Question

A report states that adults 18- to 24- years-old send and receive 128 texts every day. Suppose we take a sample of 25- to 34- year-olds to see if their mean number of daily texts differs from the mean for 18- to 24- year-olds.

(a) State the null and alternative hypotheses we should use to test whether the population mean daily number of texts for 25- to 34-year-olds differs from the population daily mean number of texts for 18- to 24-year-olds. (Enter != for ≠ as needed.)

H_0:

H_a:

(b) Suppose a sample of thirty 25- to 34-year-olds showed a sample mean of 118.9 texts per day. Assume a population standard deviation of 33.17 texts per day.

Compute the p-value. (Round your answer to four decimal places.)

p-value =

(c) With α = 0.05 as the level of significance, what is your conclusion?

(d) Repeat the preceding hypothesis test using the critical value approach.

State the null and alternative hypotheses. (Enter != for ≠ as needed.)

H₀:

H_a:

(e) Find the value of the test statistic. (Round your answer to two decimal places.)

State the critical values for the rejection rule. (Use α = 0.05. Round your answer to two decimal places. If the test is one-tailed, enter NONE for the unused tail.)

test statistic≤

test statistic≥

(f)

State your conclusion.

Do not reject H₀. We cannot conclude that the population mean daily texts for 25- to 34-year-olds differs significantly from the population mean of 128 daily texts for 18- 24-year-olds.

Reject H₀. We cannot conclude that the population mean daily texts for 25- to 34-year-olds differs significantly from the population mean of 128 daily texts for 18- 24-year-olds.   

Do not reject H₀. We can conclude that the population mean daily texts for 25- to 34-year-olds differs significantly from the population mean of 128 daily texts for 18- 24-year-olds.

Reject H₀. We can conclude that the population mean daily texts for 25- to 34-year-olds differs significantly from the population mean of 128 daily texts for 18- 24-year-olds.

Answer 1

(a)

H0: = 128

Ha: 128

(b)

SE = /

= 33.17/

= 6.0560

Test Statistic is given by:
Z = (118.9 - 128)/6.0560

= - 1.5026

Table of Area Under Standard Normal Curve gives area = 0.4332

So,

p- value = (0.5 - 0.4332) X 2 = 0.1336

(c)

Since p - value = 0.1335 is greater than = 0.05, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that their mean number of daily texts differs from the mean for 18- to 24- year-olds.

(d)

H0: = 128

Ha: 128

SE = /

= 33.17/

= 6.0560

Test Statistic is given by:
Z = (118.9 - 128)/6.0560

= - 1.50

= 0.05

From Table, critical values of Z = 1.96.

Test statistic > Critical value

Since calculated value of Z = - 1.50 is greater than critical value of Z = - 1.96, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that their mean number of daily texts differs from the mean for 18- to 24- year-olds.

(f)

Correct option:

Do not reject H₀. We cannot conclude that the population mean daily texts for 25- to 34-year-olds differs significantly from the population mean of 128 daily texts for 18- 24-year-olds.