Question

I. In the testing of a new production method, 18 employees were selected randomly and asked to try the new method. The sample mean production rate for the 18 employees was 80 parts per hour and the sample standard deviation was 10 parts per hour. Provide 90% confidence intervals for the populations mean production rate for the new method, assuming the population has a normal probability distribution.

Answer 1

Answer:

The 90% confidence interval for the mean production rate fro the new method is (75.9, 84.1).

Step-by-step explanation:

We have to calculate a 90% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=80.

The sample size is N=18.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{10}{\sqrt{18}}=\dfrac{10}{4.24}=2.36$

The degrees of freedom for this sample size are:

$df=n-1=18-1=17$

The t-value for a 90% confidence interval and 17 degrees of freedom is t=1.74.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.74 \cdot 2.36=4.1$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 80-4.1=75.9\\\\UL=M+t \cdot s_M = 80+4.1=84.1$

The 90% confidence interval for the mean production rate fro the new method is (75.9, 84.1).