Chap09_part2 - 158 Chapter 9: Two-Sample Tests with...

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Unformatted text preview: 158 Chapter 9: Two-Sample Tests with Numerical Data 9.60 (a) 1 2 : H μ μ where Populations: 1 = men and 2 = women 1 1 2 : H μ μ Decision rule: df = 113. If t > 2.3598, reject H . Test statistic: – 7.8735 D D D t S n μ = = Decision: Since t calc = 7.8735 is above the critical bound of 2.3598, reject H . There is enough evidence to conclude that the average salary for men is greater than the average salary for women. Excel output: t-Test: Paired Two Sample for Means Male Female Mean 88683.23684 74575.17544 Variance 3834187011 2243135437 Observations 114 114 Pearson Correlation 0.973734531 Hypothesized Mean Difference df 113 t Stat 7.87346338 P(T<=t) one-tail 1.14975E-12 t Critical one-tail 2.359802238 P(T<=t) two-tail 2.29949E-12 t Critical two-tail 2.620035957 (b) From the Excel output, the p-value is essentially zero. (c) The t test for the mean difference concludes that the mean salary for men is greater than the mean salary for women. 158 159 Chapter 9: Two-Sample Tests with Numerical Data 9.61 (a) 1 2 : H μ μ = where 1 is for current ad pages and 2 for last year ad pages 1 1 2 : H μ μ Decision rule: df = 18. If t > 2.1009 or t < - 2.1009, reject H . Test statistic: – 1.8536 D D D t S n μ = = - Decision: Since t calc = - 1.8536 is between the critical bounds of - 2.109 and 2.109, do not reject H . There is not enough evidence to conclude that the mean number of advertising pages in the current issues is different from that of the previous year. Excel output: t-Test: Paired Two Sample for Means Current Ad Pages Ad Pages Last Year Mean 32.09894737 42.13842105 Variance 570.5288655 1057.623614 Observations 19 19 Pearson Correlation 0.689220948 Hypothesized Mean Difference df 18 t Stat-1.853565221 P(T<=t) one-tail 0.040132628 t Critical one-tail 1.734063062 P(T<=t) two-tail 0.080265256 t Critical two-tail 2.100923666 (b) From the Excel output, the p-value = 0.0803. The probability of observing a test statistic that is further away from 0 than the current value is 0.0803. (c) 23.6091 10.0395 2.1009 19 D S D t n t = - 21.4187 1.3398 D μ- < < (d) Since 0 is contained in the 95% confidence interval for the difference in the mean number of advertising pages in the current issues compared to the previous year, the null hypothesis in (a) cannot be rejected. The conclusion drawn using the confidence interval is identical to that in (a). The t test for the mean difference concludes that there is not enough evidence that the mean number of advertising pages in the current issues is different from that of the previous year. 159 Solutions to End-of-Section and Chapter Review Problems 160 9.62 Manufacturer A Manufacturer B Minimum 684 819 First Quartile 852 943 Median 916.5 1015.5 Third Quartile 972 1096 Interquartile Range 120 153 Maximum 1093 1230 Range 409 411 Mean 909.65 1018.35 Median 916.5 1015.5 Mode 926 1077 Standard Deviation 94.3052 96.9014 Sample Variance 8893.4641 9389.8744 Count 40 40 Box-and-whisker Plot Manufacturer A Manufacturer B...
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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Chap09_part2 - 158 Chapter 9: Two-Sample Tests with...

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