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!

2-mean hypothesis testing example

*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.