In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6983subjects randomly selected from an online group involved with ears. There were 1315 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​: p=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, greater than   the significance​ level, ▼   reject fail to reject   the null hypothesis. There is ▼   sufficient 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 6983subjects randomly selected from an online group involved with ears. There were 1315 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​:
p=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,
greater than
 
the significance​ level,
 
reject
fail to reject
 
the null hypothesis. There is
 
sufficient
insufficient
 
evidence to support the claim that the return rate is less than​ 20%.
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