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Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola
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Solved: Auto Fatalities The table below lists auto

Chapter 11, Problem 1 RE

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QUESTION:

Auto Fatalities The table below lists auto fatalities by day of the week for a recent year (based on data from the Federal Highway Administration).

Minitab results are also shown. Use a 0.01 significance level to test the claim that auto fatalities occur on the different days of the week with the same frequency. Can you provide an explanation for the results?

MINITAB

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QUESTION:

Auto Fatalities The table below lists auto fatalities by day of the week for a recent year (based on data from the Federal Highway Administration).

Minitab results are also shown. Use a 0.01 significance level to test the claim that auto fatalities occur on the different days of the week with the same frequency. Can you provide an explanation for the results?

MINITAB

ANSWER:

Answer:

Step 1 of 2

The expected values for each category, Ei , could be determined. With these observed and expected numbers of cases, the hypotheses can be written as

H0 : the claim that auto fatalities occur on the different days of the week with the same frequency

H1 : the claim that auto fatalities occur on the different days of the week with the different frequency.

Use a 0.05 significance level

SL. No.

O

p = 1/7

E = np

(O - E )2

(O - E)2/E

1

3797

0.142857

4399.571

363092.3

82.52902

2

3615

0.142857

4399.571

615552.3

139.9119

3

3724

0.142857

4399.571

456396.8

103.7366

4

4004

0.142857

4399.571

156476.8

35.56636

5

4867

0.142857

4399.571

218489.5

49.66153

6

5786

0.142857

4399.571

1922184

436.9026

7

5004

0.142857

4399.571

365333.9

83.03852

Sum

30797

1

30797

4097526

931.3466

The first step in conducting the significance test is to compute the expected frequency for each outcome given that the null hypothesis is true. For example, the expected frequency of a "1" is 7 since the probability of a "1" coming up is 1/7 and there were a total of 30797.

Expected frequency (E) = np

                                       = 30797(1/7)

                                       = 4399.571

The calculation continues as follows. Letting E be the expected frequency of an outcome and O be the observed frequency of that outcome.

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