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Comparing Two Independent Population Means with Known Population Standard Deviations

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: This module provides an overview of hypothesis testing in situations where there are both two independent population means and known population standard deviations in statistics.

Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is X 1 ¯ - X 2 ¯ X 1 - X 2 . The normal distribution has the following format:

Normal distribution

X 1 ¯ X 2 ¯ ~ N [ u 1 u 2 , ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ] X 1 ¯ X 2 ¯ ~N[ u 1 u 2 , ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ]
(1)

The standard deviation is:

( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2
(2)

The test statistic (z-score) is:

z = ( x 1 ¯ x 2 ¯ ) ( μ 1 μ 2 ) ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 z= ( x 1 ¯ x 2 ¯ ) ( μ 1 μ 2 ) ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2
(3)

Example 1

independent groups, population standard deviations known: The mean lasting time of 2 competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distribution. The following table is the result.

Table 1
Wax Sample Mean Number of Months Floor Wax Last Population Standard Deviation
1 3 0.33
2 2.9 0.36

Problem 1

Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.

Solution

This is a test of two independent groups, two population means, population standard deviations known.

Random Variable: X 1 ¯ - X 2 ¯ = X 1 - X 2 = difference in the mean number of months the competing floor waxes last.

H o : μ 1 μ 2 H o : μ 1 μ 2

H a : μ 1 > μ 2 H a : μ 1 > μ 2

The words "is more effective" says that wax 1 lasts longer than wax 2, on the average. "Longer" is a ">"">" symbol and goes into HaHa. Therefore, this is a right-tailed test.

Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula above, the distribution is:

X 1 ¯ X 2 ¯ ~ N ( 0 , 0.33 2 20 + 0.36 2 20 ) X 1 ¯ X 2 ¯ ~N ( 0 , 0.33 2 20 + 0.36 2 20 )

Since μ 1 μ 2 μ 1 μ 2 then μ 1 μ 2 0 μ 1 μ 2 0 and the mean for the normal distribution is 0.

Calculate the p-value using the normal distribution: p-value = 0.1799

Graph:

Figure 1
Normal distribution curve of difference in the average number of months the competing floor waxes last with values of 0 and 0.1 on the x-axis. A vertical upward line extends from 0.1 to the curve and the p-value points to the area to the right of 0.1.

x 1 ¯ - x 2 ¯ = 3 - 2.9 = 0.1 x 1 - x 2 =3-2.9=0.1

Compare α and the p-value: α = 0.05α=0.05 and p-value = 0.1799. Therefore, α <α< p-value.

Make a decision: Since α <α< p-value, do not reject H o H o .

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.

Note:
TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μμ1: and arrow to > μμ2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again but instead of Calculate do Draw.

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