Question

I want to know which of two manufacturing methods will be better. I create 10 prototypes using the first process, and 10 using the second. There were 3 defectives in the first batch and 5 in the second. Find a 95% confidence interval for the difference in the proportion of defectives.
a. (-0.62, 0.22)
b. (-0.56, 0.22)
c. (-0.493, 0.160)

Answer 1

Answer:

Option A

Step-by-step explanation:

Given information:

$n_1=10, n_2=10$

$x_1=3, x_2=5$

Using the given information we get

$p_1=\frac{x_1}{n_1}=\frac{3}{10}=0.3$

$p_2=\frac{x_2}{n_2}=\frac{5}{10}=0.5$

The formula for confidence interval for the difference of proportions is

$C.I.=(p_1-p_2)\pm z*\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$

From the standard normal table the value of z* at 5% confidence interval is 1.96.

Substitute the given values in the above formula.

$C.I.=(0.3-0.5)\pm z*\sqrt{\frac{0.3(1-0.3)}{10}+\frac{0.5(1-0.5)}{10}}$

$C.I.=-0.2\pm (1.96)\sqrt{0.021+0.025}$

$C.I.=-0.2\pm (1.96)\sqrt{0.046}$

$C.I.=-0.2\pm 0.42$

$C.I.=(-0.2-0.42,-0.2+0.42)$

$C.I.=(-0.62,0.22)$

Therefore, the correct option is A.