In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6973subjects randomly selected from an online group involved with ears. There were 1326 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis.     A. H0​: H≠0.2      H1​: p=0.2 B. H0​: p=0.2      H1​: p≠0.2 C. H0​: p<0.2      H1​: p=0.2 D. H0​: p=0.2      H1​: p>0.2 E. H0​: p>0.2     H1​: p=0.2   F. H0​: p=0.2     H1​: p<0.2   The test statistic is z=  ​(Round to two decimal places as​ needed.)   The​ P-value is ​(Round to three decimal places as​ needed.)   Because the​ P-value is? ▼ -less than or -greater than   the significance​ level, ▼ -reject or  -fail to reject   the null hypothesis. There is ▼ -sufficient or -insufficient   evidence to support the claim that the return rate is less than​ 20%.

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In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6973subjects randomly selected from an online group involved with ears. There were 1326 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis.
 
 
A. H0​: H≠0.2
     H1​: p=0.2
B. H0​: p=0.2
     H1​: p≠0.2
C. H0​: p<0.2
     H1​: p=0.2
D. H0​: p=0.2
     H1​: p>0.2
E. H0​: p>0.2
    H1​: p=0.2
 
F. H0​: p=0.2
    H1​: p<0.2
 
The test statistic is z= 
​(Round to two decimal places as​ needed.)
 
The​ P-value is
​(Round to three decimal places as​ needed.)
 
Because the​ P-value is?
-less than or
-greater than
 
the significance​ level,
-reject or 
-fail to reject
 
the null hypothesis. There is
-sufficient or
-insufficient
 
evidence to support the claim that the return rate is less than​ 20%.
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