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Get Full Access to Elementary Statistics - 12 Edition - Chapter 11.3 - Problem 1 re
Get Full Access to Elementary Statistics - 12 Edition - Chapter 11.3 - Problem 1 re

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# Solved: Auto Fatalities The table below lists auto

ISBN: 9780321836960 18

## Solution for problem 1 RE Chapter 11.3

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition

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Problem 1 RE

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

Step-by-Step Solution:

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.

Step 2 of 2

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