Question

More time on the Internet: A researcher polled a sample of 1052 adults in the year 2010, asking them how many hours per week they spent on the Internet. The sample mean was 10.19 with a standard deviation of 13.8 A second sample of 2022 adults was taken in the year 2012. For this sample, the mean was 10.87 with a standard deviation of 14.92. Assume these are simple random samples from populations of adults. Can you conclude that the mean number of hours per week spent on the Internet increased between 2010 and 2012? Use the α = 0.01 level of significance.

1. State the null and alternative hypothesis

2. Compute the p-value and Testing Value

3. Do you reject the null hypothesis? Explain?

4. Can you conclude that the mean number of hours per week spent on the Internet increased between 2010 and 2012?

Answer 1

The test hypothesis is

This is a one-sided test because the alternative hypothesis is formulated to detect the difference from the hypothesized mean on the upper side

Now, the value of test static can be found out by following formula:

Using Excel's function =T.DIST.RT(t0,n-1), the P-value for t0 = 1.2297 in an upper-tailed t-test with 3072 degrees of freedom can be computed as

Since P = 0.10945181769070367 > 0.01, we fail to reject the null hypothesis  

Degrees of freedom on the t-test statistic are n1 + n2 - 2 = 2022 + 1052 - 2 = 3072

This implies that

Since t0 = 1.2297<2.3275623717143876=t{0.01, 3072}, we fail to reject the null hypothesis

We have insufficient evidence to claim that the mean number of hours per week spent on the Internet increased between 2010 and 2012.