Question

A machine elastic bands with breaking tension normally distributed with mean 45 N and standard deviation 4.36 N.
a. In order to test whether there is a change in the mean breaking tension, a random
sample of 50 was tested and found to have a mean breaking tension of 43.46 N.
i. Determine a 95% confidence interval for the population mean breaking tension
based on the sample mean assuming an unchanged standard deviation.

Answer 1

Sample mean=43. 46N

Population standard deviation =4.36N

95% CI =$\bar x±Z_c \frac {\sigma} {\sqrt n}$

= 43.46 ±1.96 *$\frac {4.36}{\sqrt {50}}$

=(42. 25,44.67)