Question

  You wish to see the average improvement on SOL scores after students have completed a program. A random sample of 300 students who completed the program showed an average improvement of 4.35 points with a standard deviation of 0.96 points. Calculate and interpret a 90% confidence interval for the average improvement for all students who would complete this program.

A: Verify the assumptions.

B: Calculate a 90% confidence interval for the average improvement for all students who would complete this program.

C: Interpret a 90% confidence interval for the average improvement for all students who would complete this program.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 4.35

sample standard deviation = s = 0.96

sample size = n = 300

Degrees of freedom = df = n - 1 = 300 - 1 = 299

A) Assume that the population has a normal distribution.

B) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,299 = 1.650

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

= 1.650 * ( 0.96 / 300)

Margin of error = E = 0.09

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

  ± E  

= 4.35  ± 0.09

= ( 4.26, 4.44 )

C) We are 90% confident that the true average improvement for all students who would complete this program between 4.26 and 4.44.