Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to

69726972

subjects randomly selected from an online group involved with ears. There were

13481348

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.

Upper H 0H0​:

pless than<0.2

Upper H 1H1​:

pequals=0.2

B.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pgreater than>0.2

C.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pless than<0.2

D.

Upper H 0H0​:

pnot equals≠0.2

Upper H 1H1​:

pequals=0.2

E.

Upper H 0H0​:

pequals=0.2

Upper H 1H1​:

pnot equals≠0.2

F.

Upper H 0H0​:

pgreater than>0.2

Upper H 1H1​:

pequals=0.2

The test statistic is

zequals=nothing.

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

The​ P-value is

nothing.

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

Because the​ P-value is

▼

greater than

less than

the significance​ level,

▼

reject

fail to reject

the null hypothesis. There is

▼

insufficient

sufficient

evidence to support the claim that the return rate is less than​ 20%.

Answer 1

Given that, n = 6972 and x = 1348

=> sample proportion = 1348/6972 = 0.193345

We want to test the claim that the return rate is less than 20%.

The null and alternative hypotheses are,

H0 : p = 0.20

H1 : p < 0.20

This hypothesis test is a left-tailed test.

Test statistic is,

=> Z = -1.39

p-value = P(Z < -1.39) = 1 - P(Z < 1.39) = 1 - 0.9177 = 0.0823

The p-value is 0.082

Because, the p-value is greater than the significance level, fail to reject the null hypothesis. There is insufficient evidence to support the claim that the return rate is less than​ 20%.