Question

A teacher gives a third grade class of n = 16 a reading skills test at the beginning of the school year. To evaluate the changes that occur during the year, students are tested again at the end of the year. Their test scores revealed an average improvement of M_D = 4.7 points with SS = 2160.

Construct a 99% confidence interval to estimate the average improvement for the population of third graders.

Lower and upper C.I=

Answer 1

Confidence interval for difference between two population means of paired samples is given as below:

Confidence interval = Dbar ± t*S_D/sqrt(n)

From given data, we have

M_D = Dbar = 4.7

SS = 2160

n = 16

df = n – 1 = 15

Variance = SS/(n - 1) = 2160/(16 - 1) = 2160/15 = 144

S_d = Sqrt(Variance) = Sqrt(144) = 12

Confidence level = 99%

Critical t value = 2.9467

(by using t-table)

Confidence interval = Dbar ± t*S_D/sqrt(n)

Confidence interval = 4.7 ± 2.9467*12/sqrt(16)

Confidence interval = 4.7 ± 8.8401

Lower limit = 4.7 - 8.8401 = -4.14

Upper limit = 4.7 + 8.8401 = 13.54

Lower limit = -4.17

Upper limit = 13.54

-4.14 < µ_d < 13.54