Solution to Chap 10-Business Statistic

13307 Words54 Pages
Solutions to Practice Problems for Part V 1. A random sample of 1,562 undergraduates enrolled in marketing courses was asked to respond on a scale from one (strongly disagree) to seven (strongly agree) to the proposition: "Advertising helps raise our standard of living." The sample mean response was 4.27 and the sample standard deviation was 1.32. Test at the 1% level, against a two-sided alternative, the null hypothesis that the population mean is 4.0. Formulate Hypotheses: |H0: ( |= 4 | |HA: ( |( 4 | Create Decision Rule: |Reject H0 if |-2.575 > z0.005, or | | |2.575 < z0.005 | Calculate Test Statistic: |z |[pic] | | |[pic] | Decision: Reject H0 at the 1% level. 2. A random sample of 76 percentage changes in promised pension benefits of single employer plans after the establishment of the Pension Benefit Guarantee Corporation was observed. The sample mean percentage change was 0.078 and the sample standard deviation was 0.201. Find and interpret the p-value of a test of the null hypothesis that the population mean percentage change is 0, against a two-sided alternative. Formulate Hypotheses: |H0: ( |= 0 | |HA: ( |( 0 | Calculate Test Statistic: |z |[pic] | | |[pic] | p-value: |p-value |[pic] | | |[pic] | | |[pic] | This null hypothesis can be rejected at any value of ( (significance level) greater than 0.0717%. Note: since the z value is too large to look up in the standard normal table, we need to use Excel to calculate this p-value. (Use the function:

More about Solution to Chap 10-Business Statistic

Open Document