Answer to Question #115081 in Statistics and Probability for Ayana Danah Falcasantos

Question

PROBLEM SOLVING. Perform a hypothesis test and state your decision. (20 points).
1. The manager of a TV Station claims that the average rate in Manila watching their new sitcom
is 97.7% each day with a standard deviation of 5%. A student researcher who wants statistical
evidence on this claim conducted her own surgery with a selected sample of 5 cities. Her
surgery resulted to a mean of 95%. Using of 5% significance level, can it be concluded that the
average rate watching the new sitcom is less than 97.7%? (10 points).
a. H0: a:
H1:
b. Test statistic. What will be the z-test?
c. P-value.
d. Conclusion.
e. Interpretation.
2. Last year, the mean number of bags prod

Answer 1

a. The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq97.7$

$H_1:\mu<97.7$

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

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.645.$

The rejection region for this left-tailed test is $R:\{z:z<-1.64\}$

b. The z-statistic is computed as follows:

$$z={\bar{X}-\mu \over \sigma/\sqrt{n}}={95-97.7\over 5/\sqrt{5}}\approx-1.207477$$

c. The p-value is $p=P(Z<-1.207477)=0.113624>0.05=\alpha$

d. It is concluded that the null hypothesis is not rejected.

e. Therefore, there is not enough evidence to claim that the average rate watching the new sitcom is less than 97.7%, at the 5% significance level.