Question

The data in the table is the number of absences for 7 students and their corresponding grade. Number of Absences 1 2 3 4 6 7 8 Grade 5 4.5 4 3.5 2.5 2 1

Step 1 of 3: Calculate the correlation coefficient, r. Round your answer to six decimal places. Step 2 of 3: Determine if r is statistically significant at the 0.01 level. Step 3 of 3: Calculate the coefficient of determination, r2 . Round your answer to three decimal places.

Answer 1

Number of absences (Xi)	Grade (Yi)
1 2 3 4 6 7 8	5 4.5 4 3.5 2.5 2 1	11.7649 5.9049 2.0449 0.1849 2.4649 6.6049 12.7449	3.2041 1.6641 0.6241 0.0841 0.5041 1.4641 4.8841	-6.1397 -3.1347 -1.1297 -0.1247 -1.1147 -3.1097 -7.8897
= 31	= 22.5	= 41.71	= 12.43	= -22.64

1) The correlation coefficient is given by -

  

= -0.994436

2) Null hypothesis : The population correlation coefficient is not significantly different from zero.

There is no linear relationship between the number of absences and grade.

Alternative hypothesis : The population correlation coefficient is significantly different from zero.

There is a linear relationship between the number of absences and grade.

The test statistic is given by -

Sample size = n = 7

Degrees of freedom = n-2 =7-2 = 5

Sample correlation coefficient = r = -0.994

Putting all these values in formula of test statistics, we get,

= -21.11

The P value is given by -

P(|t|   21.11) = P(t -21.11) + P(t 21.11)

~ 0.000

The probability of t 21.11 is the area under the t curve with 5 degrees of freedom on the left side of t = 21.11 and can be obtained from the t table by finding the probability corresponding to the 5 degrees of freedom and t value close to 21.11 and is found to be close to 0.

Since, P value < 0.01, so, we reject the null hypothesis. Hence, the correlation coefficient is significant.

There is relationship between the number of absences and grade.

3) Coefficient of determination,

=  0.989