Population variance known

In this and following sections we will present specific procedures for developing and implementing hypothesis test procedures with applications to business and economic problems.

We are given a random sample of n observations from a normal population with mean and known variance . If the observed sample mean is , then the test statistic is

and we can use the following tests with significance level .

1. To test either null hypothesis

or against the alternative

the decision rule is

Reject if

2. To test either null hypothesis

or against the alternative

the decision rule is

Reject if

3. To test the null hypothesis

against the two sided alternative

the decision rule is

Reject if or ,

where is the number for which

and is the standard normal distribution.

A statistical test of hypothesis procedure contains the following five steps:

1. State the null and alternative hypothesis

2. Select the distribution to use

3. Determine the rejection and nonrejection regions

4. Calculate the value of the test statistic

5. Make a decision.

Example:

A manufacturer of detergent claims that the content of boxes sold weigh on average at least 160 grams. The distribution of weights is known to be normal, with standard deviation of 14 grams. A random sample of 16 boxes yielded a sample mean weight of 158.9 grams. Test at the 10% significance level the null hypothesis that the population mean is at least 160 grams.

Solution:

Let be the mean average of all boxes and be the corresponding mean for the sample.

; ;

The significance level is is 0.1. That is, the probability of rejecting the null hypothesis when it is actually is true should not exceed 0.1. This is the probability of making a Type I error. We perform the test of hypothesis using the five steps as follows.

Step 1. State the null and alternative hypothesis

We write the null and alternative hypothesis as

grams

grams

Step 2. Select the distribution to use

Since population standard deviation is known we will use .

Step 3. Determine the rejection and nonrejection regions

The significance level is 0.1. The < sign indicates that the test is left tailed. We look for 0.9 from in the standard normal distribution table, (Table 1 of Appendix). The value of z is . (Fig. 1.4).

Step 4. Calculate the value of the test statistic

The decision to reject or not to reject the null hypothesis will depend on whether the evidence from the sample falls in the rejection or nonrejection region. If the value of the sample mean falls in rejection region, we reject . Otherwise we do not reject the null hypothesis. To locate the position of on the sampling distribution curve of in Figure 1.4 we first calculate z value for . This is called the value of the test statistic.

Step 5. Make a decision

In the final step we make a decision based on the value of the test statistic for in previous step. This value of is not less than the critical value of , and it falls in the nonrejection region. Hence we accept and conclude that based on sample information, it appears that the mean weight of all boxes is greater than 160 grams.

By accepting the null hypothesis we are stating that the difference between the sample mean and the hypothesized value of the population mean is not too large and may occurred because of the chance or sampling error. There is a possibility that the mean weight is less than 160 grams, by the luck of the draw, we selected a sample with a mean that is not too far from required mean of 160 grams.

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.

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