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. The test statistic for this hypothesis would be:

a. 2.05
b. 1.71
c. 2.25
d. 1.01

Answer 1

Answer:

b. 1.71

$z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71$  

Step-by-step explanation:

1) Data given and notation

n=827 represent the random sample taken

X=438 represent the  people that indicate that they would like to see the new show in the lineup

$\hat p=\frac{438}{827}=0.5296$ estimated proportion of people that indicate that they would like to see the new show in the lineup

$p_o=0.5$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:$p \leq 0.5$  

Alternative hypothesis:$p > 0.5$  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71$  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

$p_v =P(Z>1.71)=0.044$  

If we compare the p value obtained and the significance level assumed $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.5.