z-test
two-means
Example
The average monthly Social Security benefit for a specific year for retired workers was $954.90 and for disabled workers was $894.10. Researchers used data from the Social Security records to test the claim that that the the difference in monthly benefits between the two groups was greater than $30. Based on the following information, can the researchers’ claim be supported at the 0.05 level of significance?
Retired | Disabled | |
Sample Size | 60 | 60 |
Mean Benefit | 960.50 | 902.89 |
Pop. Standard Deviation | 98 | 101 |
Step 1:
Identify the hypothesis and claim.
Set \(\mu_1\) to be the mean benefit for retired
and set \(\mu_2\) to be the mean benefit for disabled
\[H_0:\mu_1-\mu_2\neq30\;\;\;\;\;\;\;\;\ H_1:\mu_1-\mu_2>30\;\;\;\;\text{(Claim)}\]
Step 2:
Collect the data
\[n_1=60\]
\[n_2=60\]
\[\overline{x}_1=960.50\]
\[\overline{x}_2=902.89\]
\[\sigma_1=98\]
\[\sigma_2=101\]
\[\alpha=0.05\]
Step 3:
Plug in the above data into the calculator in Excel!
*Note the newest version of this excel sheet asks for variance (not standard deviation)*
Now we have all the info to do any method we would like! Let's use P-value.
Step 4:
Analysis and conclusion.
Note that this is a two-tailed test and that
\[\frac{\alpha}{2}<P<1-\frac{\alpha}{2}\]
i.e.
\[0.025<0.936<0.975\]
we do not reject \(H_0\), thus we do not have enough evidence to support our claim.