Question

A manufacturer is comparing shipments of machine parts from two suppliers. The parts from Supplier A are less expensive, but the manufacturer is concerned they may be of lower quality than those from Supplier B.

700 parts were selected from Supplier A and 121 of them were found to be defective. 1200 parts were selected from Supplier B and 180 of them were found to be defective.

Can we conclude at a 1% significance level that Supplier A has a higher proportion of defects than Supplier 2?

Rounding: Round test statistics to two decimal places. Round all other values as specified in the problem.

Do steps 0 –5 and find the p-value. You may round your sample proportions to three decimal places .

	Number of Parts	Number of Defective units
Supplier A	700	121
Supplier B	1200	180

0.

1.

2.

3.

4.

5.

p-value

Answer 1

p1cap = X1/N1 = 121/700 = 0.1729
p1cap = X2/N2 = 180/1200 = 0.15
pcap = (X1 + X2)/(N1 + N2) = (121+180)/(700+1200) = 0.1584

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

Rejection Region
This is right tailed test, for α = 0.01
Critical value of z is 2.33.
Hence reject H0 if z > 2.33

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.1729-0.15)/sqrt(0.1584*(1-0.1584)*(1/700 + 1/1200))
z = 1.32

P-value Approach
P-value = 0.093
As P-value >= 0.01, fail to reject null hypothesis.