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 $135.00 . Assume the population standard deviation is $15.90 .
A.) the 90% confidence interval is
B.) the 95% confidence interval is
C.) Which interval is wider?
Solution :
Given that,
= $135
= $15.90
n = 60
A)At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
Z_{/2} = Z_{0.05} = 1.645
Margin of error = E = Z_{/2}* ( /n)
= 1.645 * (15.90 / 60)
= 0.436
At 90% confidence interval estimate of the population mean is,
- E < < + E
135 - 0.436 < < 135 + 0.436
134.564 < < 135.436
(134.564,135.436 )
The 90% confidence interval is $134.564 to $135.436
B)At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z_{/2} = Z_{0.025} = 1.96
Margin of error = E = Z_{/2}* ( /n)
= 1.96 * (15.90 / 60)
= 0.519
At 95% confidence interval estimate of the population mean is,
- E < < + E
135 - 0.519 < < 135 + 0.519
134.481 < < 135.519
(134.481,135.519 )
The 95% confidence interval is $134.481 to $135.519
C) The 95% interval is wider.
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