Question

A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If this is to happen, it will have to move one of its most viewed shows to another slot. The network conducts a survey asking its viewers which show they would rather watch. The network receives 827 responses, of which 438 indicate that they would like to see the new show in the lineup.

a. The test statistic for this hypothesis would be:_______
b. Calculate the critical value at α = 0.10.
c. What should the television network do?

1. Do not reject H0, the television network should keep its current lineup.
2. Reject H0, the television network should not keep its current lineup.

d. Select the hypotheses to test if the television network should give its newest television show a spot during prime viewing time at night.

H0: p = 0.50; HA: p < 0.50
H0: p > 0.50; HA: p < 0.50
H0: p < 0.50; HA: p > 0.50

Answer 1

a. The test statistic for this hypothesis would be z=1.71

b. Critical value at α = 0.10: z=1.282

c. Reject H0, the television network should not keep its current lineup.

d. H0: p ≤ 0.50; HA: p > 0.50

Explanation:

We have to performa an hypothesis test on a proportion.

The claim is that the proportion of viewers that prefer the new show is bigger than 50% (meaning that most viewers prefer the new show).

The null hypothesis will state that both shows have the same proportion.

Then, the null and alternative hypothesis are:

`H_0: \pi\leq0.50\\\\H_a: \pi>0.5`

The sample proportion is p=0.53

`p=X/n=438/827=0.53`

The standard error of the proportion is:

`\sigma_p=\sqrt{(p(1-p))/(n)}=\sqrt{(0.5*0.5)/(827)}= 0.017`

Then, the z-statistic can be calculated as:

`z=(p-\pi-0.5/n)/(\sigma_p)=(0.53-0.50-0.5/827)/(0.017)=( 0.029 )/(0.017) =1.706`

The critical value for a right tailed test at a significance level of 0.10 is zc=1.282. This value can be looked up in a standard normal distribution table.

As the statistic is bigger than the critical value, it lies in the rejection region. The null hypothesis is rejected. There is evidence to support the claim that the proportion of viewers which support the new show is larger than 0.50.