A US Web Usage Snapshot indicated a monthly average of 36 In

A U.S. Web Usage Snapshot indicated a monthly average of 36 Internet visits per user from home. A random sample of 24 Internet users yielded a sample mean of 42.1 visits with a standard deviation of 5.3. At the 0.01 level of significance, can it be concluded that this differs from the national average?

Step 1: State hypotheses by filling in the symbol (=, <, >, or not equal) and the population mean:

H_o: Blank

H₁: Blank

Step 2: Find the critical value (from the table) (example: 2.345 or -2.345 or +/- 2.345)

Critical t value is:

Step 3: Compute the test value using the formula (round to two decimal places, example 6.45):

T test value is:

Step 4: Reject the null or do not reject the null (type in either Reject the null or do not reject the null only):

Step 5: Conclusion sentence (type in either is or is not only, to reflect what you found):

There         enough evidence to support the claim that the mean number of Internet visits per user from home differs from 36.

Solution

A U.S. Web Usage Snapshot indicated a monthly average of 36 Internet visits per user from home. A random sample of 24 Internet users yielded a sample mean of 42.1 visits with a standard deviation of 5.3. At the 0.01 level of significance, can it be concluded that this differs from the national average?

Step 1: State hypotheses by filling in the symbol (=, <, >, or not equal) and the population mean:

H_o: =36

H₁: not equal 36

Step 2: Find the critical value (from the table) (example: 2.345 or -2.345 or +/- 2.345)

Critical t value is: +/- 2.807

Step 3: Compute the test value using the formula (round to two decimal places, example 6.45):

T test value is: 5.64

Step 4: Reject the null or do not reject the null (type in either Reject the null or do not reject the null only): Reject the null

Step 5: Conclusion sentence (type in either is or is not only, to reflect what you found):

There     is    enough evidence to support the claim that the mean number of Internet visits per user from home differs from 36.

t Test for Hypothesis of the Mean

Data

Null Hypothesis                m=

36

Level of Significance

0.01

Sample Size

24

Sample Mean

42.1

Sample Standard Deviation

5.3

Intermediate Calculations

Standard Error of the Mean

1.0819

Degrees of Freedom

23

t Test Statistic

5.64

Two-Tail Test

Lower Critical Value

-2.807

Upper Critical Value

2.807

p-Value

0.0000

Reject the null hypothesis

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	36
Level of Significance	0.01
Sample Size	24
Sample Mean	42.1
Sample Standard Deviation	5.3
Intermediate Calculations
Standard Error of the Mean	1.0819
Degrees of Freedom	23
t Test Statistic	5.64
Two-Tail Test
Lower Critical Value	-2.807
Upper Critical Value	2.807
p-Value	0.0000
Reject the null hypothesis