In 1991, a survey of 820 households showed that 142 of them use e-mail. Use thos

Question

In 1991, a survey of 820 households showed that 142 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions.

a. Which of the following is the hypothesis test to be conducted?

a. H0 : p < 0.15 H1 & p = 0.15

b. H0 : p = 0.15 & H1 : p < 0.15

c. H0 : p = 0.15 & H1 : p > 0.15

e. H0 : p > 0.15 & H1 : p = 0.15

b. What is the test statistic?

z = ________ (Round to two decimal places as needed.)

c. What is the P-value?

P-value = __________ (Round to three decimal places as needed.)

d. What is the conclusion?

There is suffiecient evidence to support the claim that more than 15% of households use e-mail.

There is not sufficient evidence to support the claim that more 15% of households use e-mail.

e. Is the conclusion valid today? Why or why not?

a. Yes, the conclusion is valid today because the requirements to perform the test are satisfied.

No, the conclusion is not valid today beause the populaion characteristics of the use of e-mail are changing rapidly.

c. You can make no decisions about the validty of the conclusion today.

Explanation / Answer

Below are null and alternate hypothesis
H0 : p = 0.15 & H1 : p > 0.15

(a) Option C

b)
p = 142/820 = 0.1732
SE = sqrt(p*(1-p)/n) = sqrt(0.1732 * (1-0.1732)/820) = 0.0132

Test statistics, z = (0.1732 - 0.15)/0.0132 = 1.7576

c)
p-value = 0.0394

d)
As p-value is less than the significance level of 0.05 we reject null hypothesis
There is suffiecient evidence to support the claim that more than 15% of households use e-mail

e)
You can make no decisions about the validty of the conclusion today.

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