A manufacturer of flash drives for data storage operates a production facility that runs on three 8-hour shifts per day.The following contingency table shows the number of flash drives that were defective and not defective from each shift. Based on the critical value approach and using α = 0.05,which of the following decisions and conclusions for this hypothesis test would be accurate?
A) The test statistic is greater than the critical value;we reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are independent of one another.
B) The test statistic is greater than the critical value;we fail to reject the null hypothesis and can conclude that the quality of the flash drive production and the production shift are independent of one another.
C) The test statistic is less than the critical value;we fail to reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are not independent of one another.
D) The test statistic is less than the critical value;we reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are independent of one another.

Question

A manufacturer of flash drives for data storage operates a production facility that runs on three 8-hour shifts per day.The following contingency table shows the number of flash drives that were defective and not defective from each shift.  Based on the critical value approach and using α = 0.05,which of the following decisions and conclusions for this hypothesis test would be accurate?
A) The test statistic is greater than the critical value;we reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are independent of one another. 
B) The test statistic is greater than the critical value;we fail to reject the null hypothesis and can conclude that the quality of the flash drive production and the production shift are independent of one another. 
C) The test statistic is less than the critical value;we fail to reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are not independent of one another. 
D) The test statistic is less than the critical value;we reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are independent of one another.

Quizplus · Accepted Answer

To test the independence of production shift and flash drive quality, we use chi-square goodness-of-fit test. The null hypothesis is that the quality of flash drive production is independent of the production shift. The alternative hypothesis is that the quality of flash drive production is dependent on the production shift.

The test statistic is calculated as

χ2=(Σ(O−E)2)/(E)

where O is the observed frequency and E is the expected frequency under the null hypothesis.

We calculate the expected frequency for each shift by multiplying the total number of defective and non-defective flash drives by the proportion of the shift's total working hours to the production facility's total working hours.

The expected frequencies and test statistic are shown in the table below.

Shift | Defective | Non-defective | Total | Proportion | Expected Defective | Expected Non-defective | 
|------|-----------|---------------|-------|------------|--------------------|------------------------|
|   1  |    7      |      793      |  800  |    1/3     |  (11*800)/2400 = 3.67      |    (789*800)/2400 = 263.33     |
|   2  |    9      |      791      |  800  |    1/3     |  (11*800)/2400 = 3.67      |    (789*800)/2400 = 263.33     |
|   3  |   10      |      790      |  800  |    1/3     |  (11*800)/2400 = 3.67      |    (789*800)/2400 = 263.33     |
| Total|   26      |     2374      | 2400  |     -      |           -          |                -           |
 
The degrees of freedom are (3-1)*(2-1) = 2. We find the critical value of chi-square at α=0.05 and df = 2 to be 5.99 using chi-square distribution table.

The test statistic is calculated as

χ2=(7−3.67)2/3.67+(9−3.67)2/3.67+(10−3.67)2/3.67+(793−263.33)2/263.33+(791−263.33)2/263.33+(790−263.33)2/263.33 = 3.810

The test statistic is less than the critical value (3.810 < 5.99), so we fail to reject the null hypothesis. Therefore, we cannot conclude that the quality of flash drive production and the production shift are not independent of one another. The correct answer is (C) The test statistic is less than the critical value;we fail to reject the null hypothesis and cannot conclude that the quality of the flash drive production and the production shift are not independent of one another.

A Manufacturer of Flash Drives for Data Storage Operates a Production