1. The Gibbs Baby Food Company wishes to compare the weight gain of infants using their brand versus their competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. The standard deviation of the sample was 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds, with a standard deviation of 2.9 pounds. At the 0.05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Gibbs brand sample: x1 = 7.6, s1 = 2.3, n1 = 40 Competing sample: x2 = 8.1, s2 = 2.9, n2 = 55 ) Hypothesis: Ho : 1 2 Ha : 1 < 2 (claim) B) Critical test value:

Because the sample sizes are both > 30, and the populations are assumed to be independent, we can use the z-test to compare the means.

At = 0.05, for a one-tailed test, zcrit = -1.645 C) Test value: ztest = [ (x1 x2) 0 ] / [ [ (s1)2/n1 + (s2)2/n2 ] ztest = (7.6 8.1) / [ (2.3)2/40 + (2.9)2/55 ] ztest = -0.9363 D) Decision |ztest| < |zcrit| so test is outside the rejection area. Cannot reject the null hypothesis. E) Summary There is not enough evidence to support the claim that babies using Gibbs brand

food gain less weight.

2. Ms. Lisa Monnin is the budget director for the New Process Company. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information: Sales Staff ($) 191 135 146 165 136 142 Audit Staff ($) 130 102 129 143 149 120 139 At the 0.10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than for the audit staff? Sales staff: x1 = 152.50, s1 = 19.8725, n1 = 6 Audit staff: x2 = 130.29, s2 = 14.6162, n2 = 7 ) Hypothesis: Ho : 1 2 Ha : 1 > 2 (claim) B) Critical test value:

Because the population variance is unknown and the sample sizes are both < 30, the z-test is not appropriate here. Instead, use the t-test to compare the means.

At = 0.10, with 5 degrees of freedom, tcrit = 1.476 C) Test value: ttest = [ (x1 x2) 0 ] / [ (s1)2/n1 + (s2)2/n2 ] ttest = (152.5 130.29) / [ (19.8725)2/6 + (14.6162)2/7 ] ttest = 2.2628 D) Decision ttest > tcrit, so the test is inside the rejection area. Reject the null hypothesis. E) Summary There is enough evidence to support the claim that the mean daily expenses are

greater for the sales staff than for the audit staff.

3. The research department at the home office of New Hampshire Insurance conducts on-going research on the causes of automobile accidents, the characteristics of drivers, and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies written on married persons revealed that 150 had been in at least one accident in the previous three-year period. At the 0.05 significance level, can the insurance company conclude that single persons are more prone to having accidents than married persons? This problem requires a comparison of proportions. Single drivers: x1 = 120, n1 = 400 Married drivers: x2 = 150, n2 = 600 ) Hypothesis: Ho : p1 p2 Ha : p1 > p2 (claim) B) Critical test value:

This test is a z-test, with modifications made for comparing proportions. At = 0.05, for a one-tailed test, zcrit = 1.645 C) Test value: p1 = x1/n1 = 120/400 = 0.3 p2 = x2/n2 = 150/600 = 0.25 pbar = (x1 + x2) / (n1 + n2) = (120 + 150) / (400 + 600) = 0.27 qbar = 1 pbar = 0.73 ztest = [ (p1 p2) 0 ] / [ (pbar)(qbar)(1/n + 1/n2) ] ztest = (0.3 0.25) / [ (0.27)(0.73)(1/400 + 1/600) ] ztest = 1.7447 D) Decision ztest > zcrit, so the test is inside the rejection area. Reject the null hypothesis. E) Summary There is enough evidence to support the claim that single persons are more prone

to having accidents than married persons.