Question

3. Media Metrix, Inc. tracks Internet users in seven countries: Australia, Great Britain, Canada, France, Germany, Japan, and the United State. According to resent measurement figures, American home users rank first in Internet usage with a mean of 14 hours per month. Assume that in a following-up study involving a sample of 205 Canadian Internet users, the sample mean was 12.8 hours per month and the sample standard deviation was 6.2 hours. a. Formulate the null and alterative hypotheses that can be used to determine whether the sample data support the conclusion that Canadian Internet users have a population mean less than the U.S. mean of 14 hours per month. b. What is the p-value? c. Using α =0.01, what is your conclusion? d. Test the hypotheses using the critical value approach (α =0.01).

Answer 1

a)

H0: = 14

Ha: < 14

b)

Test statistics

t = - / S / sqrt(n)

= 12.8 - 14 / 6.2 / sqrt( 205)

= -2.77

This is test statistics value.

From T table,

With test statistics t = -2.77 and df = 204,

p-value = 0.0031

c)

Since p-value < 0.01 , we have sufficient evidence to reject H0.

d)

Critical t value at 0.01 level with 204 df = -2.345

Since test statistics < -2.345, we have sufficient evidence to reject H0.