12.4: Two Means T Pooled Test

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Page ID: 105869

Anton Butenko
Mt. San Jacinto College

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Next, we will learn how to apply the Two Means $T$ Pooled Procedure to test a statistical claim about 2 population means when population standard deviations are unknown but can be assumed equal. Consider the following example.

A group of activists wants to the verify the claim that in a certain Fortune 500 company the average salary of female employees is different from the average salary of male employees. Two samples of male and female workers were obtained independently and analyzed. The sample of 19 women had the average salary $64,491 and the standard deviation $13,211. The sample of 24 men had the average salary $67,541 and the standard deviation $19,274. Test the claim at $10\%$ significance level.

Note that the sample means are $\bar{x}_1=64491$ dollars and $\bar{x}_2=67541$ dollars, the sample standard deviations are $s_1=13211$ dollars and $s_2=19274$ dollars, and the sample sizes are $n_1=19$ and $n_2=24$ for women and men respectively.

Now, let’s identify the statistical claim that needs to be tested:

“women’s average salary is different from men’s”

The claim is that women’s average salary is different from men’s, so we can symbolically express the claim as

$\mu_1\neq\mu_2$ or $\mu_1-\mu_2\neq0$

Since the claim is about the 2 population means and the population standard deviations are unknown, we will use the 2 Means $T$ Procedure but now we need to choose between pooled or non-pooled? We use the $T$ Pooled Procedure when the standard deviations can be assumed equal, otherwise we use the $T$ Non-Pooled Procedure.

To test the claim whether the standard deviations are the same we setup the 2 Variances $F$ Test with $\alpha=10\%$ (for convenience we chose the same alpha as in the problem but in general it doesn’t have to match and depends on the context).

$H_0: \frac{\sigma^2_1}{\sigma^2_2}=1$

$H_a: \frac{\sigma^2_1}{\sigma^2_2}\neq1$

$f_0=\frac{s^2_1}{s^2_2}=\frac{13211^2}{19274^2}=0.470$

$\text{p-value}=0.106>\alpha=10\%$

The conclusion is that we don’t have sufficient evidence to reject the null hypothesis, therefore we will continue assuming that the variances are the same and choose the $T$-Pooled Procedure.

Try it Yourself!

Let’s check if all necessary assumptions are satisfied:

The samples are simple random and independent
The CLT must be applicable i.e. the populations are normal, or the samples are greater than 30
The population standard deviations are unknown but are assumed equal

Try It Yourself!

To use the $T$ Pooled Procedure, we must compute the pooled standard deviation using the formula:

$s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s^2_2}{n_1+n_2-2}}$

$s_p=\sqrt{\frac{(19-1)\cdot13211^2+(24-1)\cdot19274^2}{19+24-2}}=16485.20$

Try It Yourself!

Also, the degrees of freedom of the involved $T$ distribution is

$df=n_1+n_2-2=19+24-2=41$

Try It Yourself!

We will use the following template to perform the hypothesis testing:

In step 1, we will set up the hypothesis:

Since the claim is in the form of an inequality, therefore it must be set as an alternative hypothesis, therefore the null hypothesis is $\mu_1-\mu_2=0$ and the test is two-tail, that is:

$H_0: \mu_1-\mu_2=0$

$H_a: \mu_1-\mu_2\neq0$

In step 2, we will identify the significance level:

The significance level can always be found in the text of the problem. In our case it is $10\%$, thus:

$\alpha=0.10$

In step 3, we will find the test statistic using the formula:

$t_0=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{64491-67541}{16485.20\sqrt{\frac{1}{19}+\frac{1}{24}}}=-0.602$

Try It Yourself!

In step 4, we will perform either the critical value approach or p-value approach to test the claim:

In critical value approach, we construct the rejection region using the $T$-curve with $df=41$:

RR: less than $-t_{0.05}=-1.683$ or greater than $t_{0.05}=1.683$

Try It Yourself!

In p-value approach, we compute the p-value:

P-Value: $2P(T<-0.602)=0.550$

Try It Yourself!

In step 5, we will draw the conclusion:

In the critical value approach, we must check whether the test statistic is in the rejection region or not. Our test statistic is $-0.602$ and it is between the critical values $-1.683$ and $1.683$, thus it is NOT in the rejection region.
In the p-value approach, we must check whether the p-value is less than the significance level or not. Our p-value is $0.550$ and it is greater than $\alpha=0.10$.

Both tests suggest that we DO NOT reject the null hypothesis in favor of the alternative.

In step 6, we will interpret the results:

Under $10\%$ significance level we DO NOT have sufficient evidence to suggest that women’s salaries are different from men’s salaries.

We discussed how to apply the Two Means $T$ Pooled Procedure to test a statistical claim about two population means with population standard deviations that are unknown but assumed equal.

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