Answer to Question #127384 in Statistics and Probability for Yeap Kai Wei

Question

Question 4
Three types of machine A, B and C, are used by a company to produce items which are very prone to a certain type of imperfection. A random sample of 100 items from each machine showed the number of perfect and imperfect items as follows:

Condition Machine
A B C
Perfect 33 39 48
Imperfect 67 61 52

Test whether there is evidence of significant association between machine type and proneness to imperfection. Use 5% level of significance.
(10 marks)

Answer 1

The following cross tabulation have been provided. The row and column total have been calculated and they are shown below:

A B C Total

Perfect 33 39 48 120

imperfect 67 61 52 180

Total 100 100 100 300

The expected values are computed in terms of row and column totals. In fact, the formula is $E_{ij} = \frac{R_i \times C_i}{T}$ , where $R_i$

​ corresponds to the total sum of elements in row i

$C_j$corresponds to the total sum of elements in column j and T is the grand total. The table below shows the calculations to obtain the table with expected values:

Expected values A B C

Perfect $\frac{100×120}{300} ​ =40$ $\frac{100×120}{300}=40$ $\frac{100×120}{300}=40$

Imperfect $\frac{100×180}{300}=60$ $\frac{100×180}{300}=60$ $\frac{100×180}{300}=60$

Based on the observed and expected values, the squared distances can be computed according to the following formula: $\frac{(E - O)^2}{E}$ The table with squared distances is shown below:

Squared distances A B C

Perfect $\frac{(33 - 40)^2}{40}=1.225$ $\frac{(39 - 40)^2}{40}=0.025$ $\frac{(48 - 40)^2}{40}=1.6$

Imperfect $\frac{(67 - 60)^2}{60}=0.817$ $\frac{(61 - 60)^2}{60}=0.017$ $\frac{(52 - 60)^2}{60}=1.067$

The following null and alternative hypotheses need to be tested:

$H_0$ ​: The two variables are independent

$H_a$ : The two variables are dependent

This corresponds to a Chi-Square test of independence

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(2−1)×(3−1)=2,

The Chi-Squared statistic is computed as follows:

$\chi ^{2}$ $=1.225+0.817+0.025+0.017+1.6+1.067=4.75$

The corresponding p-value for the test is p = $P(\chi_{ 2}^2 >4.75)=0.093$ ,(using chi square table)

since p value is less than 0.05, alpha value It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.