Hypothesis Tests about the Mean
of a Normally Distributed Population


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Suppose that we are studying a population that is known to be normally distributed. We again consider the following three hypothesis tests about the mean µ that were discussed in the previous section:

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


As before, we accept or reject each null hypothesis based on the value of a sample mean Xbar from a random sample of size n. We conduct the tests exactly as before with the use of the test statistic x = (Xbar - m) / ( Sx / Sqrt(n) ); however, we now let X ~ t(n - 1) (the t-distribution with n - 1 degrees of freedom) and compute the tail probabilities with X.

Since the population is known to be normally distributed, we no longer need large random samples to apply the test. Moreover, the test statistics are precisely a t(n - 1) distribution (unlike from an arbitrary population where the test statistics are only approximately standard normal for large random samples.)

If a is our pre-chosen level of significance, then we reject the first null hypothesis if P(X >= x) < a. We reject the second null hypothesis if P(X <= x) < a. We reject the third null hypothesis if P(X <= x) < a / 2 or P(X >= x) < a / 2.

We note that the first hypothesis above is equivalent to the null hypothesis Ho: µ = m with a one-sided alternative Ha: µ > m. Likewise the second hypothesis is equivalent to the test Ho: µ = m with the one-sided alternative Ha: µ < m.



Using the TMNTEST Program

The TMNTEST program can be used to perform these hypothesis tests. To execute the program, enter the value of the mean to be tested, along with the sample average, the sample 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. Cholesterol levels are often found to be normally distributed. Suppose a random sample of 40 adults yields an average cholesterol level of 213 with a sample deviation of 48.4. Test the following three hypotheses about the true average adult level µ at the 0.06 level of significance.

1. Ho: µ <= 200
2. Ho: µ >= 200
3. Ho: µ = 200


Solution. After calling up the TMNTEST program, enter 200 for TEST MEAN, enter 40 for SAMPLE SIZE, enter 213 for XBAR, enter 48.4 for SAMPLE DEV., and enter .06 for LEVEL OF SIG.

We see that we must reject only the first hypothesis since the right tail value is 0.0487. In other words, if µ <= 200, then there would be at most a 4.87% chance of Xbar being as large as 213. This chance is below the level our chosen significance, so we have enough evidence to reject the claim. The p-value for this test is this right tail value of 0.0487.

Since Xbar = 213 >= 200, we automatically accept the second null hypothesis.

For the two-sided test required for the third hypothesis, the smallest tail value of 0.0487 is not below half the level of significance, so we do not have enough evidence to reject. The p-value for this test is 2*.0487 = 0.0974, which is greater than a = 0.06. Since we accept the second hypothesis and do not reject the third hypothesis, we ultimately conclude that µ >= 200.



Exercises

1. Assume the level of tar among all brands of cigarettes is normally distributed. The chart below lists the tar levels, as produced by the Federal Trade Commision, for a sample of major brands.

Tar Level in Milligrams by Cigarette Brand

Alpine

14.1

Kool

16.6

Old Gold

17.0

Benson & Hedges

16.0

L&M

14.9

Pall Mall Light

12.8

Bull Durham

29.8

Lark Lights

13.7

Raleigh

15.8

Camel Lights

8.0

Marlboro

15.1

Salem Ultra

4.5

Carlton

4.1

Merit

7.8

Tareyton

14.5

Chesterfield

15.0

MultiFilter

11.4

True

7.3

Golden Lights

8.8

Newport Lights

9.0

Virginia Slims

15.2

Kent

12.4

Now

1.0

Winston Lights

12.0

Test the following three null hypotheses about the true average tar level µ among all cigarette brands at the 0.05 level of significance.

1. Ho: µ <= 15
2. Ho: µ >= 15
3. Ho: µ = 15


2. A brewer is trying to decide if there is a noticable increase in pH level from 3.4 after his beer has been in cold storage for two months. A random sample of 48 such bottles yielded a sample mean pH level of 3.56 with a sample deviation of 0.38. Perform an appropriate hypothesis test and state your conclusion.



Solutions

1. First, enter the data into a list and compute the summary statistics. We obtain a sample mean of Xbar = 12.36666667 and a sample deviation of 5.73628 for this sample of size 24.

Next call up the TMNTEST program and enter 15 for TEST MEAN. For the statistics, you can type them in directly or access them from the VARS screens.

We see that we must reject the second and third hypotheses since the left tail value is 0.0172 from a test statistic of about -2.249. Hence, we accept that µ <= 15 but reject that µ = 15. Therefore, based on this data we can conclude that µ < 15.

If µ >= 15 or if µ = 15, then there would be at most a 1.72% chance of Xbar being as small as 12.3667 with a random sample of size 24. Thus, we must reject both of these claims.


2. Historically, we can assume that the pH levels of brewed beer are normally distributed. Here, we shall test Ho: µ = 3.4 with a one-sided alternative Ha: µ > 3.4. Equivalently we can use the null hypothesis Ho: µ <= 3.4. We shall use a 0.03 level of significance.

Call up the TMNTEST program and enter 3.4 for TEST MEAN, 48 for SAMPLE SIZE, 3.56 for XBAR, enter .38 for SAMPLE DEV., and .03 for LEVEL OF SIG.

We receive a right-tail (p-value) of 0.0027 from a test statistic of 2.917138. If the true mean pH level after two months were 3.3 (or less) then there would be at most 0.0027 probability of receiving a sample average as high as 3.56 with a random sample of size 48. This low p-value gives significant evidence to believe that the pH level has risen.



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