Answer to Question #154650 in Statistics and Probability for chi 3

Question

The operations manager of a company that manufactures

shirts wants to determine whether there are

differences in the quality of workmanship among the

three daily shifts. She randomly selects 600 recently

made shirts and carefully inspects them. Each shirt is

classified as either perfect or flawed, and the shift that

produced it is also recorded. The accompanying table

summarizes the number of shirts that fell into each cell.

Do these data provide sufficient evidence to infer that

there are differences in quality between the three shifts?

Shirt Condition

Shift

1 2 3

Perfect 240 191 139

Flawed 10 9 11

Answer 1

The operations manager of a company that manufactures shirts and she randomly selected 600 recently made shirts and carefully inspects them. Each shirt is classified as either perfect or flawed, and the shift that produced it is also recorded in the following table.

The claim is whether there is any significant differences in quality between three shifts.

From this information we have to write the null and alternative hypotheses are as follows:

H₀: There are no differences in the quality between three shifts.

H₁: here are differences in quality between three shifts.

The level of significance α=0.05

The test statistic for testing null hypothesis is

$χ^2 = \sum^{k}_{i=1}\frac{(f_i-e_i)^2}{e_i}$

k = the number of cells in the cross-classification table

e_i = the expected frequencies

f_i = observed frequencies

The entries in the table are observed values f_i.

Expected frequencies for a contingency table

The expected frequency of the cell in row i and column j is

$e_{ij}=\frac{Row\; i \;total \times Column\; j\; total}{Sample \;size}$

The expected frequency values are produced as follows:

$e_{11}=\frac{Row\; 1 \;total \times Column\; 1\; total}{Sample \;size} \\ = \frac{570 \times 250}{600} \\ = 237.5 \\ e_{12}=\frac{Row\; 1 \;total \times Column\; 2\; total}{Sample \;size} \\ = \frac{570 \times 200}{600} \\ = 190 \\ e_{13}=\frac{Row\; 1 \;total \times Column\; 2\; total}{Sample \;size} \\ = \frac{570 \times 150}{600} \\ = 142.5 e_{21}=\frac{Row\; 2 \;total \times Column\; 1\; total}{Sample \;size} \\ = \frac{30 \times 250}{600} \\ = 12.5 \\ e_{22}=\frac{Row\; 2 \;total \times Column\; 2\; total}{Sample \;size} \\ = \frac{30 \times 200}{600} \\ = 10 \\ e_{23}=\frac{Row\; 2 \;total \times Column\; 3\; total}{Sample \;size} \\ = \frac{30 \times 150}{600} \\ = 7.5$

Thus the observed and estimated frequencies denoted are parentheses are tabulated as follows:

Now, we calculate the test statistic:

$χ^2 = \sum^{k}_{i=1}\frac{(f_i-e_i)^2}{e_i} \\ = \frac{(240-237.5)^2}{237.5} + \frac{(191-190)^2}{190} + \frac{(139-142.5)^2}{142.5}+ \frac{(10-12.5)^2}{12.5}+ \frac{(9-10)^2}{10}+\frac{(11-7.5)^2}{7.5} \\ = 0.0263+0.00526+0.085965+0.5+0.1+1.6333 \\ = 2.3508$

The test statistic χ² = 2.3508

The P-value at 2 degrees of freedom with this test statistic value is 0.3087 from excel we use CHISQ.DIST.RT (2.3508, 2).

Here we observe that, the P-value is greater than the level of the significance 0.05, so we fail to reject the null hypothesis. There is no sufficient evidence to conclude that there are significant differences in quality between three shifts.