Question

A student counsellor is concerned about students’ stress levels. S/he wants to understand whether students in different courses have similar stress levels.

The following table shows a random sample of students in different courses and their stress levels (n= 178178). Use the table to answer the questions which follow.

	Stress Levels by Enrolment
Enrolment	Low	Normal	High
Arts	33	39	26
Science	23	31	26

1. (1 mark) What proportion of the students in the study have normal stress level? (3dp) Answer

2. (1 mark) What proportion of the Science students have low stress level? (3dp) Answer

3. (1 mark) What proportion of the students have high stress levels were Arts students? (3dp) Answer

The student counsellor would like to answer the following research question: Do students in different courses experience different stress levels? S/he sets up the following null and alternative hypotheses to answer the research question.

?0:H0: There is no association between enrolled course and level of stress. ?1:H1: There is an association between enrolled course and level of stress.

4. (1 mark) What is the expected value for Arts students who have low stress levels? (3dp) Answer

5. The absolute value of the test statistic is equal to (type your answer with 3 dp) Answer

6. (1 mark) The degrees of freedom is equal to (type your answer as an integer) Answer

7. (1 mark) The p-value is smaller than 0.05 Answertruefalse

8. (1 mark) We decide that Answerp -p value is greater than 0.05, therefore we do not reject the null hypotesisp -p value is less than 0.05, therefore we reject the null hypotesis

9. (2 mark) Our conclusion is that Answer -There is no association between enrolled course and level of stress. -There is an association between enrolled course and level of stress.

10. We are 95% confident that proportion of students with normal stress level for Science is between Answer (lower limit) and Answer (upper limit). Type your answers with 3dp.

Answer 1

Let us first calculate row totals and column total s

	Low	Normal	High	Total
Arts	33	39	26	98
Science	23	31	26	80
Total	56	70	52	178

1) Proportion of students have Normal stress level = 70 /178 = 0.393

2) Proportion of science students have low stress level = 23/ 80 = 0.288

Note : There are 80 science students , we can see that out of 80 23 have low stress level.

3) Proportion of having High stress level who are Arts students = 26 / 52

Note : There are 52 students with high stress levels. Out of 52 , 26 are Arts students .

4) Expected value of arts students who have low stress level = row total *column total / grand total

= 98*56 / 178

= 30.831

5) Test statistic

where Oi : observed frequency

Ei : expected frequency

The contingency table ( of expected and observed frequency)

The expected frequencies are in the parenthesis

	Low	Normal	High	Total
Arts	33(30.831)	39(38.539)	26(28.629)	98
Science	23(25.169)	31(31.461)	26(23.371)	80
Total	56	70	52	178

Calculation of Chi square

	Oi	Ei	(Oi-Ei)^2/Ei
	33	30.381	0.2258
	39	38.539	0.0055
	26	28.629	0.2414
	23	25.169	0.1869
	31	31.461	0.0068
	26	23.371	0.2957
sum			0.9621

Thus

0.962

6) degrees of freedom = (3-1)*(2-1) = 2

Note : degrees of freedom = ( number of rows -1) *( number of columns-1)

7) P value for 0.962 with df = 2 is

P value = 0.618

Since P value > 0.05

Answer is : FALSE

Note : Excel formula for P value "=CHISQ.DIST.RT(0.962,2)"

8)We decide that

Answer : P value is greater than 0.05 , thererefore we do not reject the null hypothesis .

Our conlcusion is

Answer : there is no association between enrolled course and level of stress.

9) 95% confidence interval for population proportion is

where

31/ 80 = 0.3875 ( sample proportion of science students who have normal stress level )

For 95% confidence , zc = 1.96 ( from z table)

Thus , 95% confidence interval is

= (0.281 , 0.494 )

Answer We are 95% confident that the proportion of students with normal stress level for Science is between 0.281 and 0.494 .