Question

A production manager randomly sampled production lines at a factory that produces automobiles. She wanted to find out how many production lines caused defects in newly produced automobiles. The proportion of production lines that caused defects was 0.08, with a margin of error of 0.01. Construct a confidence interval for the proportion of production lines that caused defects

Answer 1

To construct the confidence interval for the proportion of production lines causing defects, add and subtract the margin of error from the proportion, giving a range from 0.07 to 0.09.

Step-by-step explanation:

The question asks for the construction of a confidence interval for the proportion of production lines that cause defects in automobiles. Given that the proportion is 0.08 with a margin of error of 0.01, the confidence interval can be calculated by subtracting and adding the margin of error from the given proportion. Therefore, the confidence interval is from 0.07 to 0.09 (0.08 - 0.01 to 0.08 + 0.01).

This interval means we are confident that the true proportion of defective production lines is between 7% and 9%, under the specified level of confidence which is not explicitly given but typically might be 95% confidence unless otherwise stated.

Answer 2

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

Step-by-step explanation:

A confidence interval for a population proportion is a function of the sample proportion and the margin of error.

The interval has two bounds, a lower bound and an upper bound.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this problem, we have that:

Sample proportion 0.08

Margin of error 0.01

0.08 - 0.01 = 0.07

0.08 + 0.01 = 0.09

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).