Answer to Question #262212 in Statistics and Probability for christabel

Question

0. After receiving a failing grade in ECON 306, a student complained to the professor that the examination was unfair. The student claims that ‘almost everyone’ did worse in the examination than in the Interim Assessment (IA), and that the average mark in the examination was much lower than the IA. The Professor then asked 8 students who were in his office about the marks they obtained in IA and the examination. The marks were as follows: Student s Adwoa Kofi Afua Ama Kwame Abena Yaw Esi IA 59 69 54 77 70 59 85 92 Exams 72 44 52 81 88 94 78 86 Test the claim of the student if the marks are normally distributed at 5% level of significance

Answer 1

Our sample is dependent and we should consider using matched/paired t-test for our analysis.

To perform this test, we begin by finding the difference between pairs of observation for each student. We shall use the notation $d,$ to represent the differences and use it to make a decision.

The hypotheses to be tested are,

$H_0:\mu_d=0$

$Against$

$H_1:\mu_d\not=0$, where $d$ is the difference between each pair of observations.

To test these hypotheses, we first determine the mean of the differences$(\bar{d})$ given by,

$\bar{d}=(\sum(d))/n$, where $n=8.$

Now,

$\sum(d)=-30$

Therefore,

$\bar{d}=-30/8=-3.75$

Variance of the differences is given by,

$V(d)=\sum(d-\bar{d})^2/(n-1)$

$V(d)=2335.5/7=333.6429(4\space d\space p)$.

The standard deviation $SD(d)$ is given by,

$SD(d)=\sqrt{V(d)}=\sqrt{333.6429}=18.26589$.

The test statistic is given by,

$t^*=\bar{d}/(SD(d)/\sqrt n)$

$t^*=-3.75/(18.26589/\sqrt {8})$

$t^*=-3.75/6.457968=-0.58068.$

$t^*$ is compared with the student's t distribution with $(n-1)=8-1=7$ degrees of freedom at $\alpha=0.05$.

The table value $t_{\alpha/2,7}=t_{0.05/2,7}=t_{0.025,7}=2.365$ and reject the null hypothesis if $|t^*|\gt t_{0.025,7}.$

Since $|t|=0.58068\lt t_{0.025,7}=2.365,$ we fail to reject the null hypothesis and conclude that sufficient evidence exist to show that the marks of the students are normally distributed at 5% level of significance.