Question

19. Suppose you wanted to estimate the mean age of native-born students at North Campus. How many students must you survey if you wanted to be 99% confident in your result and your error is no more than 2 years? Use the standard deviation of 8.65.

22. In a survey of 735 students at Broward College, North Campus, the mean number of hours spent on homework was 7.15 with a standard deviation of 3.20. Using these results, test the claim at the .05 significance level that the mean number of hours spent on homework by students at North Campus is equal to 8.50. Be sure to show all the steps in the hypothesis testing procedure including the null and alternative hypotheses, the calculation of the test statistic, and correct wording of the final conclusion.

Answer 1

Solution(19)
Margin of error for 99% confidence interval can be calculated as
Margin of error = Zalpha/2 * SD/sqrt(n)
Here margin of error = 2
SD = 8.65, Zalpha/2 = 2.575
2 = 2.575*8.65/sqrt(n)
n = (2.575*8.65/2)^2 = 124.029 so Sample size required is 124
Solution(22)
Null hypothesis H0: mean = 8.5
Alternate hypothesis Ha: mean is not equal to 8.5
Given no. of sample = 735
Sample mean = 7.15
Sample standard deviation = 3.2
Test stat = (7.15-8.5)/3.2/sqrt(735) = -11.43
Given alpha = 0.05 and this is two tailed test
p-value <0.00001
Here we can reject the null hypothesis as p-value is less than alpha value so we can say that we have significant evidence to say that the mean number of hours spent on homework by students at north campus is not equal to 8.5