Watching TV: In 2012, the General Social Survey asked a sample of 1326 people how much time they spent watching TV each day. The mean number of hours was 3.02 with a standard deviation of 2.64. A sociologist claims that people watch a mean of 3 hours of TV per day. Do the data provide sufficient evidence to conclude that the mean hours of TV watched per day differs from the claim? Use the = α0.05 level of significance and the P-value method with the TI-84 Plus calculator.

We fail to reject H₀ as there is insufficient evidence at 0.5% level of significance to conclude that the mean hours of TV watched per day differs from the claim.

Step-by-step explanation:

This is a two-tailed test.

We first need to calculate the test statistic. The test statistic is calculated as follows:

Z_calc = X - μ₀ / (s / √n)

where

X is the mean number of hours μ₀ is the mean that the sociologist claims is true s is the standard deviation n is the sample size

Therefore,

Z_calc = (3.02 - 3) / (2.64 / √ (1326))

= 0.2759

Now we have to calculate the z-value. The z-value is calculated as follows:

z_α/2 = z_ (0.05/2) = z_0.025

Using the p-value method:

P = 1 - α/2

= 1 - 0.025

= 0.975

Thus, using the positive z-table, you will find that the z-value is

1.96.

Therefore, we reject H₀ if | Z_calc | > z_ (α/2)

Thus, since

| Z_calc | < 1.96, we fail to reject H₀ as there is insufficient evidence at 0.5% level of significance to conclude that the mean hours of TV watched per day differs from the claim.