Barbara Illowsky; Susan Dean

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

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. 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 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 calculated as follows:

\sqrt{\frac{(s_{1})^{2}}{n_{1}} + \frac{(s_{2})^{2}}{n_{2}}}

10.1

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

\frac{({\bar{x}}_{1} - {\bar{x}}_{2}) - (μ_{1} - μ_{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 σ₁ are the unknown population standard deviations,
${\bar{x}}_{1}$ and ${\bar{x}}_{2}$ are the sample means, and
μ₁ and μ₂ are the population means.

The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student’s t-distribution with df as follows:

Degrees of freedom

d f = \frac{{(\frac{{(s_{1})}^{2}}{n_{1}} + \frac{{(s_{2})}^{2}}{n_{2}})}^{2}}{(\frac{1}{n_{1} - 1}) {(\frac{{(s_{1})}^{2}}{n_{1}})}^{2} + (\frac{1}{n_{2} - 1}) {(\frac{{(s_{2})}^{2}}{n_{2}})}^{2}}

When both sample sizes n₁ and n₂ are five or larger, the Student’s t approximation is very good. Notice that the sample variances (s₁)² and (s₂)² are not pooled. (If the question comes up, do not pool the variances.)

It is not necessary to compute this by hand. A calculator or computer easily computes it.

Example 10.1

Independent groups

The average amount of time boys and girls aged 7 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 10.1. 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 10.1

Problem

Is there a difference in the mean amount of time boys and girls aged 7 to 11 play sports each day? Test at the 5 percent level of significance.

Solution

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
H_a: μ_g ≠ μ_b H_a: μ_g – μ_b ≠ 0
The words the same tell you H₀ has an "=". Since there are no other words to indicate H_a, assume it says is different. This is a two-tailed test.

Distribution for the test: Use t_df where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances.

Calculate the p-value using a Student’s t-distribution: p-value = 0.0054

Graph:

This is a normal distribution curve representing the difference in the average amount of time girls and boys play sports all day. The mean is equal to zero, and the values -1.2, 0, and 1.2 are labeled on the horizontal axis. Two vertical lines extend from -1.2 and 1.2 to the curve. The region to the left of x = -1.2 and the region to the right of x = 1.2 are shaded to represent the p-value. The area of each region is 0.0028.

Figure 10.2

$s_{g} = 0.866$
$s_{b} = 1$
So, ${\bar{x}}_{g} - {\bar{x}}_{b}$ = 2 – 3.2 = –1.2
Half the p-value is below –1.2, and half is above 1.2.

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

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, 0.866 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is –3.14. Do the procedure again, but instead of Calculate do Draw.

Conclusion—: At the 5 percent level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged 7 to 11 play sports per day is different (mean number of hours boys aged 7 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 7 to 11 play sports per day is greater than the mean number of hours played by boys).

Try It 10.1

Two samples are shown in Table 10.2. 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 percent level of significance.

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

Figure 10.3

right

Problem

g. What is the p-value?

Solution

g. 0.1928

Problem

h. Do you reject or not reject the null hypothesis?

Solution

h. do not reject

Problem

i. Conclusion:

Solution

i. At the 1 percent 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 average, than a student who graduates from College B.

Try It 10.2

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 5 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.

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

Example 10.3

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 10.3 and Table 10.4.

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 10.3 Online 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 10.4 Face-to-Face Class

Problem

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 percent significance level. Answer the following questions:

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

(See the conclusion in Example 10.2, and write yours in a similar fashion.)

Using the TI-83, 83+, 84, 84+ Calculator

First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ₁: and arrow to < μ₂ (less than). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.

Note

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

Solution

two means
unknown
Student’s t
${\bar{X}}_{1} - {\bar{X}}_{2}$
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. H_a: 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.
left-tailed
p-value = 0.0011

Figure 10.4
Reject the null hypothesis.
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 percent 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.

Cohen’s Standards for Small, Medium, and Large Effect SizesCohen’s d is a measure of effect size based on the differences between two means. Cohen’s d, named for U.S. statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.

Size of Effect	d
Small	0.2
medium	0.5
Large	0.8

Table 10.5 Cohen’s Standard Effect Sizes

Cohen’s d is the measure of the difference between two means divided by the pooled standard deviation: $d = \frac{{\bar{x}}_{1} - {\bar{x}}_{2}}{s_{p o o l e d}}$ where $s_{p o o l e d} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}}$ .

Example 10.4

Problem

Calculate Cohen’s d for Example 10.2. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

Solution

μ₁ = 4 s₁ = 1.5 n₁ = 11
μ₂ = 3.5 s₂ = 1 n₂ = 9
d = 0.384
The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small, indicating that there is not a significant difference between them.

Example 10.5

Problem

Calculate Cohen’s d for Example 10.3. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

Solution

d = 0.834; large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the final exam scores of online students and students in a face-to-face class is large, indicating a significant difference.

Try It 10.5

Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen, while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the Northeast and in the West as identified by Nasdaq on May 24, 2013 are listed in Table 10.6 and Table 10.7, respectively.

94.2	75.2	69.6	52.0	48.0	41.9	36.4	33.4	31.5	27.6
77.3	71.9	67.5	50.6	46.2	38.4	35.2	33.0	28.7	26.5
76.3	71.7	56.3	48.7	43.2	37.6	33.7	31.8	28.5	26.0

Table 10.6 Northeast

126.0	70.6	65.2	51.4	45.5	37.0	33.0	29.6	23.7	22.6
116.1	70.6	58.2	51.2	43.2	36.0	31.4	28.7	23.5	21.6
78.2	68.2	55.6	50.3	39.0	34.1	31.0	25.3	23.4	21.5

Table 10.7 West

Is there a difference in the weighted alpha of the top 30 stocks of banks in the Northeast and in the West? Test at a 5 percent significance level. Answer the following questions:

Is this a test of two means or two proportions?
Are the population standard deviations known or unknown?
Which distribution do you use to perform the test?
What is the random variable?
What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
Is this test right-, left-, or two-tailed?
What is the p-value?
Do you reject or not reject the null hypothesis?
At the _____ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
Calculate Cohen’s d and interpret it.

10.1 Two Population Means with Unknown Standard Deviations

Independent groups

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