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= 200200). Use the table to answer the questions which follow.

	Stress Levels by Enrolment
Enrolment	Low	Normal	High
Social Science	40	42	43
Engineering	22	31	22

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

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

3. (1 mark) What proportion of the students have high stress levels were Social Science 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 Social Science 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 value is greater than 0.05, therefore we do not reject the null hypotesisp value is less than 0.05, therefore we reject the null hypotesis

9. (2 mark) Our conclusion is that AnswerThere 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 Engineering is between  Answer (lower limit) and  Answer (upper limit). Type your answers with 3dp.

Answer 1

Results
	low	Normal	High			Row Totals
Social Science	40  (38.75)  [0.04]	42  (45.62)  [0.29]	43  (40.62)  [0.14]			125
Engineering	22  (23.25)  [0.07]	31  (27.38)  [0.48]	22  (24.38)  [0.23]			75
Column Totals	62	73	65			200  (Grand Total)

1)

P(normal) = 75/200 =   0.375
2)

P(low |arts) = 22/62 = 0.3548
3)

P(SS|high) = 43/65=   0.6615

Expected frequency of a cell = sum of row*sum of column / total sum                      

4) 62*125/200 = 38.75

5)

Results
	low	Normal	High			Row Totals
Social Science	40  (38.75)  [0.04]	42  (45.62)  [0.29]	43  (40.62)  [0.14]			125
Engineering	22  (23.25)  [0.07]	31  (27.38)  [0.48]	22  (24.38)  [0.23]			75
Column Totals	62	73	65			200  (Grand Total)

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe = 1.2458

The p-value is .536381.

6)

Number of Rows =   2
Number of Columns =   3
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 3- 1 ) =   2

7)

yes

8)

null hypothesis is rejected

9)

There is an association between two variable

10)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x = 73    
Sample Size,   n = 75   
                  
Sample Proportion ,    p̂ = x/n = 0.9733
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] = 0.0003465   
margin of error , E = Z*SE =    1.960   * 0.0003465 = 0.000679
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.9733   -   0.000679   = 0.9726
Interval Upper Limit = p̂ + E =   0.9733   +   0.000679   = 0.9739
                 
95%   confidence interval is (   0.973 < p < 0.974 )

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