Question

In: Math

You are given the sample mean and the population standard deviation. Use this information to construct...

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals.

A random sample of 60 home theater systems has a mean price of $111.00. Assume the population standard deviation is $16.20.

The 90% confidence interval is(--,--)

The 99% confidence interval is(--,--)

Which interval is wider?

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 111.00

Population standard deviation =    = 16.20

Sample size = n = 60

a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645


Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 16.20 /  60)

= 3.44

At 90% confidence interval estimate of the population mean is,

  ± E

111.00 ± 3.44   

( 107.56, 114.44 )  

b) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576


Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 16.20 /  60)

= 5.39

At 99% confidence interval estimate of the population mean is,

  ± E

111.00 ± 5.39   

( 105.61, 116.39 )  

A 99% confidence interval is wider,


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