A manufacturing produces automobiles in two different systems (system A and syst

Question

A manufacturing produces automobiles in two different systems (system A and system B). However, an engineer observed that system A produces defective cars higher than system B. To support this theory, 100 cars built in system A is selected, 8 of them were defects. A random sample of 200 cars built in system B. 12 of them were defects. Let: P_a = proportion of defective cars from system A P_b = proportion of defective cars from system B Construct a 90% confidence interval for the difference in the proportions of defective cars from system A and system B. Test the hypothesis whether the proportion of defective cars from system A is higher than proportion of defective cars from system B at alpha = 0.05. Using the same hypothesis, test the hypothesis at alpha = 0.35. Is there any difference in your conclusion in (ii) and (iii)? If yes, explain.

Explanation / Answer

Solution :-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: PA >= PB
Alternative hypothesis: PA < PB

Hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the proportion of system A is sufficiently smaller than the proportion of system B.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.8 * 100) + (0.06 * 200)] / (100 + 200) = 92/300 = 0.3067

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = sqrt [ 0.3067 * 0.6933 * ( 1/100 + 1/200 ) ] = sqrt [0.0031895] = 0.056

z = (p1 - p2) / SE = (0.8 - 0.06)/0.056 = 1.3214

where p1 is the sample proportion in sample A, where p2 is the sample proportion in sample B, n1 is the size of sample A, and n2 is the size of sample B.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than 1.3214.

We use the Normal Distribution Calculator to find P(z < 1.3214) = 0.90682. Thus, the P-value = 0.9082.

P(z > 1.3214) = 1 - 0.90682 = 0.0918

Interpret results. Since the P-value (0.0918) is greater than the significance level (0.05), we cannot reject the null hypothesis.

If we test the hypothesis at significance level of 0.35, here we can reject the null hypothesis as 0.35 is greater than 0.0918.

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