Question

A materials engineer wants to test whether or not a certain device spends an average of more than 30 seconds soaking a component part in solution. Therefore, she goes out and takes a simple random sample of n =16 parts, and finds that the sample mean is 33.4 seconds with a sample standard deviation of s =8.0 seconds.

At α= 0.05, the Shapiro-Wilk Test for Normality has a p-value of p = 0.0667.

Suppose the engineer goes back and she selects a 17th sample unit. Using a 95% confidence level, provide the lower end point of a two-tailed interval that would be used to predict the range of values that this 17th sampled unit would be in with respect to seconds soaking a component part in solution.

Answer 1

Answer:

Using a 95% confidence level, the lower end point of a two-tailed interval that would be used to predict the range of values that this 17th sampled unit would be in with respect to seconds soaking a component part in solution is of 29.137 seconds.

Step-by-step 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 = 16 - 1 = 15

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.1315

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.1315\frac{8}{\sqrt{16}} = 4.263$

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 33.4 - 4.263 = 29.137 seconds.

Using a 95% confidence level, the lower end point of a two-tailed interval that would be used to predict the range of values that this 17th sampled unit would be in with respect to seconds soaking a component part in solution is of 29.137 seconds.