Question

A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no more than 3% of the boards fall outside the most rigid level of industry performance specifications. In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence limit of the 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification.

Answer 1

Answer:

The 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

Let X = number of boards that fall outside the most rigid level of industry performance specifications.

In a random sample of 300 boards the number of defective boards was 12.

Compute the sample proportion of defective boards as follows:

$\hat p =\frac{12}{300}=0.04$

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

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

The critical value of z for 95% confidence level is,

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

*Use a z-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).