Solution to the practice test ch 10 correlation reg ch 11 gof ch12 anova

Solution to the practice test ch 10 correlation reg ch 11 gof ch12 anova

• 1.
  1 Statistics, Sample Test(Exam Review) Solution Module 5: Chapters 10, 11 & 12 Review Chapters 10: Correlation & Regression Chapters 11: Goodness-of-Fit (Multinomial Experiments) & Contingency Tables) Chapter 12: Contingency Tables, Analysis of Variance Chapters 10: Correlation & Regression 1. Given the sample data: Data X 1 1 3 5 Y 2 8 6 4 a. Find the value of the linear correlation coefficient r. b. Test the claim that there is a linear correlation between the two variables x and y. Use both Method 1 & Method 2. ( = 0.05) c. Find the regression equation. d. Find the best predicted value of y, when x is equal to 2. Solution: a) 𝒓 = 𝒏 ∑ 𝒙𝒚 − (∑ 𝒙)(∑ 𝒚) √𝒏 ∑ 𝒙𝟐 − (∑ 𝒙)𝟐 √𝒏 ∑ 𝒚𝟐 − (∑ 𝒚)𝟐 = 𝟒(𝟒𝟖)−𝟏𝟎(𝟐𝟎) √𝟒(𝟑𝟔)−𝟏𝟎𝟐√𝟒(𝟏𝟐𝟎)−𝟐𝟎𝟐 = −𝟖 √𝟒𝟒√𝟖𝟎 = −𝟎. 𝟏𝟑𝟓 (Strong positive linear relationship: r will be close to +1. (0.7 or larger) Strong negative linear relationship: r will be close to −1. (− 0.7 or smaller) b) Methods 1 & 2: Method 1: T TEST Method 2: Use Pearson Correlation Coefficient r
• 2.
  2 1) State thenull and the alternative hypotheses: 𝑯𝟎: 𝝆 = 𝟎 𝑯𝟏: 𝝆 ≠ 𝟎, 𝟐𝑻𝑻, 𝑪𝒍𝒂𝒊𝒎 2) Calculate the test statistic (observed value):TS TS 𝒕 = 𝒓−𝝁𝒓 √𝟏−𝒓𝟐 𝒏−𝟐 = 𝒕 = 𝒓−𝝁𝒓 √𝟏−𝒓𝟐 𝒏−𝟐 = −.𝟏𝟑𝟓−𝟎 √𝟏−(−.𝟏𝟑𝟓)𝟐 𝟒−𝟐 = −.𝟏𝟑𝟓 𝟎.𝟕𝟎𝟎𝟔𝟒 = −𝟎. 𝟏𝟗𝟐𝟕 Method 1 : T TEST Method 2: Use r Ts: t = -0.1927 r = - 0.135 3) Select the distribution and determine the RR & NRR. Method 1 : TTest  = 0.05, df = n-2 = 2 CV: t = 4.303 CV: - 4.303 TS: - 0.1927 CV: 4.303 Method 2: Use the value for r: Correlation Coefficient Table: n =4,  = 0.05, r = 0.950 CV: r = -.950 TS: r = -.135 CV: r = .950
• 3.
  3 4) Make adecision: Don’t reject H0, The claim is false, There is no linear correlation between the 2 variables. C) Find the regression equation 𝒃𝟏 = 𝒏 ∑ 𝒙𝒚 − (∑ 𝒙)(∑ 𝒚) 𝒏 ∑ 𝒙𝟐 − (∑ 𝒙)𝟐 = 𝟒(𝟒𝟖) − 𝟏𝟎(𝟐𝟎) 𝟒(𝟑𝟔) − 𝟏𝟎𝟐 = −𝟖 𝟒𝟒 = −𝟎. 𝟏𝟖𝟐 𝒃𝟎 = 𝒚 − 𝒃𝟏𝒙, 𝒚 = ∑ 𝒚 𝒏 = 𝟐𝟎 𝟒 = 𝟓, 𝒙 = ∑ 𝒙 𝒏 = 𝟏𝟎 𝟒 = 𝟐. 𝟓 𝒃𝟎 = 𝟓 − (−𝟎. 𝟏𝟖𝟏𝟖𝟏𝟖)(𝟐. 𝟓) = 𝟓. 𝟒𝟓 0 1 ˆ ˆ 0.182 5.45 y b b x y x = + → = − + D) Find the best predicted value of y, when x is equal to 2. (If there is not a significant linear correlation, the best predicted y-value is y ) Since 𝑟 = −0.135 , and there is no linear correlation between the 2 variables, we use 𝑦 ̅ = 5 Note: If there is a strong positive linear relationship between the variables, the value of r will be close to +1. (0.7 or larger) and if there is a strong negative linear relationship between the variables, the value of r will be close to −1. (− 0.7 or smaller) 2. A researcher is studying the effects of Testosterone injections on different lab mice. She controls the concentration of Testosterone injected into the mice and measures their metabolic rates. Analyze the following dataset. Will you be running a correlation or a regression? Answer: Regression, since she controls the concentration of Testosterone injected into the mice and measures their metabolic rates. Testosterone: x Metabolic Rate: y 0.6 500 0.9 556 1.02 578 1.06 581 1.08 590 1.11 605 1.16 612 1.2 624 1.24 639 1.27 640 1.32 643 1.37 650 1.39 658 1.43 661 1.46 670 1.49 672 1.5 673
• 4.
  4 Excel: 𝑟 =0.9934, 𝑟2 = 0.9869 Coefficients Standard Error t Stat P-value Intercept 379.4928 7.300295309 51.98321 2.35738E-18 Testosterone 199.05934 5.918276011 33.6347 1.5316E-15 a. Find the value of the linear correlation coefficient r, and test the claim that there is a linear correlation between the two variables x and y. ( = 0.05) 1) H0: 𝜌 = 0 (There is NO correlation between the concentration of Testosterone in mice and their metabolic rates.) Ha: 𝜌 ≠ 0, 2TT (There is a correlation between the concentration of Testosterone in mice and their metabolic rates.), Claim 2) Test statistic (TS) Excel: Method 1: 𝑡 = 𝑟−𝜇𝑟 √1−𝑟2 𝑛−2 = 33.635 Method 2 : r = 0.9934 Excel: 3) Based on the t-Test, since p-value = 1.53163 E – 15 4) Decision: Reject H0 since p-value < 0.05 The claim is True There is a significant linear correlation between the 2 variables. Therefore, we conclude that there is a significant correlation between the concentration of Testosterone in mice and their metabolic rates. Since this is a regression study, the significant slope suggests that the increase in concentration of Testosterone in mice causes increase in their metabolic rates. Find the regression equation. Excel: Equation of the Regression line: 𝑦 ̂ = 199.059𝑥 + 379.493 Find the best predicted value of y, metabolic rates, when x is equal to 1.3, the concentration of Testosterone. Significant linear correlation: → Plug in 𝑦 ̂ = 199.059 ∙ 1.3 + 379.493 = 638.2697 Using the above pairs and the value of r, what proportion of the variation in metabolic rates can be explained by the variation in the concentration of Testosterone? Excel: 𝑟2 = 0.9869 → 98.69%
• 5.
  5 Statistics, Sample Test(Exam Review) Solution Module 5: Chapters 10, 11 & 12 Review Chapters 11: Goodness-of-Fit (Multinomial Experiments) and Contingency Tables) 1. Here are the observed frequencies from four categories: 5, 6, 8, and 13. At significance level of 0.05, test the claim that the four categories are all equally likely. a. State the null and alternative hypothesis. b. What is the expected frequency for each of the four categories? c. What is the value of the test statistic? d. Find the critical value(s). e. Make a decision Solution: Definition: A goodness-of-fit test is used to test the hypothesis that an observed frequency distribution fits (or conforms to) some claimed distribution. Goodness-of-fit hypothesis tests are always right-tailed. (If the sum of these weighted squared deviations ∑ (𝑂−𝐸)2 𝐸 is small, the observed frequencies are close to the expected frequencies and there would be no reason to reject the claim that it came from that distribution. Only when the sum is large is the reason to question the distribution. Therefore, the chi-square goodness-of-fit test is always a right tail test.) Given: Multinomial Experiment, k = 4 categories, α = 0.05, use ‘Goodness of fit Test’ a) State the null and the alternative hypotheses: 0 1 2 3 4 : , H P P P P Claim = = = 𝑯𝟏: At least one of the probabilities is different from the others. RTT b) 𝑬 = 𝒏 𝒌 = ∑ 𝒐 𝒌 = 𝟓 + 𝟔 + 𝟖 + 𝟏𝟑 𝟒 = 𝟖 c) Calculate the value of the test statistic (observed value): TS: 𝝌𝟐 = ∑ (𝑶−𝑬)𝟐 𝑬 = (5 − 8)2 + (6 − 8)2 + (8 − 8)2 + (13 − 8)2 8 = 38 8 = 4.750 d) 𝝌𝟐 - Test  = 0.05, df = k – 1 = 3 𝑪𝑽: 𝝌𝟐 (𝜶=𝟎.𝟎𝟓,𝒅𝒇=𝟑) = 𝟕. 𝟖𝟏𝟓
• 6.
  6 TS: 4.750 CV:7.815 (If we use a 2TT we get) 𝑪𝑽: 𝝌𝟐 𝑹(𝜶=𝟎.𝟎𝟐𝟓,𝒅𝒇=𝟑) = 𝟗. 𝟑𝟒𝟖 𝝌𝟐 𝑳(𝜶=𝟎.𝟗𝟕𝟓,𝒅𝒇=𝟑) = 𝟎. 𝟐𝟏𝟔 e) Make a decision: Don’t reject 𝑯𝟎, The claim is true. There is sufficient evidence to support the claim that the four categories are all equally likely. 2. (Goodness-of-fit Test in Multinomial Experiments :) A professor asked 40 of his students to identify the tire they would select as a flat tire of a car carrying 4 students who misses a test (an excuse). The following table summarizes the result, use a 0.05 significance level to test the claim that all 4 tires have equal proportions of being claimed as flat. Tire Left Front Right Front Left Rear Right Rear Number selected 11 15 8 6 Given: Multinomial Experiment, k = 4 categories, 𝜶 = 𝟎. 𝟎𝟓, Use ‘Goodness of fit Test’ 1) State the null and the alternative hypotheses: 0 : 0.25, lF lR rF rR H P P P laim P C = = = = (l = Left, r = Right, R = Rear, F = Front) 𝑯𝟏: At least one 𝒑𝒊 ≠ 𝟎. 𝟐𝟓, 𝟐𝑻𝑻 2) 𝑬 = 𝒏𝒑 = 𝟒𝟎(𝟎. 𝟐𝟓) = 𝟏𝟎 TS: 𝝌𝟐 = ∑ (𝑶−𝑬)𝟐 𝑬 = (11−10)2+(15−10)2+(8−10)2+(6−10)2 10 = 46 10 = 4.6 3) 𝝌𝟐 - Test  = 0.05, df = k – 1 = 3 𝑪𝑽: 𝝌𝟐 (𝜶=𝟎.𝟎𝟓,𝒅𝒇=𝟑) = 𝟕. 𝟖𝟏𝟓 TS: 4.6 CV: 7.815
• 7.
  7 (If we usea 2TT we get): 𝑪𝑽: 𝝌𝟐 𝑹(𝜶=𝟎.𝟎𝟐𝟓,𝒅𝒇=𝟑) = 𝟗. 𝟑𝟒𝟖, 𝝌𝟐 𝑳(𝜶=𝟎.𝟗𝟕𝟓,𝒅𝒇=𝟑) = 𝟎. 𝟐𝟏𝟔 4) Make a decision: Don’t reject 𝑯𝟎, The claim is true; There is sufficient evidence to support the claim that all 4 tires have equal proportions of being claimed as flat. 3. (Test of Independence) Using a 0.05 significance level, test the claim that when the Titanic sank, whether someone survived or died is independent of whether that person is a man, woman, boy, or girl. Table of observed values: 𝐸 = (𝑟𝑜𝑤𝑡𝑜𝑡𝑎𝑙)(𝑐𝑜𝑙𝑢𝑚𝑛𝑡𝑜𝑡𝑎𝑙) (𝐺𝑟𝑎𝑛𝑑𝑡𝑜𝑡𝑎𝑙) We will use this to find expected frequencies. For the upper left-hand cell, we find: 𝐸 = 706(1692) 2223 = 537.360 So, using the same procedure we get the following table of expected values: Men Women Boys Girls Total Survived 332 E=(537.36) 318 (134.022) 29 (20.326) 27 (14.291) 706 Died 1360 (1154.64) 104 (287.978) 35 (43.674) 18 (30.709) 1517 Total 1692 422 64 45 2223 To interpret this result for this upper left-hand cell, we can say that although 332 men actually survived, we would have expected 537.36 men to survive if survivability is independent of whether the person is a man, woman, boy, or girl. To interpret this result for the lower left-hand cell, we can say that although 1360 men actually died, we would have expected 1154.64 men to die if survivability is independent of whether the person is a man, woman, boy, or girl. Note: ∑ 𝐸 𝑖𝑛 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡.
• 8.
  8 Example: Column 1:537.36 + 1154.64 = 1692 Given: Contingency Table: Choose 𝜶 = 𝟎. 𝟎𝟓, We use a ‘Test of Independence’ Test of Independence (of variables): In a test of independence, we test the null hypothesis that in a contingency table, the row and column variables are independent. (That is, there is no dependency between the row variable and the column variable.) The Test of Independence of Variables is used to determine whether two variables are independent of or related to each other when a single sample is selected. Tests of independence are always right-tailed. (If the sum of these weighted squared deviations ∑ (𝑂−𝐸)2 𝐸 is small, the observed frequencies are close to the expected frequencies and there would be no reason to reject the claim that it came from that distribution. Only when the sum is large is the reason to question the distribution. Therefore, the chi-square goodness-of-fit test is always a right tail test.) 1) State the null and the alternative hypotheses: 𝑯𝟎: Whether a person survived is independent of whether the person is a man, woman, boy, or girl. Claim 𝑯𝟏: Surviving the Titanic and being a man, woman, boy, or girl is dependent. (𝑯𝟎: The row variable is independent of the column variable 𝑯𝟏: The row variable is dependent (related to) the column variable) Recall: Tests of Independence are always right-tailed. 2) TS: 𝝌𝟐 = ∑ (𝑶−𝑬)𝟐 𝑬 𝝌𝟐 = (𝟑𝟑𝟐 − 𝟓𝟑𝟕. 𝟔)𝟐 𝟓𝟑𝟕. 𝟔 +. . . (𝟏𝟖 − 𝟑𝟎. 𝟕𝟎𝟗)𝟐 𝟑. . 𝟕𝟎𝟗 = 507.084 3) 𝝌𝟐 - Test  = 0.05, 𝑑𝑓 = (𝑟 − 1)(𝑐 − 1) = (2 − 1)(4 − 1) = 3 𝑪𝑽: 𝝌𝟐 (𝜶=𝟎.𝟎𝟓,𝒅𝒇=𝟑) = 𝟕. 𝟖𝟏𝟓 (If we use a 2TT we get: 𝑪𝑽: 𝝌𝟐 𝑹(𝜶=𝟎.𝟎𝟐𝟓,𝒅𝒇=𝟑) = 𝟗. 𝟑𝟒𝟖, 𝝌𝟐 𝑳(𝜶=𝟎.𝟗𝟕𝟓,𝒅𝒇=𝟑) = 𝟎. 𝟐𝟏𝟔 4) Make a decision: Reject 𝑯𝟎 The claim is false. Survival and gender are Dependent.
• 9.
  9 4. (Test ofHomogeneity) Using the following table, with a 0.05 significance level, test the effect of pollster gender on survey responses by men. Chi-Square Test of Homogeneity: (Test of Homogeneity of Proportions): A chi-square test of homogeneity is a test of the claim that different populations have the same proportions of some characteristics. The Test of Homogeneity of Proportions is used to determine whether the proportions for a variable are equal when several samples are selected from different populations. Tests of Homogeneity are right-tailed. (If the sum of these weighted squared deviations ∑ (𝑂−𝐸)2 𝐸 is small, the observed frequencies are close to the expected frequencies and there would be no reason to reject the claim that it came from that distribution. Only when the sum is large is the reason to question the distribution. Therefore, the chi-square goodness-of-fit test is always a right tail test.) Table of observed values: 𝐸 = (𝑟𝑜𝑤𝑡𝑜𝑡𝑎𝑙)(𝑐𝑜𝑙𝑢𝑚𝑛𝑡𝑜𝑡𝑎𝑙) (𝐺𝑟𝑎𝑛𝑑𝑡𝑜𝑡𝑎𝑙) Table of expected values: Men Women Total Men Who agree 560 (578.67) 308 (289.33) 868 Women Who agree 240 (221.33) 92 (110.67) 332 Total 800 400 1200 Given: Contingency Table: Choose 𝜶 = 𝟎. 𝟎𝟓, We use a ‘Test of Homogeneity, 1) State the null and the alternative hypotheses: 𝑯𝟎: The proportions of agree/disagree responses are the same for the subjects interviewed by men and the subjects interviewed by women. Claim 𝑯𝟏: The proportions are different. 2) TS: 𝝌𝟐 = ∑ (𝑶−𝑬)𝟐 𝑬 𝝌𝟐 = (𝟓𝟔𝟎 − 𝟓𝟕𝟖. 𝟔𝟕)𝟐 𝟓𝟕𝟖. 𝟔𝟕 +. . . (𝟗𝟐 − 𝟏𝟏𝟎. 𝟔𝟕)𝟐 𝟏𝟏𝟎. 𝟔𝟕 = 𝟔. 𝟓𝟐𝟗
• 10.
  10 3) 𝝌𝟐 - Test = 0.05, 𝑑𝑓 = (𝑟 − 1)(𝑐 − 1) = (2 − 1)(2 − 1) = 1 𝑪𝑽: 𝝌𝟐 (𝜶=𝟎.𝟎𝟓,𝒅𝒇=𝟏) = 𝟑. 𝟖𝟒𝟏 CV: 3.841 TS: 6.529 (If we use a 2TT we get): 𝑪𝑽: 𝝌𝟐 𝑹(𝑨𝒓𝒆𝒂=𝟎.𝟎𝟐𝟓,𝒅𝒇=𝟏) = 𝟓. 𝟎𝟐𝟒, 𝝌𝟐 𝑳(𝑨𝒓𝒆=𝟎.𝟗𝟕𝟓,𝒅𝒇=𝟏) = 𝟎. 𝟎𝟎𝟏 4) Make a decision: Reject 𝑯𝟎 The claim is False. There is sufficient evidence to reject the claim of equal proportions. 5. Test of Homogeneity: Using the following table, with a 0.05 significance level, test the association between people living in a city and becoming infected with a highly resistant bacterium. Living Location City Outside City Bacterial Infected 4239 5923 Condition Not Infected 12900 18986 Step 1: H0: Whether a person is infected is independent of whether the person lives in the city or outside the city. ( p1 = p2 = p3 = … = pn), Claim H1: Whether a person is infected is dependent of whether the person lives in the city or outside the city. RTT
• 11.
  11 Step 2: 𝑻𝑺:𝜒2 = ∑ (𝑂−𝐸)2 𝐸 𝑑𝑓 = (𝑟 − 1)(𝑐 − 1) 𝐸 = (row sum)(column sum) grand total = (10162)(17139) 42048 = 4142.088 Expected Frequencies City Outside City Infected 4142.088 6019.91196 Not Infected 12996.912 18889.088 𝜒2 = (4239−4142.088)2 4142.088 +. . . + (18986−18889.088 )2 18889.088 = 5.047424 Step 3: df = (2 – 1 )(2 – 1) = 1 & α = 0.05 → CV: 𝜒2 = 3.841 CV : χ² = 3.841, TS: χ² = 5.047 Step 4: Decision: Reject H0, (Reject independence) The claim is False There is not sufficient evidence to support the claim that whether a person is infected is independent of whether the person lives in the city or outside the city. (Whether a person is infected is dependent of whether the person lives in the city or outside the city) Total 17139 24909
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  12 Statistics, Sample Test(Exam Review) Solution Module 5: Chapters 10, 11 & 12 Review Chapter 12: Contingency Tables, Analysis of Variance 1. Weights in kg of Cypress trees were obtained from trees planted in a Sandy and Dry area. They were given different treatments as indicated. Fill up the table below and use a 0.05 significance level to test the claim that the four treatment categories yield Cypress trees with the same mean weight. 𝑥𝐺𝑀 = ∑𝑥 𝑁 No Treatment Fertilizer Irrigation Fertilizer & Irrigation 0.24 0.92 0.96 1.07 1.69 0.07 1.43 1.63 1.23 0.56 1.26 1.39 0.99 1.74 1.57 0.49 1.8 1.13 0.75 0.95 𝒙𝑮𝑴 = ∑ 𝒙 𝑵 = 𝟎. 𝟐𝟒 + 𝟏. 𝟔𝟗 + ⋯ + 𝟎. 𝟗𝟓 𝟏𝟓 = 𝟐𝟏. 𝟖𝟕 𝟐𝟎 = 𝟏. 𝟎𝟗𝟑𝟓 𝑺𝑺𝑩 = ∑ 𝒏𝒊(𝑿 ̄ 𝒊 − 𝑿 ̄ 𝑮𝑴)𝟐 = 𝟓(𝟏. 𝟏𝟗 − 𝟏. 𝟎𝟗𝟑𝟓)𝟐 + 𝟓(𝟎. 𝟖𝟖𝟒 − 𝟏. 𝟎𝟗𝟑𝟓)𝟐 + 𝟓(𝟏. 𝟏𝟗𝟒 − 𝟏. 𝟎𝟗𝟑𝟓)𝟐 + 𝟓(𝟏. 𝟏𝟎𝟔 − 𝟏. 𝟎𝟗𝟑𝟓)𝟐 = 𝟎. 𝟑𝟏𝟕𝟐𝟗𝟓 𝑴𝑺𝑩 = 𝑺𝑺𝑩 𝒌 − 𝟏 = 𝟎. 𝟑𝟏𝟕𝟐𝟗𝟓 𝟑 = 𝟎. 𝟏𝟎𝟓𝟕𝟔𝟓 𝑺𝑺𝒘 = ∑(𝒏𝒊 − 𝟏)𝒔𝒊 𝟐 = 𝟒(𝟎. 𝟑𝟗𝟏𝟔) + 𝟒(𝟎. 𝟑𝟗𝟎𝟓) + 𝟒(𝟎. 𝟏𝟏𝟑𝟑) + 𝟒(𝟎. 𝟏𝟗𝟎𝟏) = 𝟒. 𝟑𝟒𝟏𝟗𝟔 𝑴𝑺𝒘 𝑶𝑹 𝑴𝑺𝑬 = 𝑺𝑺𝒘 𝑵 − 𝒌 = 𝟒. 𝟑𝟒𝟏𝟗𝟔 𝟏𝟔 = 𝟎. 𝟐𝟕𝟏𝟑𝟕𝟑 𝑭 = 𝑴𝑺𝑩 𝑴𝑺𝑾 = 𝟎. 𝟏𝟎𝟓𝟕𝟔𝟓 𝟎. 𝟐𝟕𝟏𝟑𝟕𝟑 = 𝟎. 𝟑𝟖𝟗𝟕
• 13.
  13 Step 1: H0:μ1 = μ2 = μ3 = μ4 (Claim) H1: At least one of the treatments has a population mean that is different from the others. (RTT) Step 2: Test statistic: 𝑭 = 0.389741 Step 3: Find the critical value. Number of Groups (Factors): k = 4, d.f.N. = k – 1 = 4 – 1 = 3, Total sample size: N = 20, d.f.D. = N – k = 20 – 4 = 16 α = 0.05 → CV: 𝑭 = 3.23887 Or: P-value = 0.761977 >  = 0.05 Step 4: Decision: a. Fail to Reject H0 b. The claim is True c. There IS sufficient evidence to support the claim that 4 population means are all the same. We conclude that the four treatments DO result in poplar trees with the same weight. 2. Given the readability scores summarized in the following table and a significance level of  = 0.05, use technology to test the claim that the three samples come from populations with means that are not all the same. Clancy Rowling Tolstoy N 12 12 12 x 70.73 80.75 66.15 S 11.33 4.68 7.86 TI Calculator: One Way ANOVA 1. ClrList 𝑳𝟏, 𝑳𝟐, … 2. Enter data 𝑳𝟏, 𝑳𝟐, .. 3. Stat 4. Tests 5. 𝑨𝑵𝑶𝑽𝑨 (𝑳𝟏, 𝑳𝟐,. . ) 6. Enter
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  14 Step 1: 𝑯𝟎:𝝁𝟏 = 𝝁𝟐 = 𝝁𝟑 𝑯𝟏 : At least one of the means is different from the others. Claim Step 2: TS: F = 9.4695 Step 3: Find the critical value.  = 0.05, 1 2 1 3 1 2, ( 1) 3(12 1) 33 df K df K n = − = − = = − = − = CV: F = 3.2849 Also: P-value of 0.000562 or 0.001 <  = 0.05 Reject 0 H The claim is True There is sufficient evidence to support the claim that the three-population means are not all the same. We conclude that those books have readability levels that are not all the same.