Math 207—Small-sample Inference Examples

 

Example 1

Some quality control experiments require destructive sampling in order to measure some particular characteristic of the product (i.e., the test to determine whether the item is defective destroys the item). The cost of the destructive sampling often dictates small samples.

 

A manufacturer of printers for personal computers wishes to estimate the mean number of characters printed before the print head fails. The manufacturer tests 15 randomly selected print heads and records the number of characters printed until failure for each. The 15 measurements (in millions of characters) are shown in the stem and leaf plot below. The sample mean and standard deviation are 1.25 and 0.17, respectively.

 

Stem-and-leaf of Characters,  N  = 15

Leaf Unit = 0.010

 

      9| 2

    10| 7 9

    11| 3 8

    12| 0 2 5 9

    13| 2 3 6

    14| 3 8

    15| 5

 

  1. Is it appropriate to use the small-sample t procedures?

 

 

 

  1. Find a 95% confidence interval for the mean number of characters to failure for all printers.

 

 

 

 

 

  1. Suppose specifications dictate that the mean number of characters to failure is 1.2 million. At the  level, do we have enough evidence to conclude that specifications are not being met? Why or why not?

 

 

 

 

Example 2

A study was conducted on the effect of a special class designed to improve children’s verbal skills. Each of 19 children took a verbal skills test twice, both before and after a 3-week period in the special class. From the sample, the “after – before” scores had a mean of 0.695 points and a standard deviation of 1.118 points. Also, a dotplot of the sample of differences appears fairly mound-shaped.

 

  1. Is there evidence that the mean improvement is greater than 0? Carry out the significance test (state hypotheses, check conditions, calculate test statistic and p-value, interpret the results).

 

 

 

 

 

 

 

 

  1. If there is evidence, can this improvement be solely attributed to the special class?

 

 

Example 3

Behavioral researchers developed an index designed to measure managerial success. The index (measured on a 100-point scale) is based on the manager’s length of time in the organization and his or her level within the firm; the higher the index, the more successful the manager. Suppose a researcher wants to compare the average success index of two groups of managers at a large manufacturing plant. Managers in Group 1 engage in a high volume of interactions with people outside their work unit. Managers in Group 2 rarely interact with people outside their work unit. Independent random samples of 12 and 15 managers are selected from Groups 1 and 2, respectively. The sample from Group 1 has mean and standard deviation, 65.33 points and 6.61 points, respectively. The sample from Group 2 has mean and standard deviation, 56.02 points and 9.93 points, respectively. Both sample distributions look fairly mound-shaped.

 

Is there evidence of a difference between the group means? Carry out the significance test (state hypotheses, check conditions of the test, calculate test statistic and p-value, interpret the results)

 

Hypotheses

Let  be the average success index for all Group 1 managers, and let  be the average success index for all Group 2 managers.

 

We want to test the hypotheses

 

Choice of Test and Check of Conditions

This is clearly a two-sample problem. Since the population standard deviations are unknown and we have small samples, we need to use the two-sample t test. Is the normality condition met? Yes, because we’re told both sample-data distributions look mound-shaped (indicating that the populations they came from are also mound-shaped). Is the equal-variances condition met? Yes, since , by our rule of thumb it seems reasonable that the two population variances are the same. Hence, we can proceed with the two-sample t-test with the pooled estimate of .

 

Test Statistic

The pooled estimate of the variance is , and the test statistic is

 

P-value

To find the p-value, we must use the t distribution with (12 + 15 – 2) = 25 degrees of freedom, and since our alternative hypothesis is two-sided, we must double the p-value.

 

The p-value is for a t-distribution with 25 degrees of freedom. The value 2.786 isn’t shown in Table 4, but the value 2.787 is (which is close enough, so we’ll use it). So, based on Table 4, the p-value is 2(0.005) = 0.01.

 

Definition of the P-value and Conclusion in the Context of the Problem

If the average success index is the same for managers from both groups, then there is only a 1% chance of getting our particular difference in sample average indexes or a more extreme difference. Because our data are so unlikely, we have very strong evidence that there is in fact a difference between average success indexes for the two groups. These results are statistically significant at even the stringent 0.01 significance level.

 

Practical Significance

It is also important to consider the practical significance. A 95% confidence interval for is (2.43 points, 16.19 points). We should ask the researchers if this difference is of practical importance.