Answer to Question #154525 in Statistics and Probability for Mano

Question

To verify whether a course in statistics improved performance, a similar test was given to 12 participants both before and after the course. The original grades recorded in alphabetical order of the participants were 44. 40, 61, 52, 32, 44, 70, 41, 67, 72, 58 and 72. After the course, the grader were in the same order 58. 38, 68, 87, 46, 39. 78, 48, 78, 74, 60 and 78.

(a) Was the course useful, as measured by performance on the test? Consider these 12 participants as a sample from a population.

(b) Would the same conclusion be reached if tests were not considered paired?Use 5% level of significance in both cases

Answer 1

Soln:

Let first test be $x_1$

Let second test be $x_2$

Hypothesis

$$H_0 : \mu_1 - \mu_2 =0 \space vs \space H_1 : \mu_1 - \mu_2 < 0$$

a). As paired data

$$t = { {\sum D \over n} \over \sqrt { \sum D^2 - {{(\sum D)} ^2 \over n} \over n(n-1)}}$$

$$where D= x_1 - x_2, \space n=sample \space size$$

$\sum D = - 99, \space \sum D^2 =1973$

$$now, \space t= {{-99 \over 12} \over \sqrt {1973 - {{99}^2 \over 12} \over 12 *11} }$$

$=-2.787$

$t_{0.05,11}=1.796$

Since t-calculated < t-critical, we fail to reject $H_0$

Therefore, there is sufficient evidence to support the claim that the course was useful

b). For unpaired data, we use the independent sample t-test

$$t= {(\bar{x} _1 - \bar{x} _2) - (\mu_1 - \mu_2) \over \sqrt {{S_1 ^2 \over n_1} + {S_2 ^2 \over n_2}} }$$

$\bar {x} _1 = {\sum x_1 \over n} ={653 \over 12} =54.42$

$\bar {x}_2 ={752 \over 12}=62.67$

$S_1^2 = {\sum x^2 - {{(\sum x)} ^2 \over n_1} \over n_1 - 1} = {37723 - {{653}^2 \over 12} \over 12}$

$=198.99$

$S_2^2= {50270 - {{752}^2 \over 12} \over 12} =285.88$

$$now, \space t= {54.42-62.67 \over \sqrt {{198.99 \over 12}+{285.88 \over 12}}}$$

$=-1.2978$

$t_{ \alpha, n-1}=t_{0.05,11} =1.796$

Since t-calculated < t-critical, we fail to reject $H_0$

Therefore there is sufficient evidence to support the claim that the course was useful