Question

10. My Notes Ask Your Teacher Two proposed computer mouse designs were compared by recording wrist extension in degrees for 24 people who each used both mouse designs. The difference in wrist extension was calculated by subtracting extension for mouse type B from the wrist extension for mouse type A for each person. The mean difference was reported to be 8.82 degrees. Assume that this sample of 24 people is representative of the population of computer users. (a) Suppose that the standard deviation of the differences was 13 degrees. Is there convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B? Test the relevant hypotheses using a 0.05 significance level. (Use Hd = A - MB) State the appropriate null and alternative hypotheses. Ho: Md = 0 Ha: ud < 0 Ho: 4 = 0 Ha: d = 0 Hold > 0 Ha: Ad < 0 Hoiud < 0 | Ha: Hd > 0 o Ho: 4d = 0 Hai Md > 0 Find the test statistic and P-value. (Use a table or technology. Round your test statistic to one decimal place and your p-value to three decimal places.) t = 3.323 P-value = 0.999 State the conclusion in the problem context. We fail to reject Ho. There is not convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B. . We reject Ho. There is convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type We fail to reject Ho. There is convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B. We reject Ho. There is not convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B.

Answer 1

Solution: a) Given information; Sample size; 24 Sample mean difference; 8.82 Sample standard deviation of the difference; 13 We are testing; Hi Ho = 0 versus H: 4>0 (last option is correct) The t test statistics is calculated as follows, X-Ha - 8.82-0 I star = 133.3 The P-value is calculated from t distribution table; P-value = P(t>3.324)=>0.001 Conclusion; The second is correct (we reject null hypothesis since the p value less than 5% and conclude that ud > 0).

b) When sample standard deviation of difference is 26. We are testing; Hoilo = 0 versus H1; >0 (last option is correct) The t test statistics is calculated as follows, Izar = X-la 8.82-0 -=> 26 => 1.6 The P-value is calculated from t distribution table; P-value = P(t> 3.324)=>0.055 Conclusion; The first is correct (we fail reject null hypothesis since the p value is not less than 5% and conclude that ud = 0). c) The last option is correct.