A study of 102 patients who had digestive-tract surgery found that those who chewed a stick of gum for 15 minutes at mealtimes ended up being discharged after an average of 4.4 days in the hospital, compared to an average of 5.2 days for their counterparts who were not provided with gum. Assuming (1) a sample size of 51 for each group and (2) sample standard deviations of 1.3 and 1.9 days, respectively, use a z-test and the 0.01 level of significance in examining whether the mean hospital stay for the gum-chewing patients could have been this much lower simply by chance. Determine and interpret the p-value for the test.
Answer to relevant QuestionsA study found that women of normal weight missed an average of 3.4 days of work due to illness during the preceding year, compared to an average of 5.2 days for women who were considered overweight. Assuming (1) a sample ...A study has been conducted to examine the effectiveness of a new experimental program for preparing high school students for the Scholastic Aptitude Test. Eighty students have been randomly divided into two groups of 40. The ...Summary data for two independent samples are p1 = 0.36, n1 = 150, and p2 = 0.29, n2 = 100. Use the 0.025 level of significance in testing H0: π1 ≤ π2 versus H1: π1 > π2. A sample of 40 investment customers serviced by an account manager are found to have had an average of $23,000 in transactions during the past year, with a standard deviation of $8500. A sample of 30 customers serviced by ...Given the information in Exercise 11.10, and using the 0.02 level of significance in comparing the sample standard deviations, were we justified in assuming that the population standard deviations are equal? Would your ...
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