Describe test based on small samples
Answers
Step-by-step explanation:
EXAMPLE 11
A small component in an electronic device has two small holes where another tiny part is fitted. In the manufacturing process the average distance between the two holes must be tightly controlled at 0.02 mm, else many units would be defective and wasted. Many times throughout the day quality control engineers take a small sample of the components from the production line, measure the distance between the two holes, and make adjustments if needed. Suppose at one time four units are taken and the distances are measured as
0.0210.0190.0230.020
Determine, at the 1% level of significance, if there is sufficient evidence in the sample to conclude that an adjustment is needed. Assume the distances of interest are normally distributed.
Solution:
Step 1. The assumption is that the process is under control unless there is strong evidence to the contrary. Since a deviation of the average distance to either side is undesirable, the relevant test is
H0:μ vs. Ha:μ=≠0.020.02@ α=0.01
where μ denotes the mean distance between the holes.
Step 2. The sample is small and the population standard deviation is unknown. Thus the test statistic is
T=x−−−μ0s/n−−√
and has the Student t-distribution with n−1=4−1=3 degrees of freedom.
Step 3. From the data we compute x−−=0.02075 and s = 0.00171. Inserting these values into the formula for the test statistic gives
T=x−−−μ0s/n−−√=0.02075−0.020.00171/4–√=0.877
Step 4. Since the symbol in Ha is “≠” this is a two-tailed test, so there are two critical values, ±tα∕2=−t0.005[df=3]. Reading from the row in Figure 12.3 "Critical Values of " labeled df=3 their values are ±5.841. The rejection region is (−∞,−5.841]∪[5.841,∞).
Step 5. As shown in Figure 8.13 "Rejection Region and Test Statistic for " the test statistic does not fall in the rejection region. The decision is not to reject H0. In the context of the problem our conclusion is:
The data do not provide sufficient evidence, at the 1% level of significance, to conclude that the mean distance between the holes in the component differs from 0.02 mm.
Figure 8.13
Rejection Region and Test Statistic for Note 8.43 "Example 11"
To perform the test in Note 8.43 "Example 11" using the p-value approach, look in the row in Figure 12.3 "Critical Values of " with the heading df=3 and search for the two t-values that bracket the value 0.877 of the test statistic.