Question

Let's assume our class represents a normal population with a known mean of 90 and population standard deviation 2. There are 100 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 = 3. 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,

a) Point estimate = sample mean = = 90

Population standard deviation =    = 2

Sample size = n = 100

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 * ( 2 /  100 )

= 0.39

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

  ± E

90 ± 0.39

( 89.61, 90.39 )  

b) We are 95% confident that the true mean of our class represents between 89.61 and 90.39.

c) Point estimate = sample mean = = 90

sample standard deviation = s = 3

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 * (3 / 20)

Margin of error = E = 1.40

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

  ± E  

= 90 ± 1.40

= ( 88.60, 91.40 )