Answer to Question #344317 in Statistics and Probability for Mika

Question

The following table gives the recorded grades for 10 students on a midterm test and the final examination in a statistics course.

Student

Midterm Test

Final Examination

1

84

73

2

98

63

3

91

87

4

72

66

5

86

78

6

93

78

7

80

91

8

9

92

88

10

87

77

a. Calculate the rank correlation coefficient.

b. Test the hypothesis @ .05 level of significance.

Answer 1

a. the rank correlation coefficient:

To find rank correlation coefficient we have to rank the two data sets. Ranking is achieved by giving the ranking '1' to the biggest number in a column, '2' to the second biggest value and so on. The smallest value in the column will get the lowest ranking. This should be done for both sets of measurements. Tied scores are given the mean (average) rank. Then we find the difference in the ranks (d): This is the difference between the ranks of the two values on each row of the table. 

$\rho=1-\frac{6\sum{d_i^2}}{n(n^2-1)}=\\=1-\frac{6*122}{10(100-1)}=1-0.74=0.26$

So, we have  little positive correlation between ranks.

b.

$H_0: \mu_d=0\\ H_a:\mu_d\not=0$ (Claim)

d=(score before-score after)

$\=d=\frac{\sum{d}}{n}=82/10=8.2$

$s_d=\sqrt{\frac{n(\sum{d^2})-(\sum{d})^2}{n(n-1)}}=\sqrt{\frac{10(1924)-6724}{90}}=11.8$

We can find critical value $t_0$ using t-table (in this problem $\alpha=0.05, df=10-1=9$ ):

$t_0=2.262$

The standardized test statistic is:

$t=\frac{\=d-\mu_d}{s_d/\sqrt{n}}=\frac{8.2-0}{11.8/\sqrt{10}}=0.219$

Since $t<t_0 (0.219<2.262)$, $H_0$ is not rejected.