Question

Dr. Sammy Statsnerd surveyed her statistics students (N=60) to see if there was a relationship between the number of total courses the students were taking this semester and the level of stress that they felt about her statistics course. The results of the survey are below (30 points):

Level of Stress about Statistics	Number of Courses
	Four	Five	Six
Low	9	6	1
Moderate	8	15	2
High	4	9	6
Total	21	30	9

Assess the 3 properties of bivariate relationships and justify your answers.

1. Existence

2. Strength (use maximum difference)

3. Direction

Answer 1

Dr. Sammy Statsnerd surveyed her statistics students (N=60) to see if there was a relationship between the number of total courses the students were taking this semester and the level of stress that they felt about her statistics course.

Here we have to test the hypothesis that,

H0 : The variables number of courses and level of Stress about Statistics are independent.

or there is no relationship between number of courses and level of Stress about Statistics.

H1 : The variables number of courses and level of Stress about Statistics are dependent.

or there is relationship between number of courses and level of Stress about Statistics.

Assume alpha = level of significance = 0.05

N = 60

This is Chi square test of independence.

The test statistic follows X²- distribution with (R-1)(C-1) degrees of freedoms.

where R = number of rows = 3

C = number of columns = 3

The test statistic is,

where O is observed frequency

E is expected frequency

We can do this test in MINITAB.

steps :

ENTER data into MINITAB sheet --> Tables --> Chi square test for association --> Summarized data in a two way table --> Select all the columns together --> Ok

————— 17-07-2019 20:08:55 ————————————————————

Welcome to Minitab, press F1 for help.

Chi-Square Test for Association: Worksheet rows, Worksheet columns

Rows: Worksheet rows Columns: Worksheet columns

C1 C2 C3 All

1 9 6 1 16
5.600 8.000 2.400

2 8 15 2 25
8.750 12.500 3.750

3 4 9 6 19
6.650 9.500 2.850

All 21 30 9 60

Cell Contents: Count
Expected count

Pearson Chi-Square = 9.326, DF = 4, P-Value = 0.053
Likelihood Ratio Chi-Square = 8.752, DF = 4, P-Value = 0.068

* NOTE * 3 cells with expected counts less than 5

Test statistic = 9.326

P-value = 0.053

DF = 4

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : The variables number of courses and level of Stress about Statistics are independent.