Question

In a quality control test of parts manufactured at Dabco Corporation, an engineer sampled parts produced on the first, second, and third shifts. The research study was designed to determine if the population proportion of good parts was the same for all three shifts. The test to be used is a chi-square test of homogeneity using a significance level α=0.05.

Testing the equality of population proprotions for three or more populations

Observed	Production Shift
Quality	First	Second	Third
Good	285	368	176
Defective	15	32	24

1. State the hypotheses.
2. Use Excel to calculate the expected frequencies under the assumption H₀ is true and use these and the observed frequencies to calculate the test statistic.
3. Use Excel to find the p-value.

Hint: Use CHISQ.DIST.RT

4. What is your hypothesis test decision?

Hint: There are two possible decisions: reject H₀ in favour of H_a or fail to reject H₀.

5. What is your conclusion in the context of the application?

Hint: The conclusion should relate back to the question.

6. If the conclusion in part (e) is that the population proportions are not all equal, use the Marascuilo Pairwise Comparison Procedure to determine which pairs of shifts differ in terms of quality. Which shift(s) need to improve the quality of parts produced?

Answer 1

a) Null hypothesis (Ho) : The population proportion of good parts is same for the three shifts: first, second and third

Alternative hypothesis (Ha) : The population proportion of good parts is different for the three shifts: first, second and third

b) to find expected frequency at any cell: (row sum of that cell* column sum of that cell) / 900

where 900 is total frequency

		observed frequency					expected frequency
		production shift						production shift
quality	first	second	third	row sum		quality	first	second	third
good	285	368	176	829		good	276.3333	368.4444	184.2222
defective	15	32	24	71		defective	23.66667	31.55556	15.77778
col sum	300	400	200	900

oi represent the observed frequency and ei represents the expected frequency and the value of test statisitic is 8.103071

oi	ei	(oi-ei)^2 / ei
285	276.33	0.272026
368	368.44	0.000525
176	184.22	0.366781
15	23.67	3.175703
32	31.56	0.006134
24	15.78	4.281901
		8.103071	value of test statistics

c) Degree of freedom = (number of rows -1)*(number of columns - 1) = (2-1)*(3-1) = 2 and the test statistic value = 8.103071

so, p value is given by

=CHISQ.DIST.RT(8.103071,2)

= 0.017396

d) test decision

since alpha = 0.05 and p value = 0.017396

since p value < alpha, we reject the null hypothesis Ho in favour of Ha

e) conclusion:

The population proportion of good parts is different for the three shifts: first, second and third