9.2 - Two Independent Means

9.2 - Two Independent Means

Let's explore how we can compare the means of two independent groups. If the populations are known to be approximately normally distributed, or if both sample sizes are at least 30, then the sampling distribution can be estimated using the \(t\) distribution. If this assumption is not met then simulation methods (i.e., bootstrapping or randomization) may be used.


9.2.1 - Confidence Intervals

9.2.1 - Confidence Intervals

Given that the populations are known to be normally distributed, or if both sample sizes are at least 30, then the sampling distribution can be approximated using the \(t\) distribution, and the formulas below may be used. Here you will be introduced to the formulas to construct a confidence interval using the \(t\) distribution. Minitab will do all of these calculations for you, however, it uses a more sophisticated method to compute the degrees of freedom so answers may vary slightly, particularly with smaller sample sizes. 

General Form of a Confidence Interval
\(point \;estimate \pm (multiplier) (standard \;error)\)

Here, the point estimate is the difference between the two mean, \(\overline X _1 - \overline X_2\).

Standard Error
\(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\)
Confidence Interval for Two Independent Means
\((\bar{x}_1-\bar{x}_2) \pm t^\ast{ \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\)

The degrees of freedom can be approximated as the smallest sample size minus one.

Estimated Degrees of Freedom

\(df=smallest\;n-1\)

Example: Exam Scores by Learner Type

A STAT 200 instructor wants to know how traditional students and adult learners differ in terms of their final exam scores. She collected the following data from a sample of students:

Descriptive Statistics
  Traditional Students Adult Learners

\(\overline x\)

41.48 40.79
\(s\) 6.03 6.79
\(n\) 239 138

She wants to construct a 95% confidence interval to estimate the mean difference.

The point estimate, or "best estimate," is the difference in sample means: \(\overline x _1 - \overline x_2 = 41.48-40.79=0.69\)

The standard error can be computed next: \(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{\frac{6.03^2}{239}+\frac{6.79^2}{138}}=0.697\)

To find the multiplier, we need to construct a t distribution with \(df=smaller\;n-1=138-1=137\) to find the t scores that separate the middle 95% of the distribution from the outer 5% of the distribution:

Minitab Express was used to construct a t distribution with 137 degrees of freedom to find the t scores that separate the middle 95% from the outer 5%. Those t scores are plus and minus 1.97743

\(t^*=1.97743\)

Now, we can combine all of these values to construct our confidence interval: 

\(point \;estimate \pm (multiplier) (standard \;error)\)

\(0.69 \pm 1.97743 (0.697)\)

\(0.69 \pm 1.379\) The margin of error is 1.379

\([-0.689, 2.069]\) 

We are 95% confident that the mean difference in traditional students' and adult learners' final exam scores is between -0.689 points and +2.069 points. 


9.2.1.1 - Minitab: Confidence Interval Between 2 Independent Means

9.2.1.1 - Minitab: Confidence Interval Between 2 Independent Means

Minitab can be used to construct a confidence interval for the difference between two independent means. Note that the confidence intervals given in the Minitab output assume that either the populations are normally distributed or that both sample sizes are at least 30.

Minitab®  – Confidence Interval Between 2 Independent Means

Let's estimate the difference between the mean weight (in pounds) of females and the mean weight of males. Both sample sizes are at least 30 so the sampling distribution can be approximated using the t distribution.  

  1. Open the Minitab file: class_survey.mpx
  2. Select Stat > Basic Statistics > 2 Sample t
  3. Select Both samples are in one column from the dropdown
  4. Double click the variable Weight in the box on the left to insert the variable into the Samples box
  5. Double click the variable Biological Sex in the box on the left to insert the variable into the Sample IDs box
  6. Click OK

This should result in the following output:

Method

\(\mu_1\): mean of Weight when Biological Sex = Female

\(\mu_2\): mean of Weight when Biological Sex = Male

Difference: \(\mu_1-\mu_2\)

Equal variances are not assumed for this analysis.

Descriptive Statistics: Weight
Gender N Mean StDev SE Mean
Female 126 136.7 23.4 2.1
Male 99 172.7 27.3 2.7
Estimation for Difference
Difference 95% CI for Difference
-35.99 (-42.79, -29.20)

Interpret

I am 95% confident that in the population the mean weight of females is between 29.202 pounds and 42.787 pounds less than the mean weight of males.


9.2.1.1.1 - Video Example: Mean Difference in Exam Scores, Summarized Data

9.2.1.1.1 - Video Example: Mean Difference in Exam Scores, Summarized Data

9.2.2 - Hypothesis Testing

9.2.2 - Hypothesis Testing

The formula for the test statistic follows the same general format as the others that we have seen this week:

Test Statistic
\(test\; statistic = \dfrac{sample \; statistic - null\;parameter}{standard \;error}\)

Minitab will compute the test statistic for you! You will just need to determine if equal variances should be assumed or not. There is one example below walking through these procedures by hand, but you are strongly encouraged to use Minitab whenever possible.

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 \(n_2 \geq 30\). If the second assumption is not met then you can conduct a randomization test.

Below are the possible null and alternative hypothesis pairs:

Research Question Are the means of group 1 and group 2 different? Is the mean of group 1 greater than the mean of group 2? Is the mean of group 1 less than the mean of group 2?
Null Hypothesis, \(H_{0}\) \(\mu_1 = \mu_2\) \(\mu_1 = \mu_2\) \(\mu_1 = \mu_2\)
Alternative Hypothesis, \(H_{a}\) \(\mu_1 \neq \mu_2\) \(\mu_1 > \mu_2\) \(\mu_1 < \mu_2\)
Type of Hypothesis Test Two-tailed, non-directional Right-tailed, directional Left-tailed, directional
2. Calculate an appropriate test statistic.

Standard Error

\(\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}\)

Test Statistic for Independent Means

\(t=\dfrac{\bar{x}_1-\bar{x}_2}{ \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\)

Estimated Degrees of Freedom

\(df=smallest\;n - 1\)

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

The \(t\) test statistic found in Step 2 is used to determine the p-value. 

4. Decide between the null and alternative hypotheses.

If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.

5. State a "real world" conclusion.

Based on your decision in Step 4, write a conclusion in terms of the original research question.


9.2.2.1 - Minitab: Independent Means t Test

9.2.2.1 - Minitab: Independent Means t Test

Here we will use Minitab to conduct an independent means t-test. Note that Minitab uses a more complicated formula for computing the degrees of freedom for this test.

Within Minitab, the procedure for obtaining the test statistic and confidence interval for independent means is identical.

Minitab®  – Conducting an Independent Means t Test

Let's compare the mean SAT-Math scores of students who have and have not ever cheated. Both sample sizes are at least 30 so the sampling distribution can be approximated using the \(t\) distribution. 

  1. Open the Minitab file: class_survey.mpx
  2. Select Stat > Basic Statistics > 2 Sample t...
  3. Enter the variable SATM into the Samples box
  4. Enter variable Ever_Cheat into the Sample IDs box
  5. Click OK

This should result in the following output:

2-Sample t: SATM by Ever Cheat
Method

\(\mu_1\): mean of SATM when Ever_Cheat = No
\(\mu_2\): mean of SATM when Ever_Cheat = Yes
Difference: \(\mu_1-\mu_2\)

Equal variances are not assumed for this analysis.

Descriptive Statistics: SATM
Ever_Cheat N Mean StDev SE Mean
No 163 604.0 86.9 6.8
Yes 53 583.7 79.2 11
Estimation of Difference
Difference 95% CI for Difference
20.3 (-5.2, 45.8)
Test
Null hypothesis \(H_0\): \(\mu_1-\mu_2=0\)
Alternative hypothesis \(H_1\): \(\mu_1-\mu_2\neq0\)
T-Value DF P-Value
1.58 95 0.117

The result of our two independent means t test is \(t(95) = 1.58, p = 0.117\). Our p-value is greater than the standard alpha level of 0.05 so we fail to reject the null hypothesis. There is not evidence to state that the mean SAT-Math scores of students who have and have not ever cheated are different. 

Note that we could also interpret the confidence interval in this output. We are 95% confident that the mean difference in the population is between -5.16 and 45.78.

The example above uses a dataset. The following examples show how you can conduct this type of test using summarized data.


9.2.2.1.1 - Example: Summarized Data

9.2.2.1.1 - Example: Summarized Data

Example: Weight by Treatment

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
Method

\(\mu_1\): 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
Test
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.


9.2.2.1.3 - Example: Height by Sex

9.2.2.1.3 - Example: Height by Sex

Research Question: In the population of all college students, is the mean height of females less than the mean height of males?

Data concerning height (in inches) were collected from 99 females and 126 males.

This example uses the following Minitab file: class_survey.mpx

1. Check assumptions and write hypotheses

We have two independent groups: females and males. Height in inches is a quantitative variable. This means that we will be comparing the means of two independent groups.

There are 126 females and 99 males in our sample. The sampling distribution will be approximately normally distributed because both sample sizes are at least 30.

This is a left-tailed test because we want to know if the mean for females is less than the mean for males. 

(Note: Minitab will arrange the levels of the explanatory variable in alphabetical order. This is why "females" are listed before "males" in this example.)

\(H_{0}:\mu_f = \mu_m \)
\(H_{a}: \mu_f < \mu_m \)

2. Calculate the test statistic
  1. Open the file and select Stat > Basic Statistics > 2 Sample t...
  2. Enter variable Height into the Samples box
  3. Enter the variable Biological Sex in the box into the Sample IDs box
  4. Click OK

This should result in the following output:

Method

\(\mu_1\): mean of Height when Biological Sex = Female
\(\mu_2\): mean of Height when Biological Sex = Male
Difference: \(\mu_1-\mu_2\)

Equal variances are not assumed for this analysis.

Descriptive Statistics: Height
Gender N Mean StDev SE Mean
Female 126 65.62 6.53 0.58
Male 99 70.24 3.63 0.37
Estimation for Difference
Difference 95% CI for Difference
-4.6234 (-5.978, -3.269)
Test
Null hypothesis

\(H_0\): \(\mu_1-\mu_2=0\)

Alternative hypothesis \(H_1\): \(\mu_1-\mu_2<0\)
T-Value DF P-Value
-6.73 202 0.000

The test statistic is t = -6.73

3. Determine the p-value

From the output given in Step 2, the p value is 0.000

4. Make a decision

\(p\leq.05\), therefore we reject the null hypothesis.

5. State a "real world" conclusion

There is evidence that the mean height of female students is less than the mean height of male students in the population. 


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