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 three defectives in the first batch and five in the second. What is a 95% confidence interval for the difference in the proportion of defectives?

Answer 1

Answer: (-0.620, 0.220)

Step-by-step explanation:

The formula to find the confidence interval for the difference in true proportion of the two groups. is given by :-

$\hat{p}_1-\hat{p}_2\pm z^* \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

, where $n_1$ = Sample size for first group.

$n_2$ = Sample size for second group.

$\hat{p}_1$ = Sample proportion for first group.

$\hat{p}_2$ = Sample proportion for second group.

z* = critical z-value.

Let first group be "group of prototypes by first process " and second group be "group of prototypes by second process".

$p_1$ = Proportion of defectives in the first batch.

$p_2$ = Proportion of defectives in the second batch.

From question , we have

$n_1=10$ , $n_2=10$  , $\hat{p}_1=\dfrac{3}{10}=0.3$  , $\hat{p}_2=\dfrac{5}{10}=0.5$

By z-table , Critical value for 95% confidence interval is z* =1.96.

Substitute all values in formula , we get

$0.3-0.5\pm (1.96)\sqrt{\dfrac{0.3(1-0.3)}{10}+\dfrac{0.5(1-0.5)}{10}}$

$-0.2\pm (1.96)\sqrt{0.021+0.025}$

$-0.2\pm (1.96)\sqrt{0.046}$

$-0.2\pm (1.96)(0.21448)$

$-0.2\pm 0.4203808$

$=(-0.2-0.4203808,\ -0.2+0.4203808)\\\\=(-0.6203808,\ 0.2203808)\\\\\approx(-0.620,\ 0.220)$

Hence,  a 95% confidence interval for the difference in the proportion of defectives is (-0.620, 0.220).