Explanation Given info: The data represents the deaths that involve a bicycle, recorded on each day of the week. Calculation: The test hypotheses are given below: Null hypothesis: H 0 : The bicycle facilities have equal frequency on each day. Alternative hypothesis: H 1 : The bicycle facilities do not have equal frequency on each day. Decision rule: If χ 0 2 > χ α 2 , then reject the null hypothesis. The requirements that must be satisfied to perform a goodness of fit test are as follows: All the expected frequencies are greater than or equal to 1. That is, all E i ≥ 1 where E i represents the expected counts of category i. Not more than 20% of the expected frequencies are less than 5. The expected count for each category is obtained by multiplying the given proportion value of each category with the sum of the observed values. That is, the expected value is 28.57 for all categories. The requirements for goodness of fit test are satisfied. The test statistic for goodness of fit test is given below: χ 0 2 = ∑ ( O i − E i ) 2 E i The test statistic can be obtained as follows: Day of week O i E i ( O i − E i ) 2 ( O i − E i ) 2 E i Sunday 16 28.57 158.0049 5.530448 Monday 35 28

MyLab Statistics with Pearson eText -- Standalone Access Card -- for Fundamentals of Statistics

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ISBN: 9780134743295

Author: Michael III Sullivan

Publisher: PEARSON

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Chapter 12.1, Problem 20AYU

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To test: Whether the bicycle facilities have equal frequency on each day at α=0.05 level of significance.

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Chapter 12.1, Problem 19AYU

Chapter 12.1, Problem 21AYU

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MyLab Statistics with Pearson eText -- Standalone Access Card -- for Fundamentals of Statistics

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To test: Whether the bicycle facilities have equal frequency on each day at α = 0.05 level of significance.

Solution Summary: The author explains that there is enough evidence to infer that bicycle facilities do not have equal frequency on each day.

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To test: Whether the bicycle facilities have equal frequency on each day at α=0.05 level of significance.

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Chapter 12 Solutions