9.1: Two Population Means (Independent)

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Two Population Means (Independent)

jkesler

[latexpage]

1. The two independent samples are simple random samples from two distinct populations.
2. For the two distinct populations:
   • if the sample sizes are small, the distributions are important (should be normal)
   • if the sample sizes are large, the distributions are not important (need not be normal)

NOTE

The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, $\bar X_1 – \bar X_2$, and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, $\bar X_1 – \bar X_2$.

The standard error is:

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

The test statistic (t-score) is calculated as follows:

$$t=\frac{(\bar x_1 – \bar x_2) – (\mu_1-\mu_2)}{\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}}$$

where:

• s₁ and s₂, the sample standard deviations, are estimates of σ₁ and σ₂, respectively.
• σ₁ and σ₂ are the unknown population standard deviations.
• $\bar x_1$ and $\bar x_2$ are the sample means.
• μ₁ and μ₂ are the population means, and the null hypothesis will always be H₀: μ₁ = μ₂

Note

μ₁ and μ₂ are the population means, and the null hypothesis will always be H₀: μ₁ = μ₂ . This means that the calculation $(\mu_1-\mu_2)$ in our test statistic will always be zero, so we can simplify the test statistic calculation as

$$t=\frac{\bar x_1 – \bar x_2}{\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}}$$

The number of degrees of freedom (df) requires a somewhat complicated calculation. However, for this introductory course, we will be using a simplified calculation. The test statistic calculated previously is approximated by the Student’s t-distribution with df that is 1 less than the smaller of $n_1$ and $n_2$ from your two samples. For example, if $n_1=14$ and $n_2=9$, we would use $df=8$.

Notice that the sample variances (s₁)² and (s₂)² are not pooled, meaning we don’t consider the standard deviation of the samples as if they were both just one big sample together. (If the question comes up, do not pool the variances for means.)

Example 9.1

Independent groups

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table 9.1. Each populations has a normal distribution.

	Sample Size	Average Number of Hours Playing Sports Per Day	Sample Standard Deviation
Girls	9	2	0.866
Boys	16	3.2	1.00

Table 9.1

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

Solution 9.1

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μ_g is the population mean for girls and μ_b is the population mean for boys. This is a test of two independent groups, two population means.

Random variable: $\bar X_g-\bar X_b=$ difference in the sample mean amount of time girls and boys play sports each day.

H₀: μ_g = μ_b  H₀: μ_g – μ_b = 0

H₁: μ_g ≠ μ_b  H₁: μ_g – μ_b ≠ 0

The words “the same” tell you H₀ has an “=”. Since there are no other words to indicate H₁, assume it says “is different.”
This is a two-tailed test (from the fact that H₁: μ_g ≠ μ_b),.

Calculate the test statistic:

$$t=\frac{\bar x_1 – \bar x_2)}{\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}}=\frac{2- 3.2}{\sqrt{\frac{(0.866)^2}{9}+\frac{(1)^2}{16}}}=−3.142390292$$

Distribution for the test:
Use t_df where df = min($n_g, n_b$)-1 = min(9,16)-1 = 9-1=8.

Get the area into the tail beyond the test statistic:

Since our test statistic is negative, we’ll get the area to the left using Google Sheets.

=TDIST( 3.142390292, 8, 2 )

The first value given to the TDIST function is the positive version of our test statistic, the second value is our degrees of freedom, and the third is the number of tails.

p-value = 0.0138

Make a decision: Since α > p-value, reject H₀. This means you reject μ_g = μ_b. The means are different.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).

If we used our student-t distribution table instead of Google Sheets, we look in the 8th row, and find between which two columns our test statistic, rounded to 3.142, would be.

We see that 3.142 is less than 3.355, but greater than 2.896, so we find that our p-value is greater than 0.01 and less than 0.02.
0.01 < p-value < 0.02
Since the significance level is 0.05, we know that our p-value is less than the significance level (if our p-value is less than 0.02, it must be less than 0.05), so the conclusion is the same as using Google Sheets.

 

Try It 9.1

Two samples are shown in Table 9.2. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

	Sample Size	Sample Mean	Sample Standard Deviation
Population A	25	5	1
Population B	16	4.7	1.2

Table 9.2

Example 9.2

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples 10 graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.

 

1. Is this a test of two means or two proportions?
2. Are the populations standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
6. Is this test right-, left-, or two-tailed?
7. What is the p-value?
8. Do you reject or not reject the null hypothesis?
9. Conclusion:

Solution 9.2

1. two means
2. unknown
3. Student’s t
4. $\bar X_A -\bar X_B$ (or $\bar X_1 -\bar X_2$ )
5. $H_0: \mu_A = \mu_B$
   $H_1: \mu_A > \mu_B$
6. right tail
7. 0.1942 (by Google Sheets with the following function)
   =TDIST(0.906032848,9,1)
   (0.906032848 is the test statistic)
8. Do not reject H₀
9. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

Try It 9.2

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

1. Are the population standard deviations known?
2. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?

Example 9.3

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table 9.3 and Table 9.4.

67.6	41.2	85.3	55.9	82.4	91.2	73.5	94.1	64.7	64.7
70.6	38.2	61.8	88.2	70.6	58.8	91.2	73.5	82.4	35.5
94.1	88.2	64.7	55.9	88.2	97.1	85.3	61.8	79.4	79.4

Table 9.3Online Class

77.9	95.3	81.2	74.1	98.8	88.2	85.9	92.9	87.1	88.2
69.4	57.6	69.4	67.1	97.6	85.9	88.2	91.8	78.8	71.8
98.8	61.2	92.9	90.6	97.6	100	95.3	83.5	92.9	89.4

Table 9.4Face-to-face Class

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

1. Is this a test of two means or two proportions?
2. Are the population standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
6. Is this test right, left, or two tailed?
7. What is the p-value?
8. Do you reject or not reject the null hypothesis?
9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(See the conclusion in Example 9.2, and write yours in a similar fashion)

Note

Be careful not to mix up the information for Group 1 and Group 2!

Solution 9.3

Before you begin, you should copy and paste data from each sample into a Google Sheet. Then use the AVERAGE, STDEV.S, and COUNT functions to get the sample mean, sample standard deviation, and sample size for each sample.

You can then use the information above to calculate the test statistic as -3.229

1. two means
2. unknown
3. Student’s t with df = 30-1 = 29
4. $\bar X_1-\bar X_2$
   1. H₀: μ₁ = μ₂ Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
   2. H₁: μ₁ < μ₂ Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
5. left-tailed
6. p-value = 0.0015 using the Google Sheets function
   =TDIST(3.229,29,1)
7. Reject the null hypothesis
8. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class.
   
   At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.