Question

In: Statistics and Probability

A recent study examined the effects of carbon monoxide exposure on a group of construction workers....

A recent study examined the effects of carbon monoxide exposure on a group of construction workers. The following table presents the numbers of workers who reported various symptoms, along with the shift (morning,  evening,  or night) that they worked.

Morning Shift Evening Shift Night Shift
Influenza

15

18

20
Headache

32

26

5
Weakness

16

11

6
Shortness of Breath

7

11

11
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Test the hypothesis of independence. Use the a=0.10

level of significance and the P-value method with the TI-84 Plus calculator.What do you conclude?

Part 1 of 4

Your Answer is correct

State the null and alternate hypotheses.

H0: The shift and reported symptoms ▼are independent.

H1: The shift and reported symptoms ▼are not independent.This hypothesis test is a ▼right-tailed test.

Part: 1 / 4

1 of 4 Parts Complete

Part 2 of 4

Find the P-value. Round your answer to four decimal places.

The P-value is .

Solutions

Expert Solution

a.

Null and alternative hypotheses:

The null and alternative hypotheses to test whether the experimental method is superior to conventional method are independent are as follows:

H0: the work shift and various symptoms are independent.

H­a: the work shift and various symptoms are not independent.

The formula for expected frequency is,

E = (Row total) (Column total) / Grand total.

E1=(53*70)/178=208427

The value of test statistics is obtained below:

The formula for chi-square test statistics is

Decision rule:

If p-value ≤ significance level, reject the null hypothesis.

If p-value > significance level, do not reject the null hypothesis.

p-value for the test is

Night Shift Night Shift Total Observed(0) Influenza Morning Shift 15 5 Evening Shift 18 26 Headache Weakness 16 11 Shortness of Breath 11 Total 66 42

Morning Shift Observed(0) Evening Shift Total Night Shift Influenza Headache 20.8427 19.65169 12.50562 24.77528 23.35955 14.86517 12.97753 12.23596 7.786517 Weakness Shortness of Breath 11.40449 10.75281 6.842697 Total 70 66 42 178

Observed (O) Expected Chi-square=(Observed count - Expected Count count)^2/Expected count 20.84 1.636545106 19.65 0.138549618 15 20 12.51 4.484420464 24.78 23.36 14.87 2.103648103 0.298356164 6.551237391 12.4 1.04516129 11 12.24 0.125620915 7.79 0.411309371 11.4 1.698245614 10.75 0.005813953 6.84 2.53005848 Chi - squar 21.02896647 0.033072 CHITEST(M10:M21, N10:N21) P value

Night Shift Night Shift Total Observed(0) Influenza Morning Shift 15 5 Evening Shift 18 26 Headache Weakness 16 11 Shortness of Breath 11 Total 66 42

Morning Shift Observed(0) Evening Shift Total Night Shift Influenza Headache 20.8427 19.65169 12.50562 24.77528 23.35955 14.86517 12.97753 12.23596 7.786517 Weakness Shortness of Breath 11.40449 10.75281 6.842697 Total 70 66 42 178

Observed (O) Expected Chi-square=(Observed count - Expected Count count)^2/Expected count 20.84 1.636545106 19.65 0.138549618 15 20 12.51 4.484420464 24.78 23.36 14.87 2.103648103 0.298356164 6.551237391 12.4 1.04516129 11 12.24 0.125620915 7.79 0.411309371 11.4 1.698245614 10.75 0.005813953 6.84 2.53005848 Chi - squar 21.02896647 0.033072 CHITEST(M10:M21, N10:N21) P value

We were unable to transcribe this image

We were unable to transcribe this image

Morning Shift Observed(0) Evening Shift Total Night Shift Influenza Headache 20.8427 19.65169 12.50562 24.77528 23.35955 14.86517 12.97753 12.23596 7.786517 Weakness Shortness of Breath 11.40449 10.75281 6.842697 Total 70 66 42 178

Observed (O) Expected Chi-square=(Observed count - Expected Count count)^2/Expected count 20.84 1.636545106 19.65 0.138549618 15 20 12.51 4.484420464 24.78 23.36 14.87 2.103648103 0.298356164 6.551237391 12.4 1.04516129 11 12.24 0.125620915 7.79 0.411309371 11.4 1.698245614 10.75 0.005813953 6.84 2.53005848 Chi - squar 21.02896647 0.033072 CHITEST(M10:M21, N10:N21) P value

We were unable to transcribe this image

orchestra answered 2 years ago
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