# 6b.2 - Minitab: One-Sample Mean Hypothesis Test

6b.2 - Minitab: One-Sample Mean Hypothesis Test

## Minitab® – Conduct a One-Sample Mean t-Test

Note that these steps are very similar to those for one-mean confidence interval. The differences occur in steps 4 through 8.

To conduct the one sample mean t-test in Minitab...

1. Choose Stat > Basic Stat > 1 Sample t.
2. In the drop-down box use "One or more samples, each in a column" if you have the raw data, otherwise select "Summarized data" if you only have the sample statistics.
3. If using the raw data, enter the column of interest into the blank variable window below the drop down selection. If using summarized data, enter the sample size, sample mean, and sample standard deviation in their respective fields.
4. Choose the check box for "Perform hypothesis test" and enter the null hypothesis value.
5. Choose Options.
6. Enter the confidence level associated with alpha (e.g. 95% for alpha of 5%).
7. From the drop down list for "Alternative hypothesis" select the correct alternative.
8. Click OK and OK.

## Example 6-8: Emergency Room Wait Time Recall our emergency room wait time example where an administrator at your local hospital states that on weekends the average wait time for emergency room visits is 10 minutes. From our random sample of 40 patients, the average wait time for these 40 patients was 11 minutes with a standard deviation of 3 minutes. We conducted the test at a 5% level of significance and wanted to demonstrate that the average time exceeded 10 minutes. Also, recall in that example we found by hand a test statistic of t* = 2.11 and p-value with a range between 0.01 to 0.025

Our hypotheses were: $H_0 \colon \mu=10$ and $H_a\colon \mu>10$

Conduct the same test using Minitab.

Using Minitab...

1. Select Stat > Basic Stat > 1 Sample t.
2. Choose the summarized data option and enter 40 for "Sample size", 11 for the "Sample mean", and 3 for the "Standard deviation".
3. Check the box for "Perform Hypothesis Test" and enter the null value of 10
4. Click Options .
5. With our stated alpha value of 5% we keep the default confidence level of 95.
6. Select "Mean> hypothesized mean" from the "Alternative Hypothesis" list.
7. Click OK and OK again.

The output is:

#### One-Sample T

Test of $\mu$ = 10 vs $\mu$ > 10

N Mean StDev SE Mean 95% Lower Bound T P
40 11.000 3.000 0.474 10.201 2.11 0.021

Again, as the output indicates, our hand calculations were quite good. Notice that Minitab provides a more exact p-value of 0.021 which corresponds to our results as it falls within our calculated range of 0.01 to 0.025.

## Finding Exact Critical Value for a One-Sample Mean t-Test

Since the t-table is not as detailed as the z-table, we can only estimate the critical value when the degrees of freedom are not found on the table. In order to obtain the exact critical value to use in order to conduct the rejection region approach, we can use a statistical package such as Minitab.

Minitab commands to obtain critical value:

1. Calc > Probability Distributions > t-distribution
2. Choose the radio button for 'Inverse Cumulative Distribution' (this finds the t-value that produces the entered probability to the left of it).
3. Enter the correct degrees of freedom
4. Choose the radio button for 'Input constant' and enter the alpha value (if one-side alternative) or alpha/2 (if two-sided alternative).
5. Click Ok

## 6-8 Cont'd... Find the exact critical value for our emergency room example. Recall by hand that we had to use the row with 35 degrees of freedom instead of the correct df of 39. In that example our critical value for alpha of 5% was 1.69.

1. Go to Calc > Probability Distributions > t-distribution .
2. Choose the radio button for 'Inverse Cumulative Distribution.'
3. Enter 39 for 'degrees of freedom.'
4. Choose the radio button for 'Input Constant' and enter 0.05

The output is as follows:

Student's t distribution with 39 DF

P(X$\le$x) x
0.05 -1.68488

This is where you need to be a little careful. Remember that our alternative was "greater than" or a right-tailed test. The output is the critical value for a left-tailed test. However, since the t-distribution is symmetrical, the area to the left of -1.68488 would be the same as the area to the right of 1.68488. Therefore, the critical value for out test with 39 degrees of freedom would be 1.68488, which doesn't differ much from the 1.69 we estimated using 35 degrees of freedom. This is why the table skips going one by one after 30; there is little difference between the values when increasing by only one degree of freedom.

  Link ↥ Has Tooltip/Popover Toggleable Visibility