on the lower limit implies that the first mean is less than the second mean. The positive sign on the upper limit implies that the first mean is greater than the second mean.) Realize that many of assumptions are not valid in this example, so the interpretation may be invalid. ] ] === Homework #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In each problem show all steps of the hypothesis test or confidence interval. If some of the assumptions are not met, note that the results of the test or interval may not be correct and then continue the process of the hypothesis test or confidence interval. Unless directed by your instructor, do not assume the variances are equal (except in problems 11 through 16). + The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in Example #math.equation(block: false, alt: "3")[$3$], and the income of females is given in table \#9.3.4 ("Median income of," 2013). Is there enough evidence to show that the mean income of males is more than of females? Test at the 1% level. #figure(table( columns: 7, align: left, inset: 6pt, table.header([\$42,951], [\$52,379], [\$42,544], [\$37,488], [\$49,281], [\$50,987], [\$60,705]), [\$50,411], [\$66,760], [\$40,951], [\$43,902], [\$45,494], [\$41,528], [\$50,746], [\$45,183], [\$43,624], [\$43,993], [\$41,612], [\$46,313], [\$43,944], [\$56,708], [\$60,264], [\$50,053], [\$50,580], [\$40,202], [\$43,146], [\$41,635], [\$42,182], [\$41,803], [\$53,033], [\$60,568], [\$41,037], [\$50,388], [\$41,950], [\$44,660], [\$46,176], [\$41,420], [\$45,976], [\$47,956], [\$22,529], [\$48,842], [\$41,464], [\$40,285], [\$41,309], [\$43,160], [\$47,573], [\$44,057], [\$52,805], [\$53,046], [\$42,125], [\$46,214], [\$51,630], [], [], [], [], )) #figure(table( columns: 8, align: left, inset: 6pt, table.header([\$31,862], [\$40,550], [\$36,048], [\$30,752], [\$41,817], [\$40,236], [\$47,476], [\$40,500]), [\$60,332], [\$33,823], [\$35,438], [\$37,242], [\$31,238], [\$39,150], [\$34,023], [\$33,745], [\$33,269], [\$32,684], [\$31,844], [\$34,599], [\$48,748], [\$46,185], [\$36,931], [\$40,416], [\$29,548], [\$33,865], [\$31,067], [\$33,424], [\$35,484], [\$41,021], [\$47,155], [\$32,316], [\$42,113], [\$33,459], [\$32,462], [\$35,746], [\$31,274], [\$36,027], [\$37,089], [\$22,117], [\$41,412], [\$31,330], [\$31,329], [\$33,184], [\$35,301], [\$32,843], [\$38,177], [\$40,969], [\$40,993], [\$29,688], [\$35,890], [\$34,381], [], [], [], [], )) + The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in Example #math.equation(block: false, alt: "3")[$3$], and the income of females is given in Example #math.equation(block: false, alt: "4")[$4$] ("Median income of," 2013). Compute a 99% confidence interval for the difference in incomes between males and females in the U.S. + A study was conducted that measured the total brain volume (TBV) (in #math.equation(block: false, alt: "m m cubed")[$m m^(3)$]) of patients that had schizophrenia and patients that are considered normal. Example #math.equation(block: false, alt: "5")[$5$] contains the TBV of the normal patients and Example #math.equation(block: false, alt: "6")[$6$] contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013). Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal? Test at the 10% level. #figure(table( columns: 6, align: left, inset: 6pt, table.header([1663407], [1583940], [1299470], [1535137], [1431890], [1578698]), [1453510], [1650348], [1288971], [1366346], [1326402], [1503005], [1474790], [1317156], [1441045], [1463498], [1650207], [1523045], [1441636], [1432033], [1420416], [1480171], [1360810], [1410213], [1574808], [1502702], [1203344], [1319737], [1688990], [1292641], [1512571], [1635918], [], [], [], [], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([1331777], [1487886], [1066075], [1297327], [1499983], [1861991]), [1368378], [1476891], [1443775], [1337827], [1658258], [1588132], [1690182], [1569413], [1177002], [1387893], [1483763], [1688950], [1563593], [1317885], [1420249], [1363859], [1238979], [1286638], [1325525], [1588573], [1476254], [1648209], [1354054], [1354649], [1636119], [], [], [], [], [], )) + A study was conducted that measured the total brain volume (TBV) (in #math.equation(block: false, alt: "m m cubed")[$m m^(3)$]) of patients that had schizophrenia and patients that are considered normal. Example #math.equation(block: false, alt: "5")[$5$] contains the TBV of the normal patients and Example #math.equation(block: false, alt: "6")[$6$] contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013). Compute a 90% confidence interval for the difference in TBV of normal patients and patients with Schizophrenia. + The length of New Zealand (NZ) rivers that travel to the Pacific Ocean are given in Example #math.equation(block: false, alt: "7")[$7$] and the lengths of NZ rivers that travel to the Tasman Sea are given in Example #math.equation(block: false, alt: "8")[$8$] ("Length of NZ," 2013). Do the data provide enough evidence to show on average that the rivers that travel to the Pacific Ocean are longer than the rivers that travel to the Tasman Sea? Use a 5% level of significance. #figure(table( columns: 5, align: left, inset: 6pt, table.header([209], [48], [169], [138], [64]), [97], [161], [95], [145], [90], [121], [80], [56], [64], [209], [64], [72], [288], [322], [], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([76], [64], [68], [64], [37], [32]), [32], [51], [56], [40], [64], [56], [80], [121], [177], [56], [80], [35], [72], [72], [108], [48], [], [], )) + The length of New Zealand (NZ) rivers that travel to the Pacific Ocean are given in Example #math.equation(block: false, alt: "7")[$7$] and the lengths of NZ rivers that travel to the Tasman Sea are given in Example #math.equation(block: false, alt: "8")[$8$] ("Length of NZ," 2013). Estimate the difference in mean lengths of rivers between rivers in NZ that travel to the Pacific Ocean and ones that travel to the Tasman Sea. Use a 95% confidence level. + The number of cell phones per 100 residents in countries in Europe is given in Example #math.equation(block: false, alt: "9")[$9$] for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in Example #math.equation(block: false, alt: "10")[$10$] also for the year 2010 ("Population reference bureau," 2013). Is there enough evidence to show that the mean number of cell phones in countries of Europe is more than in countries of the Americas? Test at the 1% level. #figure(table( columns: 6, align: left, inset: 6pt, table.header([100], [76], [100], [130], [75], [84]), [112], [84], [138], [133], [118], [134], [126], [188], [129], [93], [64], [128], [124], [122], [109], [121], [127], [152], [96], [63], [99], [95], [151], [147], [123], [95], [67], [67], [118], [125], [110], [115], [140], [115], [141], [77], [98], [102], [102], [112], [118], [118], [54], [23], [121], [126], [47], [], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([158], [117], [106], [159], [53], [50]), [78], [66], [88], [92], [42], [3], [150], [72], [86], [113], [50], [58], [70], [109], [37], [32], [85], [101], [75], [69], [55], [115], [95], [73], [86], [157], [100], [119], [81], [113], [87], [105], [96], [], [], [], )) + The number of cell phones per 100 residents in countries in Europe is given in Example #math.equation(block: false, alt: "9")[$9$] for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in Example #math.equation(block: false, alt: "10")[$10$] also for the year 2010 ("Population reference bureau," 2013). Find the 98% confidence interval for the difference in mean number of cell phones per 100 residents in Europe and the Americas. + A vitamin K shot is given to infants soon after birth. Nurses at Northbay Healthcare were involved in a study to see if how they handle the infants could reduce the pain the infants feel ("SOCR data nips," 2013). One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods (Example #math.equation(block: false, alt: "11")[$11$]), and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot (Example #math.equation(block: false, alt: "12")[$12$]). Is there enough evidence to show that infants cried less on average when they are held by their mothers than if held using conventional methods? Test at the 5% level. #figure(table( columns: 6, align: left, inset: 6pt, table.header([63], [0], [2], [46], [33], [33]), [29], [23], [11], [12], [48], [15], [33], [14], [51], [37], [24], [70], [63], [0], [73], [39], [54], [52], [39], [34], [30], [55], [58], [18], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([0], [32], [20], [23], [14], [19]), [60], [59], [64], [64], [72], [50], [44], [14], [10], [58], [19], [41], [17], [5], [36], [73], [19], [46], [9], [43], [73], [27], [25], [18], )) + A vitamin K shot is given to infants soon after birth. Nurses at Northbay Healthcare were involved in a study to see if how they handle the infants could reduce the pain the infants feel ("SOCR data nips," 2013). One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods (Example #math.equation(block: false, alt: "11")[$11$]), and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot (Example #math.equation(block: false, alt: "12")[$12$]). Calculate a 95% confidence interval for the mean difference in mean crying time after being given a vitamin K shot between infants held using conventional methods and infants held by their mothers. + Redo problem 1 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example #math.equation(block: false, alt: "3")[$3$]). + Redo problem 2 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example #math.equation(block: false, alt: "3")[$3$]). + Redo problem 7 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example #math.equation(block: false, alt: "3")[$3$]). + Redo problem 8 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example #math.equation(block: false, alt: "3")[$3$]). + Redo problem 9 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example #math.equation(block: false, alt: "3")[$3$]). + Redo problem 10 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example #math.equation(block: false, alt: "3")[$3$]). Answer For all hypothesis tests, just the conclusion is given. For all confidence intervals, just the interval using technology is given. See solution for the entire answer. + Reject Ho + #math.equation(block: false, alt: "$ 65443.80 less than μ sub 1 minus μ sub 2 less than $ 13340.80")[$\$ 65443.80 < μ_(1) − μ_(2) < \$ 13340.80$] + Fail to reject Ho + #math.equation(block: false, alt: "minus 51564.6 m m cubed less than μ sub 1 minus μ sub 2 less than 75656.6 m m cubed")[$− 51564.6 attach(upright(m) upright(m), t: 3) < μ_(1) − μ_(2) < 75656.6 attach(upright(m) upright(m), t: 3)$] + Reject Ho + #math.equation(block: false, alt: "23.2818 k m less than μ sub 1 minus μ sub 2 less than 103.67 k m")[$23.2818 upright(k) upright(m) < μ_(1) − μ_(2) < 103.67 upright(k) upright(m)$] + Reject Ho + #math.equation(block: false, alt: "4.3641 less than μ sub 1 minus μ sub 2 less than 37.5276")[$4.3641 < μ_(1) − μ_(2) < 37.5276$] + Fail to reject Ho + #math.equation(block: false, alt: "minus 10.9726 s less than μ sub 1 minus μ sub 2 less than 11.3059 s")[$− 10.9726 s < μ_(1) − μ_(2) < 11.3059 s$] + Reject Ho + #math.equation(block: false, alt: "$ 6544.98 less than μ sub 1 minus μ sub 2 less than $ 13339.60")[$\$ 6544.98 < μ_(1) − μ_(2) < \$ 13339.60$] + Reject Ho + #math.equation(block: false, alt: "4.8267 less than μ sub 1 minus μ sub 2 less than 37.0649")[$4.8267 < μ_(1) − μ_(2) < 37.0649$] + Fail to reject Ho + #math.equation(block: false, alt: "minus 10.9713 s less than μ sub 1 minus μ sub 2 less than 11.3047 s")[$− 10.9713 s < μ_(1) − μ_(2) < 11.3047 s$] ]