Question

According to a survey, typical American spends 154.8 minutes per day watching TV. A survey of 50 Internet users results in a mean time watching TV per day of 128.7 minutes, with a standard deviation of 46.5 minutes. Which appropriate test we should use to determine if Internet users spend less time watching TV

Answer 1

Z > ± 1.645

z= 3.968

Explanation:

We formulate the null and alternate hypotheses as

H0 =μ2 ≥ μ1 Ha: μ2 <μ1 one sided

Let α= 0.05

Since the sample sizes are large therefore the test statistic used under H0 is

The critical region for α= 0.05 for a one tailed test Z > ± 1.645

Z = (x`2- x`1) /s/ √n

Z= 154.8-128.746.5/√50

z= 26.1/6.577

z= 3.968

Since the calculated value of z lies in the critical region we reject H0 that internet users spend more time or equal time.