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Question 1 (6) A study is conducted to examine the effects of educational level on the productivity performance level of a group of employees in a company. The data is shown in Table Q1(b) below. You are required to carry out a chi-square test to determine whether there is sufficient evidence to conclude that the operators' educational level and productivity performance are two independent variables. Educational level Productivity Performance "O' levels or lower 'A' levels Poly & beyond Below average Average Good 15 19 Table 010) (1) State the hypotheses. (2 marks) (ii) Compute the expected frequencies. (10 marks) (iii) At a = 0.01, determine the value of the test statistic. Analyse the results and draw your conclusions. (8 marks)

Answer 1

i )

Hypothesis :

H₀ : The 'Educational level' and 'Productivity performance' are two independent variables.

H₁: The 'Educational level' and 'Productivity performance' are two dependent variables.

ii)

Expected frequencies :

Rix Cj Eij = N

Where , where Ri corresponds to the total sum of elements in row i

Cj corresponds to the total sum of elements in column j

The row and column total have been calculated and they are shown below :

'O' levels or lower 'A' levels / poly &beyond Total Below average Average Good Total

The table below shows the calculations to obtain the table with expected values :

Expected Values 'O' levels or lower 'A' levels / poly &beyond Total Below average 60x30 = 16.364 110 IU.Du Average 50x30 = 13.636 50x38 = 17.727 50x41 = 18.636 110 = 21.273 Good 60x41 = 22.364 110 10.UUU Total 50 60

iii) Test statistic

Formula :

(14103

Where , Oij is observed frequencies.

Eij is Expected frequencies.

Based on the observed and expected values, the squared distances can be computed according to the following formula: (E−O)²/E

Squared Distances 'O' levels or lower 'A' levels / poly &beyond (16-13.636) Below average (14-16.364) = 0.341 = 0.41 13.636 = 0.41 16.364 = 0.341 Average (15–17.727) 17.727" = 0.42 = 0.42 (24–21.273)2 = 0.35 21.273 Good (19-18.636) 18.636 = 0,007 (22–22.364) = 0.006 22.364

So test statistic is ,

(14103

x = 0.41 +0.42 +0.007 +0.341 +0.35 +0.006

X = 1.533

Critical value :

Given significance level =$\alpha$ = 0.01

degrees of freedom( df ) = ( # of rows - 1 ) ( # columns - 1 ) = ( 3 - 1)( 2 - 1 ) = 2

So chi-square critical value is,

            

Xả,dj = 301,2 = 9.2103

{ Using Excel function , =CHIINV( $\alpha$ , df ) ,   =CHIINV( 0.01 , 2 ) = 9.2103 }

Decision about null hypothesis :

Rule : Reject null hypothesis if test statistic greater than critical value

It is observed that test statistics ( 1.533 ) is less than critical value ( 9.2103 )

So, fail to reject null hypothesis.

Conclusion :

There is sufficient evidence to conclude that the operators' education level and productivity performance are two independent variables.