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:
Ho: μ Blank
H1: μ 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.
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:
Ho: μ =36
H1: μ 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 |
A U.S. Web Usage Snapshot indicated a monthly average of 36 Internet visits per user from...
A U.S. Web Usage Snapshot indicate a monthly average of 36 internet visits per user from home. A random sample of 24 Internet user yielded a sample mean of 42.1 visitswith a standard deviation of 5.3. At the .01 level of significance can it be concluded that this differs from the national average?
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