Question

An instructor believes that students do not retain as much information from a lecture on a Friday compared to a Monday. To test this belief, the instructor teaches a small sample of college students some preselected material from a single topic on statistics on a Friday and on a Monday. All students received a test on the material. The differences in exam scores for material taught on Friday minus Monday are listed in the following table. Difference Scores (Friday − Monday) +4.4 +3.3 −1.6 +1.1 +6.2 (a) Find the confidence limits at a 95% CI for these related samples. (Round your answers to two decimal places.) to (b) Can we conclude that students retained more of the material taught in the Friday class? Yes, because 0 lies outside of the 95% CI. No, because 0 is contained within the 95% CI.

Answer 1

SOLUTION:

From given data,

An instructor believes that students do not retain as much information from a lecture on a Friday compared to a Monday. To test this belief, the instructor teaches a small sample of college students some pre selected material from a single topic on statistics on a Friday and on a Monday. All students received a test on the material. The differences in exam scores for material taught on Friday minus Monday are listed in the following table. Difference Scores (Friday − Monday) +4.4 +3.3 −1.6 +1.1 +6.2

Difference Scores (Friday − Monday) d +4.4 +3.3 −1.6 +1.1 +6.2

(  - )²    (4.4-2.68)² (3.3-2.68)²   (−1.6-2.68)²   (1.1-2.68)²   (6.2-2.68)²

We calculate sample mean and standard deviation from given data,

Sample size = n =5

Sample Mean , =

= 13.4 / 5

= 2.68

Sample Variance , s² =

= 36.548 / (5-1)

= 36.548/4

= 9.137

Sample standard deviation , s = =

= 3.0227

(a) Find the confidence limits at a 95% CI for these related samples.

95% CI for using t- dist

Significance level = = 1-0.95 = 0.05

Degree of freedom for t- distribution , df = n-1 = 5-1 = 4

Critical value = t_{/2 , df} = t _0.025,4 = 2.776 (from t- table , two-tails , df =4)

Margin of Error =E = t_{/2 , df}   s_{d /}   

_{= 2.776} 3.0227 /

= 2.776 1.35179

= 3.752569

Limits of 95% confidence interval are given by:

Lower limit = - E =  2.68 - 3.752569

= -1.072569 -1.073

Lower limit = + E =  2.68+ 3.752569

=6.432569 6.433

95% confidence interval is E = 2.68 3.752569

= ( -1.073 , 6.433)

(b) Can we conclude that students retained more of the material taught in the Friday class? Yes, because 0 lies outside of the 95% CI. No, because 0 is contained within the 95% CI.

Since 95% confidence interval contains 0, thus we can conclude that null hypothesis is not rejected and we can not conclude that students remained more of te material taught in friday's class.