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.)

Step-by-step 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

estimated proportion of subscribers indicated that they planned to renew their subscriptions

is the value that we want to test

represent the significance level

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

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

(1)  

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

Calculate the statistic  

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

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:  

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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