Question

Exercise 1. The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the measurements of the sound intensity of a random sample of 9 such food processors showed a sample mean of 50.3 decibels, find a 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type.

Exercise 2. What is the answer to Exercise 1 if the standard deviation of the population is unknown but sample standard deviation of the random sample of 9 food processors is s=2.9 decibels? Compare the length of the confidence interval obtained here with the length obtained in Exercise 1.

i need help with exercise 2.thanks

Answer 1

Solution:-

2) 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type is C.I = ( 48.071, 52.529).

if the population standard deviation is unknown then we will use t-distribution and assume that population is normally distributed.

C.I = 50.3 + 2.306 × 0.96667

C.I = 50.3 + 2.2291

C.I = ( 48.071, 52.529)

The length of the confidence interval obtained here will be greater than the the length obtained in Exercise 1, because the t-critical value is always greater than z-critical value for the same confidence level.