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Math Statistics Elementary Statistics Books a la carte Plus NEW MyLab Statistics with Pearson eText - Access Card Package (6th Edition)

a. To identify: The claim. To state: The hypothesis H 0 and H a . Answer: The claim is σ 1 2 ≠ σ 2 2 . The hypothesis H 0 and H a are, H 0 : σ 1 = σ 2 H a : σ 1 ≠ σ 2 Explanation: Given info: The level of significance is 0.01, the sample statistic for Location A is s = 0.95 , n = 16 and for Location B is s = 0.78 , n = 22 . Calculation: Here, the pH levels of the soil in two geographic locations have equal standard deviations is tested. Hence, the claim is that the pH levels of the soil in two geographic locations have equal standard deviations. The hypotheses are given below: Null hypothesis: H 0 : σ 1 2 = σ 2 2 Alternative hypothesis: H a : σ 1 2 ≠ σ 2 2 b. To identify: The level of significance. Answer: The level of significance is 0.01. Explanation: From the question is clear that the level of significance is 0.01. c. To identify: The degrees of freedom for the numerator and the denominator. Answer: The degrees of freedom for the numerator and the denominator is, d .f . N = 15 and d .f . D = 21 Explanation: Degrees of freedom: The degrees of freedom are d .f . N = n 1 − 1 and d .f . D = n 2 − 1 . That is, d .f . N = n 1 − 1 =16-1 =15 d .f . D = n 2 − 1 = 22 − 1 = 21 d. To find: The critical value. To identify: The rejection region. Answer: The critical value is 3.08. The rejection region is F > 3.08 . Explanation: Given info: The level of significance is 0.01. Calculation: Degrees of freedom: The degrees of freedom are d .f . N = n 1 − 1 and d .f . D = n 2 − 1 . That is, d .f . N = n 1 − 1 =16-1 =15 d .f . D = n 2 − 1 = 22 − 1 = 21 Critical value: The level of significance is α = 0.01 . The critical F -value for a right-tailed test is obtained using the Table 7: F -Distribution with the level of significance 0.01. Procedure: Locate 21 in the degrees of freedom, denominator row of the Table 7. Obtain the value in the corresponding degrees of freedom, numerator column below 15. That is, F 0 = 3.03 Rejection region: The null hypothesis would be rejected if F > 3.03 . Thus, the rejection region is F > 3.03 . e. To obtain: The test statistic. Answer: The test statistic is 1.483. Explanation: Calculation: Test statistic: The formula to find the F -statistic is, F = s 1 2 s 2 2 Substitute s 1 2 = 0.9025 and s 2 2 = 0.6084 F = 0.9025 0.6084 = 1.483 Thus, the F -statistic value is 1.483. f. To check: Whether the null hypothesis is rejected or fails to reject. Answer: The null hypothesis is not rejected. Explanation: Conclusion: From the result of (c), it can be observed that the test-statistic value is 0.46. Here, the F -statistic value is lesser than the critical value. That is, 0.46 < 3.08 . Thus, it can be concluded that the null hypothesis is not rejected. g. To interpret: The decision in the context of the original claim. Answer: There is no evidence to reject the claim that the pH levels of the soil in two geographic locations have equal standard deviations. Explanation: Interpretation: Here, the F - value is lesser than the critical value. That is, 1.483 < 3.03 . Thus, it can be concluding that the null hypothesis is fails to reject. Hence, it can be conclude that there is no evidence to reject the claim that the pH levels of the soil in two geographic locations have equal standard deviations.

a. To identify: The claim. To state: The hypothesis H 0 and H a . Answer: The claim is σ 1 2 ≠ σ 2 2 . The hypothesis H 0 and H a are, H 0 : σ 1 = σ 2 H a : σ 1 ≠ σ 2 Explanation: Given info: The level of significance is 0.01, the sample statistic for Location A is s = 0.95 , n = 16 and for Location B is s = 0.78 , n = 22 . Calculation: Here, the pH levels of the soil in two geographic locations have equal standard deviations is tested. Hence, the claim is that the pH levels of the soil in two geographic locations have equal standard deviations. The hypotheses are given below: Null hypothesis: H 0 : σ 1 2 = σ 2 2 Alternative hypothesis: H a : σ 1 2 ≠ σ 2 2 b. To identify: The level of significance. Answer: The level of significance is 0.01. Explanation: From the question is clear that the level of significance is 0.01. c. To identify: The degrees of freedom for the numerator and the denominator. Answer: The degrees of freedom for the numerator and the denominator is, d .f . N = 15 and d .f . D = 21 Explanation: Degrees of freedom: The degrees of freedom are d .f . N = n 1 − 1 and d .f . D = n 2 − 1 . That is, d .f . N = n 1 − 1 =16-1 =15 d .f . D = n 2 − 1 = 22 − 1 = 21 d. To find: The critical value. To identify: The rejection region. Answer: The critical value is 3.08. The rejection region is F > 3.08 . Explanation: Given info: The level of significance is 0.01. Calculation: Degrees of freedom: The degrees of freedom are d .f . N = n 1 − 1 and d .f . D = n 2 − 1 . That is, d .f . N = n 1 − 1 =16-1 =15 d .f . D = n 2 − 1 = 22 − 1 = 21 Critical value: The level of significance is α = 0.01 . The critical F -value for a right-tailed test is obtained using the Table 7: F -Distribution with the level of significance 0.01. Procedure: Locate 21 in the degrees of freedom, denominator row of the Table 7. Obtain the value in the corresponding degrees of freedom, numerator column below 15. That is, F 0 = 3.03 Rejection region: The null hypothesis would be rejected if F > 3.03 . Thus, the rejection region is F > 3.03 . e. To obtain: The test statistic. Answer: The test statistic is 1.483. Explanation: Calculation: Test statistic: The formula to find the F -statistic is, F = s 1 2 s 2 2 Substitute s 1 2 = 0.9025 and s 2 2 = 0.6084 F = 0.9025 0.6084 = 1.483 Thus, the F -statistic value is 1.483. f. To check: Whether the null hypothesis is rejected or fails to reject. Answer: The null hypothesis is not rejected. Explanation: Conclusion: From the result of (c), it can be observed that the test-statistic value is 0.46. Here, the F -statistic value is lesser than the critical value. That is, 0.46 < 3.08 . Thus, it can be concluded that the null hypothesis is not rejected. g. To interpret: The decision in the context of the original claim. Answer: There is no evidence to reject the claim that the pH levels of the soil in two geographic locations have equal standard deviations. Explanation: Interpretation: Here, the F - value is lesser than the critical value. That is, 1.483 < 3.03 . Thus, it can be concluding that the null hypothesis is fails to reject. Hence, it can be conclude that there is no evidence to reject the claim that the pH levels of the soil in two geographic locations have equal standard deviations.

Elementary Statistics Books a la carte Plus NEW MyLab Statistics with Pearson eText - Access Card Package (6th Edition)

Elementary Statistics Books a la carte Plus NEW MyLab Statistics with Pearson eText - Access Card Package (6th Edition)

6th Edition

ISBN: 9780133876239

Author: Ron Larson, Betsy Farber

Publisher: PEARSON

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Elementary Statistics Books a la carte Plus NEW MyLab Statistics with Pearson eText - Access Card Package (6th Edition)

Ch. 10.1 - Prob. 1TY Ch. 10.1 - Prob. 2TY Ch. 10.1 - Prob. 3TY Ch. 10.1 - What is a multinomial experiment?Ch. 10.1 - What conditions are necessary to use the...Ch. 10.1 - Finding Expected Frequencies In Exercises 36, find...Ch. 10.1 - Finding Expected Frequencies In Exercises 36, find...Ch. 10.1 - Finding Expected Frequencies In Exercises 36, find...Ch. 10.1 - Finding Expected Frequencies In Exercises 36, find...Ch. 10.1 - Prob. 7E

Ch. 10.1 - Prob. 8E Ch. 10.1 - Performing a Chi-Square Goodness-of-Fit Test In...Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Performing a Chi-Square Goodness-of-Fit Test In...Ch. 10.1 - Performing a Chi-Square Goodness-of-Fit Test In...Ch. 10.1 - Performing a Chi-Square Goodness-of-Fit Test In...Ch. 10.1 - Prob. 16E Ch. 10.1 - In Exercises 17 and 18, (a) find the expected...Ch. 10.1 - In Exercises 17 and 18, (a) find the expected...Ch. 10.2 - Prob. 1TY Ch. 10.2 - Prob. 2TY Ch. 10.2 - Prob. 3TY Ch. 10.2 - Prob. 1E Ch. 10.2 - Explain the difference between marginal...Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - True or False? In Exercises 5 and 6, determine...Ch. 10.2 - Finding Expected Frequencies In Exercises 712, (a)...Ch. 10.2 - Finding Expected Frequencies In Exercises 712, (a)...Ch. 10.2 - Prob. 9E Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - Performing a Chi-Square Independence Test In...Ch. 10.2 - Performing a Chi-Square Independence Test In...Ch. 10.2 - Performing a Chi-Square Independence Test In...Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Performing a Chi-Square Independence Test In...Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - Prob. 21E Ch. 10.2 - Prob. 22E Ch. 10.2 - Prob. 23E Ch. 10.2 - Prob. 24E Ch. 10.2 - Prob. 25E Ch. 10.2 - Prob. 26E Ch. 10.2 - Prob. 27E Ch. 10.2 - Prob. 28E Ch. 10.2 - Prob. 29E Ch. 10.2 - Prob. 30E Ch. 10.2 - Prob. 31E Ch. 10.2 - Prob. 32E Ch. 10.2 - Prob. 33E Ch. 10.2 - Prob. 34E Ch. 10.2 - Prob. 35E Ch. 10.2 - In your opinion, how safe is the food you buy? CBS...Ch. 10.2 - In your opinion, how safe is the food you buy? CBS...Ch. 10.2 - In your opinion, how safe is the food you buy? CBS...Ch. 10.2 - In your opinion, how safe is the food you buy? CBS...Ch. 10.2 - In your opinion, how safe is the food you buy? CBS...Ch. 10.2 - In your opinion, how safe is the food you buy? CBS...Ch. 10.3 - Prob. 1TY Ch. 10.3 - Prob. 2TY Ch. 10.3 - Prob. 3TY Ch. 10.3 - Prob. 4TY Ch. 10.3 - Explain how to find the critical value for an...Ch. 10.3 - List five properties of the F-distribution.Ch. 10.3 - List the three conditions that must be met in...Ch. 10.3 - Explain how to determine the values of d.f.N and...Ch. 10.3 - Prob. 5E Ch. 10.3 - Prob. 6E Ch. 10.3 - Prob. 7E Ch. 10.3 - Finding a Critical F-Value for a Right-Tailed Test...Ch. 10.3 - Prob. 9E Ch. 10.3 - Finding a Critical F-Value for a Right-Tailed Test...Ch. 10.3 - Finding a Critical F-Value for a Right-Tailed Test...Ch. 10.3 - Finding a Critical F-Value for a Right-Tailed Test...Ch. 10.3 - In Exercises 1318, test the claim about the...Ch. 10.3 - In Exercises 1318, test the claim about the...Ch. 10.3 - Prob. 15E Ch. 10.3 - Prob. 16E Ch. 10.3 - Prob. 17E Ch. 10.3 - In Exercises 1318, test the claim about the...Ch. 10.3 - Performing a Two-Sample F-Test In Exercises 1926,...Ch. 10.3 - Prob. 20E Ch. 10.3 - Prob. 21E Ch. 10.3 - Performing a Two-Sample F-Test In Exercises 1926,...Ch. 10.3 - Performing a Two-Sample F-Test In Exercises 1926,...Ch. 10.3 - Prob. 24E Ch. 10.3 - Prob. 25E Ch. 10.3 - Performing a Two-Sample F-Test In Exercises 1926,...Ch. 10.3 - Prob. 27E Ch. 10.3 - In Exercises 27 and 28, find the right- and...Ch. 10.3 - In Exercises 29 and 30, construct the confidence...Ch. 10.3 - In Exercises 29 and 30, construct the confidence...Ch. 10.4 - Prob. 1TY Ch. 10.4 - Prob. 2TY Ch. 10.4 - Slate the null and alternative hypotheses for a...Ch. 10.4 - What conditions are necessary in order to use a...Ch. 10.4 - Describe the difference between the variance...Ch. 10.4 - Prob. 4E Ch. 10.4 - Performing a One-Way ANOVA Test In Exercises 514,...Ch. 10.4 - Prob. 6E Ch. 10.4 - Prob. 7E Ch. 10.4 - Prob. 8E Ch. 10.4 - Prob. 9E Ch. 10.4 - Prob. 10E Ch. 10.4 - Prob. 11E Ch. 10.4 - Prob. 12E Ch. 10.4 - Prob. 13E Ch. 10.4 - Prob. 14E Ch. 10.4 - Prob. 15E Ch. 10.4 - Prob. 16E Ch. 10.4 - Prob. 17E Ch. 10.4 - Prob. 18E Ch. 10.4 - The Scheff Test If the null hypothesis is rejected...Ch. 10.4 - Prob. 20E Ch. 10.4 - Prob. 21E Ch. 10.4 - Prob. 22E Ch. 10 - In Exercises 14. (a) identify the claim and state...Ch. 10 - Prob. 10.1.2RE Ch. 10 - In Exercises 14, (a) identify the claim and state...Ch. 10 - Prob. 10.1.4RE Ch. 10 - Prob. 10.2.5RE Ch. 10 - Prob. 10.2.6RE Ch. 10 - Prob. 10.2.7RE Ch. 10 - Prob. 10.2.8RE Ch. 10 - Prob. 10.3.9RE Ch. 10 - Prob. 10.3.10RE Ch. 10 - Prob. 10.3.11RE Ch. 10 - Prob. 10.3.12RE Ch. 10 - Prob. 10.3.13RE Ch. 10 - Prob. 10.3.14RE Ch. 10 - Prob. 10.3.15RE Ch. 10 - Prob. 10.3.16RE Ch. 10 - Prob. 10.3.17RE Ch. 10 - Prob. 10.3.18RE Ch. 10 - Prob. 10.3.19RE Ch. 10 - Prob. 10.3.20RE Ch. 10 - Prob. 10.4.21RE Ch. 10 - Prob. 10.4.22RE Ch. 10 - Prob. 1CQ Ch. 10 - Prob. 2CQ Ch. 10 - Prob. 3CQ Ch. 10 - Prob. 4CQ Ch. 10 - Prob. 1CT Ch. 10 - Prob. 2CT Ch. 10 - Prob. 3CT Ch. 10 - Prob. 4CT Ch. 10 - Prob. 5CT Ch. 10 - Prob. 6CT Ch. 10 - Prob. 1RSRD Ch. 10 - Independence The contingency table below shows the...Ch. 10 - Prob. 1T Ch. 10 - Prob. 2T Ch. 10 - Prob. 3T Ch. 10 - Prob. 4T Ch. 10 - Prob. 5T Ch. 10 - Prob. 1CR Ch. 10 - Prob. 2CR Ch. 10 - Prob. 3CR Ch. 10 - Prob. 4CR Ch. 10 - Prob. 5CR Ch. 10 - Reviewing a Movie The contingency table shows how...Ch. 10 - Prob. 7CR

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