Question

9.

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6978 subjects randomly selected from an online group involved with ears. There were 1286 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

▼

greater than

less 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

Solution :

Given that ,

n = 6978

x = 1286

The null and alternative hypothesis is

H₀ : p = 0.2

H₁ : p < 0.2

This is the left tailed test .

= x / n = 1286 / 6978 = 0.1843

P₀ = 20% = 0.2

1 - P₀ = 1 - 0.2 = 0.8

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.1843 - 0.2 / [0.2 (1 - 0.2) / 6978]

= -3.28

The test statistic = -3.28

P-value = 0.0005

= 0.01  

0.0005 < 0.01

P-value <

The​ P-value is less than the significance​ level,

Reject the null hypothesis .

Conclusion :- Reject the null hypothesis. There is sufficient evidence to support the claim that the return rate is less than​ 20%.