Question

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.8 hours. (a) The sample standard deviation was not reported, but suppose that it was 6 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.6 hours. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.) P-value State the conclusion in the problem context O Reject Ho. We have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.6 hours. ⓔ Do not reject Ho, we do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.6 hours. Do not reject Ho- We have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.6 hours ● Reject Ho, we do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.6 hours. b) Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.6 hours. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places) P-value - State the conclusion in the problem context O Do not reject Ho. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.6 hours. ⓔ Reject Ho we have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.6 hours Do not reject Ho. We have convincinq evidence that the mean weekly time spent using the Internet by Canadians is qreater than 12.6 hours

Answer 1

Question 1:

The test statistic here is computed as:

$\small t^* = \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{12.8 - 12.6}{\frac{6}{\sqrt{1000}}} =1.05$

Therefore 1.05 is the test statistic value here.

For n - 1 = 999 degrees of freedom, we get from the t distribution tables that:

p = P(t₉₉₉ > 1.05) = 0.147

Therefore 0.147 is the required p-value here.