Answer:
We <em>fail to reject H₀ </em>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 <em>fail to reject H₀ </em>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.
