Question

A company that makes tile flooring routinely takes samples from the manufacturing process to check for defects. In a random sample of 609 tiles from the process, the quality control department finds that 45 of the tiles showed some kind of defect (either cracking, an uneven finish, or some other problem). Calculate the upper bound of a 99% confidence interval for the true proportion of defective tiles the manufacturing process will make

Answer 1

Answer: 0.0845

Step-by-step explanation:

Formula to find the upper bound of confidence interval for population proportion is given by :-

$\hat{p}+ z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$  = sample proportion.

z* = Critical value

n= Sample size.

Let p be the true proportion of defective tiles the manufacturing process will make.

Given : Sample size =  609

Number of tiles showed some kind of defect= 45

Then,  sample proportion =$\hat{p}=\dfrac{45}{609}\approx0.0739$

The critical value for 99% confidence interval = z*=2.576  (By z-table)

The upper bound of a 99% confidence interval for the true proportion of defective tiles the manufacturing process will make will be

$0.0739+(2.576)\sqrt{\dfrac{0.0739(1-0.0739)}{609}}$

$0.0739+ (2.576)\sqrt{0.000112378965517}$

$\approx0.0739+0.0106\\\\= 0.0845$

Hence,  the upper bound of a 99% confidence interval for the true proportion of defective tiles the manufacturing process will make = 0.0845