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.**

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