Question

A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective. Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. Is there evidence of a difference in the proportion of defective parts produced by the two suppliers? To make this determination, you test the hypotheses H0 : p1 = p2 and Ha : p1 ≠ p2. The P-value of your test is
a) 0.1164.

b) 0.0301.

c) 0.0602.

Answer 1

Answer:

Option C) 0.0602  

Step-by-step explanation:

We are given the following in the question:

$x_1 = 20\\n_1 = 400\\x_2 = 10\\n_2 = 100$

Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective.

$p_1 = \dfrac{x_1}{n_1} = \dfrac{20}{400} = 0.05\\\\p_2 = \dfrac{x_2}{n_2} = \dfrac{x_2}{n_2} = \dfrac{10}{100}= 0.1$

First, we design the null and the alternate hypothesis

$H_{0}: p_1 = p_2\\H_A: p_1 \neq p_2$

We use Two-tailed z test to perform this hypothesis.

Formula:

$\text{Pooled P} = \dfrac{x_1+x_2}{n_1+n_2}\\\\Q = 1 - P\\\\Z_{stat} = \dfrac{p_1-p_2}{\sqrt{PQ(\frac{1}{n_1} + \frac{1}{n_2})}}$

Putting all the values, we get,

$\text{Pooled P} = \dfrac{20+10}{400+100} = 0.06\\\\Q = 1 - 0.06 = 0.94\\\\Z_{stat} = \dfrac{0.05-0.1}{\sqrt{0.06\times 0.94(\frac{1}{400} + \frac{1}{100})}} = -1.883$

Now, we calculate the p-value from the table at 0.05 significance level.

P-value = 0.0602

Thus, the correct answer is

Option C) 0.0602