Answer:
Option C) 0.0602
Step-by-step explanation:
We are given the following in the question:
[tex]x_1 = 20\\n_1 = 400\\x_2 = 10\\n_2 = 100[/tex]
Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective.
[tex]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[/tex]
First, we design the null and the alternate hypothesis
[tex]H_{0}: p_1 = p_2\\H_A: p_1 \neq p_2[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]\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})}}[/tex]
Putting all the values, we get,
[tex]\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[/tex]
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
