Question

In early January 2020, the average household watched on average 57 hours of TV per week. However, social scientist suspects watching trends during COVID-19 has increased. He randomly selects 30 households and finds out that they watch 62 hours of TV per week on average and the standard deviation is 8. Is there evidence to conclude the TV hours watch has significantly increased?

1. State the Null and Alternative

2. Find The Standard error and Test statistics (show work for credit ) include the decoding.

3. Draw the standard normal Distribution and decide how likely is the test statistics.

4. Decide to reject or accept the Null hypothesis ~ Use the Test statistics and or the p-value. Show or describe where did you get the p-value if use it.

5. Conclusion

Answer 1

The provided sample mean is Xˉ=62

and the standard deviation is σ=8, a

and the sample size is n = 30.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=57

Ha: μ>57

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05,

and the critical value for a right-tailed test is

z_c = 1.64

The rejection region for this right-tailed test is

R={z:z>1.64}

(3) Test Statistics

Standard error =

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

z=3.423>zc​=1.64,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0.0003,

and since p = 0.0003<0.05,

it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected.

Therefore, there is enough evidence to claim that the population mean μ is greater than 57, at the 0.05 significance level.or we conclude the TV hours watch has significantly increased.

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