Question

A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of 46 cables and apply weights to each of them until they break. The 46 cables have a mean breaking weight of 779 lb. The standard deviation of the breaking weight for the sample is 15.4 lb. Find the 99% confidence interval to estimate the mean breaking weight for this type cable.

Answer 1

The 99% confidence interval to estimate the mean breaking weight for this type cable is between 772.9 lb and 785.1 lb.

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 46 - 1 = 45

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 45 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 2.69

The margin of error is:

`M = T(s)/(√(n)) = 2.69(15.4)/(√(46)) = 6.1`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 779 - 6.1 = 772.9 lb

The upper end of the interval is the sample mean added to M. So it is 779 + 6.1 = 785.1 lb

The 99% confidence interval to estimate the mean breaking weight for this type cable is between 772.9 lb and 785.1 lb.