Example: Weight by Treatment Section
Research question: Do patients who receive our treatment weigh less than participants who do not receive our treatment?
Participants were randomly assigned to the treatment condition or a control group. After our intervention, their weights were measured in pounds. Weight is a quantitative variable, so we are going to be comparing means in this example. If assumptions are met, we’ll be conducting a two independent means t test.
Our treatment group has a sample size of 45, mean of 140 pounds, and standard deviation of 20 pounds. Our control group has a sample size of 40, sample mean of 150 pounds, and standard deviation of 25 pounds.
Follow the 5 step hypothesis testing procedure to analyze this data in Minitab.
There are two assumptions: (1) the two samples are independent and (2) both populations are normally distributed or \(n_1 \geq 30\) and \(n_2 \geq 30\). The participants were randomly assigned to one of the two groups. They are in no way matched or paired so they are independent. Both groups have sample size of at least 30.
Our hypotheses is based on the research question "Do patients who receive our treatment weigh less than participants who do not receive our treatment?." This indicates a left tail test. (T = treatment group, C = control group)
\(H_0\): \(\mu_T = \mu_C\)
\(H_a\): \(\mu_T < \mu_C\)
Use Minitab to perform the t-test.
2-Sample independent t-test using summarized data
- Open Minitab
- Select Stat > Basic Statistics > 2 Sample t...
- Select Summarized data in the dropdown at the top
- Enter the summary statistics in the table with the treatment group as Sample 1 and the control group as Sample 2.
Sample 1 Sample 2 Sample size: 45 40 Sample means: 140 150 Standard deviation: 20 25
- Select the Options button
- For the Alternative hypothesis choose Difference < hypothesized difference
- OK and OK
And we get the following output:
\(\mu_1\): population mean of Sample 1
\(\mu_2\): population mean of Sample 2
Equal variances are not assumed for this analysis.
Estimation of Difference
|Difference||95% CI for Difference|
|Null hypothesis||\(H_0\): \(\mu_1-\mu_2=0\)|
|Alternative hypothesis||\(H_1\): \(\mu_1-\mu_2\lt0\)|
The t-value is -2.02.
The p-value is 0.024.
\(p \leq \alpha\), reject the null hypothesis.
There is evidence that patients who receive our treatment weigh less than participants who do not receive our treatment in the population.