Question

A random sample of ?n measurements was selected from a population with standard deviation ?=14.6 and unknown mean ?. Calculate a 99% confidence interval for ? for each of the following situations:

(a) ?=45, ?⎯⎯⎯=76.3

____ ≤?≤ _______

(b)  ?=65, ?⎯⎯⎯=76.3

____  ≤?≤ ______

(c)  ?=85, ?⎯⎯⎯=76.3

______ ≤?≤ _______

Answer 1

Solution :

Given that,

Population standard deviation =    = 14.6

(a)

Point estimate = sample mean = = 76.3

Sample size = n = 45

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

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

= 2.576 * ( 14.6 /  45 )

= 5.61

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

- E < < + E

76.3 - 5.61 <   < 76.3 + 5.61

70.69 <   < 81.91

The 99% confidence interval for : 70.69 <   < 81.91

(b)

Point estimate = sample mean = = 76.3

Sample size = n = 65

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

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

= 2.576 * ( 14.6 /  65 )

= 4.66

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

- E < < + E

76.3 - 4.66 <   < 76.3 + 4.66

71.64 <   < 80.96

The 99% confidence interval for : 71.64 <   < 80.96

(c)

Point estimate = sample mean = = 76.3

Sample size = n = 85

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

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

= 2.576 * ( 14.6 /   85)

= 4.08

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

- E < < + E

76.3 - 4.08 <   < 76.3 + 4.08

72.22 <   < 80.38

The 99% confidence interval for : 72.22 <   < 80.38