Question

When the components arrive, the company selects a random sample from the shipment and subjects the selected components to a rigorous set of tests to determine if the components in the shipments conform to their specifications. From a recent large shipment, a random sample of 250 of the components was tested, and 24 units failed one or more of the tests. What is the 95% confidence interval estimate for the true proportion of components, p, that fail to meet the specifications

Answer 1

Answer:

(0.059, 0.133)

Step-by-step explanation:

Sample size = n = 250

Number of units which failed the test = x = 24

Proportion of units which failed the test = $\frac{x}{n} = \frac{24}{250} =\frac{12}{125}$ = 0.096

Proportion of units which did not fail the test = q = 1 - p = 1 - 0.096 = 0.904

Confidence level = 95%

z-value for the confidence level = z = 1.96

The true proportion of the components that fail to meet the specification would be:

$(p-z\sqrt{\frac{p \times q}{n}} , p+z\sqrt{\frac{p \times q}{n}})$

Using the values, we get:

$(0.096-1.96 \times \sqrt{\frac{0.096 \times 0.904}{250}} , 0.096+1.96 \times \sqrt{\frac{0.096 \times 0.904}{250}})\\\\ =(0.059,0.133)$

Thus, 95% confidence interval estimate for the true proportion of components, p, that fail to meet the specifications is (0.059, 0.133)