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 40 cables and apply weights to each of them until they break. The 40 cables have a mean breaking weight of 775.3 lb. The standard deviation of the breaking weight for the sample is 14.9 lb. Find the 90% confidence interval to estimate the mean breaking weight for this type cable.

Answer 1

The correct answer is "771.44; 779.16".

Explanation:

Given:

Number of samples,

`n = 40`

Mean,

`\bar x = 775.3`

Standard deviation,

`\sigma = 14.9`

At 90% confidence interval,

`\alpha = 1-0.90`

`=0.10`

`(\alpha)/(2) = 0.05`

From normal distribution at 90% confidence level,

`Z_{(\alpha)/(2) } = 1.64`

Now,

Margin of error will be:

⇒
`E=Z_{(\alpha)/(2) }* (\sigma)/(√(n) )`

By substituting the values, we get

`=1.64* (14.9)/(√(40) )`

`=3.86`

hence,

The mean at 90% CI will be:

=
`\bar X \pm \ E`

=
`775.3\ \pm \ 3.86`

=
`771.44; 779.16`