Question

In 1997, a survey of 880 households showed that 139 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions.

In 1997, a survey of 880 households showed that 139 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions. a. Which of the following is the hypothesis test to be conducted? O A. Ho: p=0.15 OB. Ho:p>0.15 Hyp<0.15 Hy: p=0.15 OC. Hop#0.15 OD. Ho: p=0.15 Hyp=0.15 HD 0.15 O E. Hop<0.15 OF Ho p=0.15 Hyp=0.15 Hy p>0.15 b. What is the P value? P.value (Round to four decimal places as needed.) c. What is the conclusion? There is not sufficient evidence to support the claim that more than 15% of households use e-mail There is sufficient evidence to support the claim that more than 15% of households use e-mail

Answer 1

Option F) is correct.

$H_{0}:p=0.15$

$H_{1}:p>0.15$

We have for given example,                  
Population proportion value is =0.15              
x=139              
                  
n=880              
          
Estimate for sample proportion $= \hat{p}=0.158$
                  
Z test statistic formula for proportion                  
$Z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}$
$=\frac{0.158-0.15}{\sqrt{\frac{0.15*(1-0.15)}{880}}}$
=0.665  

P value = 0.2530.......................by using Z table or by using Excel command =1-NORMSDIST(0.665)

P value is = 0.2530 > 0.05

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