1.5. Tests of the mean of a normal distribution:

Population variance unknown. Small samples

Many times the size of a sample that is used to make test of hypothesis about is small, that is, . If the population is (approximately) normally distributed, the population standard deviation is not known and the sample size is small ( ), then the normal distribution is replaced by the Student’s t distribution to make a test of hypothesis about . In such a case the random variable

has a Student’s t distribution with degrees of freedom.

The value of test statistic t for the sample mean is computed as

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 ,

Here, is the number for which

where the random variable follows a Student’s t distribution with degrees of freedom.

Example:

The company that produces auto batteries claims that its batteries are good, for an average, for at least 64 days. A consumer protection agency tested 15 such batteries to check this claim. It found the mean life of these 15 batteries to be 62 days with a standard deviation of 3 days. At the 5% significance level, can you conclude that the claim of the company is true? Assume that the life of such a battery has and approximate normal distribution.

Solution:

Let be the mean life of all batteries and be the corresponding mean for the sample. Then from the given information,

; days; days

The mean life of all batteries is supposed to be at least 64 days. 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 write the null and alternative hypothesis as

days (The mean life is at least 64 days)

days (The mean life is less than 64 days)

Step 2. Select the distribution to use

The sample size is small ( ), and the life of a battery is approximately normally distributed. Since population standard deviation is unknown, we use the Student’s t distribution to make the test.

Step 3. Determine the rejection and nonrejection regions

The significance level is 0.05. The sign in the alternative test indicates that the test is left tailed with the rejection region in the left tail of the t distribution curve.

Area in the left tail=

Degree of freedom=

From the Student’s t distribution table (Table 2 of Appendix), the critical value of t for 14 degrees of freedom and an area 0.05 in the left tail is . (Fig.1.9).

Step 4. Calculate the value of the test statistic

As is not known, and sample size is small, we calculate the t value as follows

Step 5. Make a decision

The value of is less than the critical value of , and it falls in the rejection region. Therefore, we reject and conclude that the sample mean is too small compared to 62 days (company’s claimed value of ) and the difference between the two may not be attributed to chance alone. We can conclude that the mean life of company’s batteries is less than 62 days.

Remark: The conclusion of a t-test can also be strengthened by reporting

the significance probability (p- value) of the observed statistic. Since the t table provides only a few selected percentage points, we can get an idea about the p-value but not its exact determination. For instance, the data in example above gave an observed value with degree of freedom=14. Scanning the t table for , we notice that that 2.50 lies between and . Therefore, the p-value of is higher than 0.025 but not as great as 0.010.

Exercises

1. For each of the following examples of tests of hypothesis about , show the rejection and nonrejection regions on the t distribution curve.

a) A two tailed test with and

b) A left tailed test with and

c) A right tailed test with and

2. Consider the null hypothesis about the mean of a population that is normally distributed. Suppose a random sample of 20 observations is taken from this population to make this test. Using show the rejection and nonrejection regions and find critical value(s) for t for

a) left tailed test; b) two tailed test; c) right tailed test

3. Consider versus for a population that is normally distributed.

a) A random sample of 16 observations taken from this population produced a sample mean of 45 and a standard deviation of 5. Using , would you reject the null hypothesis?

b) Another random sample of 16 observations taken from the same population produced a sample mean of 41.9 and a standard deviation of 7. Using , would you reject the null hypothesis?

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

4. Assuming that respective populations are normally distributed, make the following hypothesis tests.

a) ; ; ; ; ;

b) ; ; ; ; ;

c) ; ; ; ; ;

5. A business school claims that students who complete a three month course of typing course can type on average, at least 1200 words an hour.

A random sample of 25 students who completed this course typed, on average, 1130 words an hour with a standard deviation of 85 words. Assume that the typing speeds for all students who complete this course have an approximate normal distribution.

Using the 5% significance level, can you conclude that the claim of the business school is true?

6. The supplier of home heating furnaces of a new model claims that the average efficiency of the new model is at least 60. Before buying these heating furnaces, a distributor wants to verify the supplier’s claim is valid. To this end, the distributor chooses a random sample of 9 heating furnaces of a new model and measures their efficiency. The data are

63; 72; 64; 69; 59; 65; 66; 64; 65

Determine the rejection region of the test with . Apply the test and state your conclusion.

7. A past study claims that adults spend an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. He took a sample of 10 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) were as follows

14; 25; 22; 38; 16; 26; 19; 23; 41; 33

Assume that the time spent on leisure activities by all adults is normally distributed. Using the 5% significance level, can you conclude that the claim of earlier study is true?

8. According to the department of Labor, private sector workers earned, on average $354.32 a week in 2001. A recently taken random sample of 400 private sector worker showed that they earn, on average, $362.50 a week with a standard deviation of $72. Find p-value for the test with an alternative hypothesis that the current wean weekly salary of private sector workers is different from $354.32.

9. A manufacturer of a light bulbs claims that the mean life of these bulbs is at least 2500 hours. A consumer agency wanted to check whether or not this claim is true. The agency took a random sample of 36 such bulbs and tested them. The mean life for the sample was found to be 2447 hours with a standard deviation of 180 hours.

a) Do you think that the sample information supports the company’s claim?

Use .

b) What is the Type I error in this case? Explain. What is the probability of making this error?

c) Will your conclusion of part a) change if the probability of making a

Type I error is zero?

10. Given the eight sample observations 31, 29, 26, 33, 40, 28, 30, and 25, test the null hypothesis that the mean equals 35 versus the alternative that it does not. Let .

Answers

1.a) reject if or ;b) reject if ;c)reject if ; 2. a) reject if ; b) reject if either

or ; c) reject if ; 3. a) ; reject ;

b) ;accept ;4.a) ;accept ;b) ;

reject ; c) ; accept ;5. ;reject ;

6. ; ; is rejected at ;7. ;

reject ;8. ; 9.a) ;accept ;

b) 0.025; c) no; 10. ; is not rejected.