Question

In: Statistics and Probability

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.

From a random sample of 74 ​dates, the mean record high daily temperature in a certain city has a mean of 84.20°F. Assume the population standard deviation is

15.06°F.

The​ 90% confidence interval is

The​ 95% confidence interval is

Solutions

Expert Solution

Solution :

Given that,

= 84.20

= 15.06

n = 74

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645* (15.06 / 74 )

= 2.88

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

- E < < + E

84.20 - 2.88 < < 84.20 + 2.88

81.32 < < 87.08

(81.32 , 87.08)

b ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 * (15.06 / 74 )

= 3.43

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

- E < < + E

84.20 - 3.43 < < 84.20 + 3.43

80.77 < < 87.63

( 80.77 , 87.63)


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