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.

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