Question

Consider a production process that produces batteries. A quality engineer has taken 20 samples each containing 100 batteries. The total number of defective batteries observed over the 20 samples is 200.
Construct a 95% confidence interval of the proportion of defectives.


Another sample of 100 was taken and 15 defectives batteries were found. What is your conclusion?

Answer 1

Answer:

The 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).

It is better to take a larger sample to derive conclusion about the true parameter value.

Step-by-step explanation:

The (1 - α) % confidence interval for proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Given:

n = 2000

X = 200

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{200}{2000}=0.10$

The critical value of z for 95% confidence interval is:

$z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.10\pm1.96\times\sqrt{\frac{0.10(1-0.10)}{2000}}\\=0.10\pm0.0034\\=(0.0966, 0.1034)$

Thus, the 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).

Now if in a sample of 100 batteries there are 15 defectives, the the 95% confidence interval for this sample is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.15\pm1.96\times\sqrt{\frac{0.15(1-0.15)}{100}}\\=0.15\pm0.0706\\=(0.0794, 0.2206)$

It can be observed that as the sample size was decreased the width of the confidence interval was increased.

Thus, it can be concluded that it is better to take a larger sample to derive conclusion about the true parameter value.