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Question 12 (1 point) Exhibit 13-17 Jimmy Kimmel Live! on ABC, The Tonight Show Starring Jimmy Fallon on NBC, and The Late Show with Stephen Colbert on CBS are three popular late-night talk shows. The following table shows the number of viewers in millions for a 8-week period during the spring for each of these shows (TV by the Numbers website). The Tonight Jimmy Show Late Show Week Kimmel Live Starring with Stephen (ABC) Jimmy Fallon Colbert (CBS) (NBC) April 11 - April 15 2.21 3.24 2.16 April 18 - April 22 2.10 2.77 2.56 April 25 - April 29 2.24 3.15 2.45 May 02 - May 06 2.12 3.20 2.57 May 09 - May 16 2.21 3.31 2.45 May 16 - May 20 1.97 3.10 2.31 May 23 - May 27 2.47 3.30 2.07 May 30 - June 03 2.64 2.66 2.08April 11 - April 15 2.21 3.24 2.16 April 18 - April 22 2.10 2.77 2.56 April 25 - April 29 2.24 3.15 2.45 May 02 - May 06 2.12 3.20 2.57 May 09 - May 16 2.21 3.31 2.45 May 16 - May 20 1.97 3.10 2.31 May 23 - May 27 2.47 3.30 2.07 May 30 - June 03 2.64 2.66 2.08 Refer to Exhibit 13-17. At the .05 level of significance, do we reject the null hypothesis of no difference in the mean number of viewers per week for the three late-night talk shows? Yes, reject Ho No, do not reject Ho The result is inconclusive None of the answers is correct.Question 13 (1 point) Exhibit 10-14 The following information was obtained from independent random samples. Assume normally distributed populations with equal variances Sample 1 Sample 2 Sample mean 38 36 Sample variance 76 73 Sample size 13 9 Refer to Exhibit 10-14. The standard error of X -X, is 3.74 3.43 9.373 19.48Question 14 (1 point) Exhibit 10-30 In order to determine whether or not there is a significant difference between the hourly wages of workers at Ford and Ferrari, the following data have been accumulated. Ford Ferrari 71 = 50 72 = 70 x , = $20.50 X , = $10.25 01= $2.50 72= $2.00 Refer to Exhibit 10-30. The alternative hypothesis for this test is HI - H2 = 0 OMI - M250 OMI - 12= 0Question 15 (1 point) Exhibit 10-30 In order to determine whether or not there is a significant difference between the hourly wages of workers at Ford and Ferrari, the following data have been accumulated. Ford Ferrari 71 = 50 72 = 70 X , = $20.50 *, = $10.25 1= $2.50 72= $2.00 Refer to Exhibit 10-30. The test statistic has a value of O 24 O 2.7 O 1.645 2.576 Question 16 (1 point) Exhibit 10-6 The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card. You are given the following information. Assume the samples were selected randomly.Question 16 (1 point) Exhibit 10-6 The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card. You are given the following information. Assume the samples were selected randomly. Store's Card Major Credit Card Sample size 64 49 Sample mean $140 $125 Population standard deviation $10 $8 Refer to Exhibit 10-6. At 95% confidence, the margin of error is O 1.694 O 3.32 1.96 O15 Question 17 (1 point) Exhibit 10-4 The following information was obtained from independent random samples. Assume normally distributed populations with equal variances. Sample 1 Sample 2 Sample mean 45 42 Sample variance 85 90 Sample size 10 12 Refer to Exhibit 10-4. The degrees of freedom for the f distribution are O 22 21 19.48 18.12Question 18 (1 point) Exhibit 10-6 The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card. You are given the following information. Assume the samples were selected randomly. Store's Card Major Credit Card Sample size 64 49 Sample mean $140 $125 Population standard deviation $10 $8 Refer to Exhibit 10-6. A point estimate for the difference between the mean purchases of the users of the two credit cards (Store's Card - Major Credit Card) is O: O O 265 O 15Question 10 (1 point) Exhibit 13-21 Sum of Degrees of Mean Source of Variation F Squares Freedom Square Between Groups (Treatment) 6,000.0 5 1,200 Within Groups (Error) 24 288 Total 29 Refer to Exhibit 13-21. The null hypothesis for this ANOVA problem is MI = My = M3 = MA= US =146 OM = M2 =...=/20 OH1 = My = M3 = MA = MS OH = H2 = M3 = HA Previous Page Next Page Page 5 of 9