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 47 cables and apply weights to each of them until they break. The 47 cables have a mean breaking weight of 777.4 lb. The standard deviation of the breaking weight for the sample is 15.5 lb.

Find the 90% confidence interval to estimate the mean breaking weight for this type cable. ( , ) Your answer should be rounded to 2 decimal places.

Answer 1

`777.4-1.679(15.5)/(√(47))=773.60`

`777.4+1.679(15.5)/(√(47))=781.20`

We are 90% confident that the true mean for the breaking weigth for this typpe of cable is between 773.60 and 781.20

Explanation:

Data provided from the problem

`\bar X=777.4` represent the sample mean for the breaking weights

`\mu` population mean

s=15.5 lb represent the sample standard deviation

n=47 represent the sample size

Confidence interval

The confidence interval for the true population mean is given by:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom, given by:

`df=n-1=47-1=46`

The Confidence level is 0.90 or 90%, the significance would be
`\alpha=1-0.9=0.1` and
`\alpha/2 =0.05` and the critical value for this case is
`t_(\alpha/2)=1.679`

Replacing into the confidence formula we got:

`777.4-1.679(15.5)/(√(47))=773.60`

`777.4+1.679(15.5)/(√(47))=781.20`

We are 90% confident that the true mean for the breaking weigth for this typpe of cable is between 773.60 and 781.20

Answer 2

The 90% confidence interval to estimate the mean breaking weight for this type cable is between 751.38 lb and 803.42 lb.

Explanation:

We are in posession of the sample's standard deviation, so we use the student t-distribution 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 = 47 - 1 = 46

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 46 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95`. So we have T = 1.6787

The margin of error is:

M = T*s = 1.6787*15.5 = 26.02 lb

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 777.4 - 26.02 = 751.38

The upper end of the interval is the sample mean added to M. So it is 777.4 + 26.02 = 803.42

The 90% confidence interval to estimate the mean breaking weight for this type cable is between 751.38 lb and 803.42 lb.