Question

Question 14 of 22 (3 points) 7.2 Section Exercise 16 For a sample of 10 operating rooms taken in a hospital study, the mean noise level was 40.21 decibels and the standard deviation was 11.5. Find the 90% confidence interval of the true mean of the noise levels in the operating rooms. Assume the variable is normally distributed. Round your answers to two decimal places.

Answer 1

Here, number of sample (n) =10

Mean = 40.21

Standard Deviation (sd) =11.5

For 90% confidence interval , from the standard normal (Z-) distribution table , the value of z-score = 1.645

The formula for a confidence interval (CI) is given by

  $mean\pm z\frac{sd}{\sqrt{n}}$

So,

$40.21\pm 1.645\frac{11.5}{\sqrt{10}}$

The lower end of the interval

$=40.21- 1.645\frac{11.5}{\sqrt{10}}$

$=34.23$

The Upper end of the interval

   $=40.21+1.645\frac{11.5}{\sqrt{10}}$

$=46.19$

Therefore,

  $34.23<\mu <46.19$