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= 214214). Use the table to answer the questions which follow.

	Stress Levels by Enrolment
Enrolment	Low	Normal	High
Human Sciences	40	43	46
Science	25	38	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 Science students have low stress level? (3dp) Answer

3. (1 mark) What proportion of the students have high stress levels were Human Sciences 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.

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

4. (1 mark) What is the expected value for Human Sciences 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 Science is between Answer (lower limit) and Answer (upper limit). Type your answers with 3dp.

Answer 1

Enrolment	Low	Normal	High	total
Human Sciences	40	43	46	129
Science	25	38	22	85
total	65	81	68	214

  proportion of the students in the study have normal stress level = 81/214= 0.379

proportion of the Science students have low stress level = 25/85 = 0.294

proportion of the students have high stress levels were Human Sciences students= 46/68= 0.676

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H0: There is no association between enrolled course and level of stress.

H1:H1: There is an association between enrolled course and level of stress.

Chi-Square Test of independence
	Observed Frequencies
Enrolment	Low	Normal	High				Total
Human Sciences	40	43	46				129
Science	25	38	22				85
Total	65	81	68				214
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	Low	Normal	High				Total
Human Sciences	65*129/214=39.182	81*129/214=48.827	68*129/214=40.991				129
Science	65*85/214=25.818	81*85/214=32.173	68*85/214=27.009				85
Total	65	81	68				214
	(fo-fe)^2/fe
Human Sciences	0.0171	0.6954	0.612
Science	0.0259	1.0554	0.929

expected value for Human Sciences students who have low stress level= 65*129/214=39.182

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   3.3350
Number of Rows =   2
Number of Columns =   3
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 3- 1 ) =   2

The p-value is smaller than 0.05= FALSE

We decide that Answerp value is greater than 0.05, therefore we do not reject the null hypotesis

Our conclusion is that AnswerThere is no association between enrolled course and level of stress

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Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   38          
Sample Size,   n =    214          
                  
Sample Proportion ,    p̂ = x/n =    0.1776          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.026123          
margin of error , E = Z*SE =    1.960   *   0.02612   =   0.0512
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.17757   -   0.05120   =   0.1264
Interval Upper Limit = p̂ + E =   0.17757   +   0.05120   =   0.2288
                  
95%   confidence interval is (   0.126   < p <    0.229   )