Question

Let's assume our class represents a normal population with a known mean of 95 and population standard deviation 3. There are 120 students in the class.

a. Construct the 95% confidence interval for the population mean.

b. Interpret what this means.

c. A few students have come in. Now we cannot assume normality and we don't know the population standard deviation. Let the sample mean = 90 and sample standard deviation = 2. Let's make the sample size 20. We can assume alpha to be .05. Construct the 95% confidence interval assuming this new information.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 95

Population standard deviation =    = 3

Sample size = n =120

a) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

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

= 1.96 * (3 /  120 )

= 0.54

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

  ± E

95 ± 0.54   

( 94.46, 95.54 )  

b) We are 95% confident that the true mean of our class represents is between 94.46 and 95.54.

c) Point estimate = sample mean = = 90

sample standard deviation = s = 2

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,19 = 2.093

Margin of error = E = t/2,df * (s /n)

= 2.093 * ( 2 / 20)

Margin of error = E = 0.94

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

  ± E  

= 90 ± 0.94

= ( 89.06, 90.94 )