Hypothesis Tests about the Mean
of an Arbitrary Population

ZMNTEST.83p
ZMNTEST.86p
zmntest.89p

Suppose someone has made the claim that the mean µ of an arbitrary population is less than or equal to some value m. We denote this claim, called a null hypothesis, by Ho: µ <= m. We test the claim by computing the sample average Xbar of a random sample (usually of size n >= 30). If Xbar is "too large," then we must reject Ho in favor of the alternative hypothesis Ha: µ > m. (This test is equivalent to a null hypothesis Ho: µ = m with a one-sided alternative Ha: µ > m.)

The decision on whether or not Xbar is "too large" is based on a desired level of significance a, such as a = 0.01, 0.025, 0.03, or 0.05. We do not want to reject Ho if Ho is true. The pre-assigned level of significance a designates the maximum allowable probability of rejecting Ho given that Ho is true.

To determine if Xbar is too large for us to accept Ho, we compute the probability of an arbitrary sample average being as large as this one value of Xbar given that Ho is true. If the probability is too small, that is below a, then we must reject Ho. In other words, if Ho: µ <= m were true, then it would be too unlikely for a sample average to be as large as this Xbar. Thus we must reject Ho.

To compute the probability, we define a test statistic by

x = (Xbar - m) / (Sx / Sqrt(n)),

where Sx is the sample deviation (or true standard deviation if it is known). If we have a smaller population of size N, then we use the finite population correction factor by multiplying Sx by Sqrt( (N-n) / (N-1) ).

We then let Z ~ N(0,1) and compute the right tail probability P(Z >= x) (also called the p-value), which gives the maximum probability of a sample average being as large as Xbar given that Ho were true. If this probability is less than a, then it is too unlikely for a sample average to be as large as Xbar. In other words, Xbar is too large and we have significant evidence to reject Ho.


Similarly, we can test a different null hypothesis such as Ho: µ >= m. In this case, we reject Ho, in favor of the alternative Ha: µ < m, if Xbar is "too small." Now we look at the left tail probability P(Z <= x) (also called the p-value), where Z and x again are defined as above. If P(Z <= x) < a, then it is too unlikely for a sample average to be as small as Xbar if Ho were true; hence, we have significant evidence to reject Ho. (This test is equivalent to a null hypothesis Ho: µ = m with a one-sided alternative Ha: µ < m.)


Lastly, we can test the null hypothesis Ho: µ = m versus the two-sided alternative Ha: µ does not equal m. Now we must reject Ho if Xbar is too large or too small. In this case, we consider both tail probabilities P(Z <= x) and P(Z >= x) and the level of significance is divided into two portions. Therefore if either tail probability is less than a / 2, then we reject Ho. In this case, the p-value is given by twice the smallest tail value.



Using the ZMNTEST Program

The ZMNTEST program can be used to perform these hypothesis tests. To execute the program, first enter 1 or 2 to designate a "large" population or a finite population of known size N which you then enter. Next, enter the value of the mean to be tested, along with the sample average, the sample deviation (or known standard deviation), and the level of significance. The program displays the conclusion for all three of these tests, along with the value of the test statistic and the left and right tail values.



Example. A random survey of 80 homeowners in a subdivision of 512 homes found that the average yearly insurance premium on their homes was $584.50 with a sample deviation of $64.85. Test the following three claims about the true average premium µ each at the 0.05 level of significance.

1. Ho: µ <= 600
2. Ho: µ >= 600
3. Ho: µ = 600


Solution. After calling up the ZMNTEST program, enter 2 for a finite population, then enter the population size of 512. Next, enter 600 for TEST MEAN, enter 80 for SAMPLE SIZE, enter 584.50 for XBAR, enter 64.85 for SAMPLE DEV., and enter .05 for LEVEL OF SIG.

The conclusions to the three tests are stated. We obtain a test statistic of -2.325 with left and right tail-values of 0.01 and 0.99. We accept the first hypothesis but reject the second and third.

Since the left tail probability of 0.01 is less than a = 0.05, we have rejected the second claim that µ >= 600. Since 0.01 is also less than a / 2 = 0.025, we have rejected the third claim that µ = 600. Since we have accepted the first hypothesis, but rejected the third, we conclude that µ < 600. That is, the average premium is less than $600.

We can further interpret the tail values as follows:

1. For the hypothesis Ho: µ <= 600: Since Xbar = $584.50 <= 600, we automatically accept this null hypothesis.

2. For the hypothesis Ho: µ >= 600: (This test is equivalent to the test Ho: µ = 600 with a one-sided alternative Ha: µ < 600.) Now Xbar may be too small to accept this hypothesis. The left tail value implies that if the claim µ >= 600 were true, then there would be at most a 1% chance of Xbar being as small as 584.50 with a random sample of size 80 from the population of size 512. This chance is too low; thus we reject the claim. (In this case, we also say that the p-value is 0.01, which is lower than the desired level of significance.)

3. For the hypothesis Ho: µ = 600: With a two-tail test, we always use the smallest tail value to draw the conclusion. Again, there is only a 1% chance of Xbar being as small as it is if the claim µ = 600 were true. This chance is too low; thus we reject the hypothesis.

For a two-tail test, the p-value is always twice the smallest tail-value. Here it is 2*.01 = 0.02, which is still lower than the desired level of significance. In terms of this p-value we can say: If µ = 600 were true, then there would be only a 2% chance of Xbar being as far away (in either direction) as $584.50 with a random sample of size 80 from the population of size 512.



Exercises

1. The following measurements are percentages of body fat from a random sample of men aged 30 to 39.

Percentages of Body Fat from Men Aged 30 -39
20.5
28.1
17.6
8.4
12.8
21.4
16.8
24.6
16.5
20.8
22.4
8.5
6.4
22
16.8
25.8
15.2
4.1
21.7
16.5
22.4
10.1
14.8
13.4
28.8
20
1.9
20.5
15.7
12.3
21.4
26.5
22
10.4
34.7
20.2

Using a 0.04 level of significance, test and explain the following three null hypotheses about the true mean µ of the percentage of body fat of all men age 30 - 39.

1. Ho: µ <= 17
2. Ho: µ >= 17
3. Ho: µ = 17


2. A sociologist does not believe a report stating that high school freshmen watch an average of 20 hours of TV per week. She feel that the average is higher. Her own random survey of 120 freshmen in a large city yields a sample average of 22.4 hours with a standard deviation of 9.26. State and explain a proper hypothesis test which may refute the original report.



Solutions

1. First, enter the data into a list and compute the summary statistics. We obtain a sample mean of Xbar = 17.8333 and a sample deviation of 7.19337 for this sample of size 36.

Next, call up the ZMNTEST program, enter 1 for a large population (all men age 30 - 39), and enter 17 for TEST MEAN. For the statistics, you can type them in directly or access them from the VARS screens.

We find that we accept all three hypotheses, which means that based on this data, we do not have significant evidence to reject that µ = 17. Let's now use the right tail value of 0.2435 to interpret the results:

For the hypothesis Ho: µ <= 17 (or equivalently Ho: µ = 17 with Ha: µ > 17): Here Xbar may be too large to accept the hypothesis. But the right-tail value implies that if the claim µ <= 17 were true, then there is as much as a 24.35% chance (or a p-value of 0.2435) of Xbar being as large as 17.8333 from a random sample of size 36. This chance is not too low (it is well above the level of significance); thus we do not reject the hypothesis.

For the hypothesis Ho: µ >= 17: Since Xbar = 17.83333 . . >= 17, we automatically accept this null hypothesis.

For the hypothesis Ho: µ = 17: For a two-sided test, we always use the smallest tail value to draw the conclusion. Again, there is a 24.35% chance of Xbar being as large as 17.8333 froma sample of size 36 if the claim µ = 17 were true. This chance is not too low since it is not less than half of the level of significance; thus, we do not reject the hypothesis.

For this two-sided test, the p-value is 2*.2435 = 0.487. If µ = 17 were true, then with a sample of size 36 there would be a 48.7% chance of Xbar being as far away (in either direction) as 17.8333.


2. Here we can use Ho: µ = 20 with a one-sided alternative Ha: µ > 20. Equivalently we can use the null hypothesis Ho: µ <= 20. We shall use a 0.05 level of significance.

Call up the ZMNTEST program and enter 1 for a large population, enter 20 for TEST MEAN, enter 120 for SAMPLE SIZE, enter 22.4 for XBAR, enter 9.26 for SAMPLE DEV., and enter .05 for LEVEL OF SIG.

We receive a right-tail (p-value) of 0.0023 from a test statistic of 2.83917. If the true average were 20 (or less) then there would be at most 0.0023 probability of receiving a sample average as high as 22.4 with a random sample of size 120. This low p-value gives significant evidence to refute the original report.



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