Question

During the independent research 30 women were chosen to measure their weight. The
mean value of weight is 66 kg and it is known from the previous experience that the weight is normally
distributed with ? = 10 kg.

a) Find a 95% two-sided confidence interval on the mean weight.

b) Find a 90% two-sided confidence interval on the mean weight.

c) Which interval is wider?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 66

Population standard deviation = = 10

Sample size = n = 30

b)

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z/2 = 1.960  

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

= 1.960 * ( 10 /  30 )

= 3.58

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

- E < < + E

66 - 3.58 <   < 66+ 3.58

62.42 <   < 69.58

( 62.42 ,69.58 )

b)

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z/2 = 1.645  
Margin of error = E = Z/2 * ( /n)

= 1.645 * (10 /  30 )

= 3.00

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

- E < < + E

66 - 3.00 <   < 66 + 3.00

63<   < 69

( 63 ,69 )

c) A 95 % confidence interval IS WIDER.