Question

1. You randomly select 36 restaurants and measure the temperature of coffee sold at each. The sample mean temperature is 162.0°F. Temperatures historically have had a population standard deviation of 10.0°F. Which of the following intervals is the correct 95% confidence interval for the true mean? (T interval or Z interval)

2. If all other quantities remain the same, how does the indicated change affect the width of a confidence interval? Decrease the level of confidence from 95% to 90%?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 162.0

Population standard deviation =    = 10.0

Sample size = n = 36

1) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

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

= 1.96 * ( 10.0 /  36 )

= 3.27

At 95% confidence interval estimate of the population mean is,

  ± E

162.0 ± 3.27

( 158.73, 165.27 )  

2) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 * ( 10.0 /  36 )

= 2.74

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

  ± E

162.0 ± 2.74

( 159.26, 164.74 )

A confidence level decreases, the width of interval is decreases