- Example: Summarized Data

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.

1. Check any necessary assumptions and write null and alternative hypotheses.

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)

\(H_0\): \(\mu_T = \mu_C\)

\(H_a\): \(\mu_T < \mu_C\)

2. Calculate an appropriate test statistic.

Use Minitab to perform the t-test.

   2-Sample independent t-test using summarized data

  1. Open Minitab
  2. Select Stat > Basic Statistics > 2 Sample t...
  3. Select Summarized data in the dropdown at the top
  4. 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
  5. Select the Options button
  6. For the Alternative hypothesis choose Difference < hypothesized difference
  7. OK and OK

And we get the following output:

2-Sample t: SATM by Ever Cheat

\(\mu_1\): population mean of Sample 1
\(\mu_2\): population mean of Sample 2
Difference: \(\mu_1-\mu_2\)

Equal variances are not assumed for this analysis.

Descriptive Statistics
Sample N Mean StDev SE Mean
Sample 1 45 140.0 20.0 3.0
Sample 2 40 150.0 25.0 4.0
Estimation of Difference
Difference 95% CI for Difference
-10.00 -1.75
Null hypothesis \(H_0\): \(\mu_1-\mu_2=0\)
Alternative hypothesis \(H_1\): \(\mu_1-\mu_2\lt0\)
T-Value DF P-Value
-2.02 74 0.024

The t-value is -2.02.

3. Determine the p-value associated with the test statistic.

The p-value is 0.024.

4. Decide between the null and alternative hypotheses.

\(p \leq \alpha\), reject the null hypothesis.

5. State a "real world" conclusion.

There is evidence that patients who receive our treatment weigh less than participants who do not receive our treatment in the population.