A recent study examined the effects of carbon monoxide exposure on a group of construction workers. The following table presents the number odf workers who reported various symptoms, along with the shift (morning, evening or night) that they worked.                                                         Shift                               Morning            Evening             NightInfluenza                    16                        13                    18Headache                   24                        33                      6Weakness                   11                        16                      5Shortness of Breath     7                           9                       9a. Compute the expected frequencies under the null hypothesis.b. Compute the value of chi-square statistic.c. Test the hypothesis of independence. Use the a=0.05 level of significance. What do you conclude?

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Asked Dec 1, 2019
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A recent study examined the effects of carbon monoxide exposure on a group of construction workers. The following table presents the number odf workers who reported various symptoms, along with the shift (morning, evening or night) that they worked.

Shift

Morning            Evening             Night

Influenza                    16                        13                    18

Headache                   24                        33                      6

Weakness                   11                        16                      5

Shortness of Breath     7                           9                       9

a. Compute the expected frequencies under the null hypothesis.

b. Compute the value of chi-square statistic.

c. Test the hypothesis of independence. Use the a=0.05 level of significance. What do you conclude?

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Step 1

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.

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

Appropriate test:

The appropriate test for the given scenario is chi-square test for independence.

Decision rule:

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

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

The expected frequency is obtained below:

Step 2

The formula for expected frequency is,

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

Step 3

b.

The value of test statistics is obtained below:

The formula for chi-square test statistics is

= (Observ...

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