Question

A college conducts both face to face and online classes that are intended to be identical. Given below are
the distributions of the two groups in terms of marks scored in a statistics examination.
Face to face Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of Students 5 10 20 25 15 15 10
On-line Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of Students 10 25 30 40 30 20 15
Test the hypothesis that the mean marks are statistically equal at the 5% level of significance and write a 5%
confidence interval estimate of the difference

Answer 1

Solution:

Face to Face Class

$$Mean, \mu_1 = {\sum f1x \over \sum f1}$$ $$= {3700 \over 100} = 37$$

$$var , \sigma_1 ^2 = {\sum f1{(x- \mu)}^2 \over \sum f1}$$ $$={26600 \over 100}= 266$$

On-line Class

$$=Mean, \mu_2 = {\sum f2x \over \sum f2}$$ $$= {6000 \over 170} = 35.2941$$

$$var , \sigma_2 ^2 = {\sum f2{(x- \mu)}^2 \over \sum f2}$$ $$= {46485.29412 \over 170} = 273.4429$$

Hypothesis Testing.

$H_0 : \mu_1 = \mu_2$ vs $H_1: \mu_1 \not = \mu_2$ at 5% level of significance

Test statistic:

$$= {\mu_1 - \mu_2 \over \sqrt{{\sigma_1^2 \over n_1} + {\sigma_2^2 \over n_2}}}$$

$$= {37 -35.2941 \over \sqrt{{266 \over 100}+ {273.4429 \over 170}}} = 0.8257$$

Z_0.025=1.96

Conclusion:

Since the calculated value of Z is less than the table value of Z, we fail to reject the null hypothesis. Therefore, the mean marks are statistically equal.

Confidence Interval.

$$=(\mu_1 - \mu_2) \pm Z_{\alpha \over2}{\sqrt{{\sigma_1^2 \over n_1} + {\sigma_2^2 \over n_2}}}$$

$$=(37-35.2941) \pm 1.96 \sqrt{{266 \over 100}+{273.4429 \over 170}}$$

$$= 1.7059 \pm 4.0494$$

Ans: $-2.3435 \le \mu_1 -\mu_2 \le 5.7553$