A sociologist surveys 50 randomly selected citizens in each of two countries to compare the mean number of hours of volunteer work done by adults in each. Among the 50 inhabitants of Lilliput, the mean hours of volunteer work per year was 52, with standard deviation 11.8. Among the 50 inhabitants of Blefuscu, the mean number of hours of volunteer work per year was 37, with standard deviation 7.2. Construct the 99% confidence interval for the difference in mean number of hours volunteered by all residents of Lilliput and the mean number of hours volunteered by all residents of Blefuscu. Test, at the 1% level of significance, the claim that the mean number of hours volunteered by all residents of Lilliput is more than ten hours greater than the mean number of hours volunteered by all residents of Blefuscu. Compute the observed significance of the test in part (b).

Answer:a:  6.765 < µ1 -  µ2 < 15.235  b: There is not enough evidence to support the claimStep-by-step explanation:The question is asking us to first create a 99% confidence interval for the difference in population mean number of volunteered hours.  The data for the first sample is:n = 50, x = 52, s = 11.8The data for the second sample is:n = 50, x = 37, s = 7.2We don't know the population standard deviations, and they weren't assumed as equal (the question didn't state that they were).  So our degrees of freedom are one less that the smaller sample size.  Since the samples are equal, we just subtract 1, giving degrees of freedom of 49.  The t-score for the confidence level of 99% is t = 2.678 (49 wasn't listed on my chart, so we take the closest value, in this case the t-score for 50)See attached photo 1 for construction of the confidence interval...For the hypothesis test we have:H0:  µ1 - µ2 = 10Ha:  µ1 - µ2 > 10  (claim)Its says test that the mean difference is more than ten.  See attached photo 2 for the hypothesis test...