Question

A publisher reports that 65% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 340 found that 60% of the readers owned a laptop. State the null and alternative hypotheses. Answer

Answer 1

Answer:

$z=\frac{0.60 -0.65}{\sqrt{\frac{0.65(1-0.65)}{340}}}=-1.933$  

The p value for this case can be calculated with this probability:

$p_v =2*P(z<-1.933)=0.0532$  

For this case is we use a significance level of 5% we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is different from 0.65 or 65%. We need to be careful since if we use a value higher than 65 for the significance the result would change

Step-by-step explanation:

Information given

n=340 represent the random sample taken

$\hat p=0.60$ estimated proportion of readers owned a laptop

$p_o=0.65$ is the value that we want to test

z would represent the statistic

$p_v{/tex} represent the p value

Hypothesis to test

We want to check if the true proportion of readers owned a laptop if different from 0.65

Null hypothesis:[tex]p=0.65$  

Alternative hypothesis:$p \neq 0.65$  

The statistic is given by:

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

Replacing we got:

$z=\frac{0.60 -0.65}{\sqrt{\frac{0.65(1-0.65)}{340}}}=-1.933$  

The p value for this case can be calculated with this probability:

$p_v =2*P(z<-1.933)=0.0532$  

For this case is we use a significance level of 5% we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is different from 0.65 or 65%. We need to be careful since if we use a value higher than 65 for the significance the result would change