Question

A publisher of a newsmagazine has found through past experience that 60% of subscribers renew their subscriptions. In a random sample of 100 subscribers, 57 indicated that they planned to renew their subscriptions. What is the p-value associated with the test that the current rate of renewals differs from the rate previously experienced? (Round your answer to four decimal places.)

Answer 1

`z=\frac{0.57 -0.6}{\sqrt{(0.6(1-0.6))/(100)}}=-0.612`

`p_v =2*P(z<-0.612)=0.5405`

Explanation:

Data given and notation

n=100 represent the random sample taken

X=57 represent the subscribers indicated that they planned to renew their subscriptions

`\hat p=(57)/(100)=0.57` estimated proportion of subscribers indicated that they planned to renew their subscriptions

`p_o=0.6` 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 (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the current rate of renewals differs from the rate previously experienced, so the system of hypothesis are:

Null hypothesis:
`p=0.6`

Alternative hypothesis:
`p \\eq 0.6`

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

`z=\frac{\hat p -p_o}{\sqrt{(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`.

Calculate the statistic

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

`z=\frac{0.57 -0.6}{\sqrt{(0.6(1-0.6))/(100)}}=-0.612`

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.

Since is a bilateral test the p value would be:

`p_v =2*P(z<-0.612)=0.5405`