Question

0.7
In an industry, 200 workers, employed for a specific job, were classified according to their pertama ad
tramag verewed / not received to test independence of a specific training and performance the data is tien
below
Performance
Not good
Good
Total
Traines
Untrained
24
120
Total
80
120
Green Chawuare at level of Agnsficance for 1 degee of freedom is 3.184​

Answer 1

Answer:

check in solutions

of science book

thank you

Answer 2

The correct question: In an industry, 200 workers, employed for a specific job, were classified according to their performance and training received/ not received to test independence of a specific training and performance. The data is summarised as follows.

Use Chi - Square test of independence at 5% level of significance and write your conclusion.

The correct answer is given below.

Given:

The data about the performance of workers are provided in table 1.

To Find:

We have to find out the result of Chi - Square test of independence at 5% level of significance.

The basic format for reporting a chi-square test result is:

X² (degrees of freedom, N = sample size) = chi-square statistic value, p = p value.

Solution:

The Chi-Square is denoted by X². The chi-square formula is:

$X^{2} = \frac{Σ(O_{i}-E_{i})^{2} }{E_{i} }$.

where $O_{i}$ is the observed value (actual value) and $E_{i}$ is the expected value.

Using the given data,

$X^{2} = \frac{(100-90)^{2} }{90} + \frac{(50-60)^{2} }{60} + \frac{(20-30)^{2} }{30} + \frac{(30-20)^{2} }{20}$ï

$X^{2} = 11.111$

Here, P-value equals 0.001392. The p value < .05 indicates that the chance of type1 error (rejecting a correct hypothesis) is small~ 0.001392 (0.14%).

The Chi-square test result is:

X² (1, N = 200) = chi-11.111, p = 0.001392.

Hence, the result of Chi - Square test of independence at 5% level of significance is $X^{2} = 11.111$.