Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6989 subjects randomly selected from an online group involved with ears. There were

1291surveys 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.

A:Identify the null hypothesis and alternative hypothesis.

A.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pless than<0.2

B.

Upper H 0H0​:

pless than<0.2

Upper H 1H1​:

pequals=0.2

C.

Upper H 0H0​:

pgreater than>0.2

Upper H 1H1​:

pequals=0.2

D.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pnot equals≠0.2

E.

Upper H 0H0​:

pnot equals≠0.2

Upper H 1H1​:

pequals=0.2

F.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pgreater than>0.2

B. The test statistic is

z=.

​(Round to two decimal places as​ needed.)

C. The​ P-value =

​(Round to three decimal places as​ needed.)

D. 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%.

Answer 1

A.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pless than<0.2

P = X / n = 1291/6989 = 0.1847

Test Statistic :-
Z = ( P - P0) / √(P0 * q0 / n)
Z = ( 0.1847 - 0.2 ) / √(( 0.2 * 0.8) /6989)
Z = -3.19

P value = P ( Z < -3.1937 ) = 0.0007 ≈ 0.001

Reject null hypothesis if P value < α = 0.01
Since P value = 0.001 < 0.01, hence we reject the null hypothesis
Conclusion :- We Reject H0

D. Because the​ P-value is (Less than) the significance​ level (reject) the null hypothesis. There is (sufficient) evidence to support the claim that the return rate is less than​ 20%.