Question

Question 4 of 5 (1 point) For a sample of 5 operating rooms taken in a hospital study, the mean noise level was 38.02 decibels and the standard deviation was 10.9. 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 7.2 Section Exercise 16 Ru

Answer 1

solution:-

given that mean = 38.02 , standard deviation = 10.9

and n = 5 then df = n - 1 = 5 - 1 = 4

90% confidence with df = 4 is t = 2.132

confidence interval formula

=> x +/- t * s/sqrt(n)

=> 38.02 +/- 2.132 * 10.9/sqrt(5)

=> 38.02 +/- 10.39

=> (27.63 , 48.41)

then

=> 27.63 < μ < 48.41