1.3. Tests of the mean of a normal distribution:

Population variance unknown (Large sample size)

When the population standard deviation is unknown, we simply estimate with the value of the sample standard deviation . We must consider separately the large sample ( ) and small sample size ( cases.

If the sample size n is large, the test procedure developed for the case when population variance is known can be employed when it is unknown, replacing by the observed sample variance . All the hypotheses and decision rules are stated in the same way as before (i.e. when is known).

Example:

When a machine that is used to make bolts is working properly, the mean length of these bolts 2.5 cm. However, from time to time this machine falls out of alignment and produces bolt that have a mean length of either less than or 2.5 cm or more than 2.5 cm. When this happens, the process is stopped and the machine is adjusted. To check whether or not the machine is producing bolts with a mean length of 2.5 cm, the quality control department at the company takes a sample of bolts each week and makes a test of hypothesis. One such a sample of 49 bolts produced a mean length of

2.49 cm and a standard deviation of 0.021 cm. Using the 5% significance level, can we conclude that the machine needs to be adjusted?

Solution:

Let be the mean length bolts made on this machine and be the corresponding mean for the sample.

; cm; cm

The mean length of all bolts is supposed to be 2.5 cm. The significance level is is 0.05. That is, the probability of rejecting the null hypothesis when it is actually is true should not exceed 0.05.

Step 1. State the null and alternative hypothesis

We are testing to find whether or not the machine needs to be adjusted. The machine will need an adjustment if the mean length of these bolts is either less than 2.5 cm or more than 2.5.

We write the null and alternative hypothesis as

cm (The machine does not need adjustment)

cm (The machine needs an adjustment)

Step 2. Select the distribution to use

Because the sample size is large (n >30), the sampling distribution of is (approximately) normal. Consequently we will use to make the test.

Step 3. Determine the rejection and nonrejection regions

The significance level is 0.05. The sign indicates that the test is two tailed with two rejection regions, one in each tail of the normal distribution curve of . Because the total area of both rejection regions is 0.05 (the significance level), the area of the rejection region in each tail is 0.025.

These areas are shown in Fig.1.5. To find the z values for these critical points, we look for 0.975 in the standard normal distribution table.

Hence, the z values of the two critical points as shown in Fig.1.5, are and 1.96.

Step 4. Calculate the value of the test statistic

The value of from the sample is 2.49. As is not known, we calculate the z value as follows

is the value of the test statistic.

Step 5. Make a decision

The value of is less than the critical value of , and it falls in the rejection region in the left tail. Hence we reject and conclude that based on sample information, it appears that the mean length of all bolts produced on this machine is not equal to 2.5 cm. Therefore, the machine needs to be adjusted

By rejecting the null hypothesis we are stating that the difference between the sample mean and the hypothesized value of the population mean is too large and may not have occurred because of the chance or sampling error. This difference seems to be real and, hence the mean length of bolts is different from 2.5 cm. Note that the rejection of the null hypothesis does not necessarily indicate that the mean length of bolts is definitely different from 2.5 cm. It simply indicates that there is strong evidence (from sample) that the mean length of bolts is not equal to 2.5 cm.

There is a possibility that the mean length of bolts equal to 2.5 cm. If so, we have wrongfully rejected the null hypothesis . This is Type I error and probability of making such an error in this case is 0.05.

Exercises

1. Make the following tests of hypotheses.

a) ; ; ; ; ;

b) ; ; ; ; ;

c) ; ; ; ; ;

2. Consider ; against the two sided alternative .

a) A random sample of 64 observations produced a sample mean of 98 and a standard deviation of 12. Using , would you reject the null hypothesis?

b) Another random sample of 64 observations taken from the same population produced a sample mean of 104 and a standard deviation of 10. Using , would you reject the null hypothesis?

Comment on the results of parts a) and b).

3. A survey showed that people with a bachelor’s degree earned average of $2116 a year in 2001. A sample of 900 persons with a bachelor’s degree taken recently by a researcher showed that the persons in this sample earned on average of $2345 a year with a standard deviation of $210. Test at 5% significance level whether people with a bachelor’s degree currently earn an average of $2116 against the alternative that it is more than $2116 in a year.

4. The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 month. A consumer protection agency that wants to check this claim took a random sample of 36 such batteries and found the mean life for this sample is 43.75 month with a standard deviation of 4 month. Using the 2.5% significance level, test the manufacturer claim against the alternative that the mean life of batteries is less than 45 month.

5. A random sample of 100 observations from a population with standard deviation 60 yielded a sample mean of 110.

a) Test the null hypothesis that against the alternative hypothesis that using . Interpret the results of the test.

b) Test the null hypothesis that against the alternative hypothesis that using . Interpret the results of the test.

c) Compare the results of the two tests you conducted. Explain why the results differ.

6. In a random sample of 250 observations, the mean and standard deviation are found to be 169.8 and 31.6, respectively. Is the claim that larger than 169 substantiated by these data at the 10% level of significance?

7. From records, it is known that the duration of treating a disease by a standard therapy has a mean of 15 days. It is claimed that a new therapy can reduce the treatment time. To test this claim, the new therapy is tried on 70 patients, and from the data of their times to recovery, the sample mean and standard deviation are found to be 14.6 and 3.0 days, respectively.

Perform the hypothesis test using a 2.5% level of significance.

8. Suppose that you are to verify the claim that on the basis of a random sample of size 70, and you know that .

a) If you set the rejection region to be , what is the level of significance of your test?

b) Find the numerical value of c so that the test has a 5% level of significance.

Answers

1. a) ; reject ; b) ; do not reject ;c)

reject ; 2. a) ; do not reject ; b) ; reject ;

3. ; reject ; 4. ; accept ; 5. a) ; reject ; b) ; accept ; 6. ; accept ;7. ; is not rejected at ; 8. a) ; b) c=21.10.