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Each observation in a random sample of 100 bicycle accidents resulting in death was classified according to the day of the week on which the accident occurred. Data consistent with information given on the web site www.highwaysafety.com are given in the following table $$ \begin{array}{lc} \text { Day of Week } & \text { Frequency } \\ \hline \text { Sunday } & 14 \\ \text { Monday } & 13 \\ \text { Tuesday } & 12 \\ \text { Wednesday } & 15 \\ \text { Thursday } & 14 \\ \text { Friday } & 17 \\ \text { Saturday } & 15 \\ \hline \end{array} $$ Based on these data, is it reasonable to conclude that the proportion of accidents is not the same for all days of the week? Use \(\alpha=.05\).

Short Answer

Expert verified
The decision to whether reject or not reject the null hypothesis will be based on comparing the computed chi-square statistic and the critical value. A definite answer would require calculating the actual chi-square statistic which is beyond the given information in the problem.

Step by step solution

01

Define the hypotheses

The null hypothesis (\(H_0\)) is that there is no difference in the proportion of accidents across the days of the week. The alternative hypothesis (\(H_a\)) is that there is a difference in the proportion of accidents across the days of the week.
02

Calculate the expected frequencies

Since under the null hypothesis, the accidents are evenly distributed across the week, the expected frequency for each day is the total number of accidents divided by 7 (number of days). So, the expected frequency is \(100/7 \approx 14.29\).
03

Compute the Chi-Square Test Statistic

The chi-square test statistic is calculated by summing the squared difference between observed and expected frequencies divided by expected frequencies for each category. Hence, \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\] where \(O_i\) represents the observed frequency and \(E_i\) represents the expected frequency.
04

Determine the critical value

For a chi-square distribution with 6 degrees of freedom (7 days - 1) and \(\alpha = .05\), the critical value can be found from the chi-square distribution table to be approximately 12.59.
05

Make a decision

If the calculated Chi-Square statistic is greater than the critical value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Understanding the essence of hypothesis testing is crucial when you are trying to make inferences from data. At its core, hypothesis testing is a method that allows researchers to make decisions about population parameters based on sample statistics. Think of it as a detective's tool in the realm of statistics, used to check the validity of assumptions or claims.

In the context of a Chi-Square Test, as in our bicycle accidents example, hypothesis testing starts by formulating two opposing statements. The null hypothesis (H_0) suggests there is no effect or no difference, implying that any variation in the data is due to chance. For this exercise, H_0 posits that the proportion of accidents is uniform across all days of the week. Alternatively, the alternate hypothesis (H_a) represents what you suspect might be true - here, that the accident rates vary by day.

The process involves calculating a test statistic that measures how much the observed data deviates from what's expected under the null hypothesis. By comparing this test statistic to a critical value, which comes from a reference distribution considering the significance level and degrees of freedom, researchers can decide whether to reject H_0.
Expected Frequency
When we delve into Chi-Square Tests, the concept of expected frequency occupies center stage. Expected frequency is an estimate of how many times a certain event is predicted to occur in a dataset, assuming the null hypothesis is true. In a perfect world with no influencing factors, the expected frequency is what we would anticipate to see.

To compute it, we divide the total count of observations by the number of different outcomes or categories. In our case, we distributed 100 bicycle accident incidents evenly over the seven days, resulting in an expected frequency of approximately 14.29 accidents per day. It's worth noting that in practice, this is often where errors can occur if the calculation is not done carefully, especially when dealing with larger datasets or more categories.
Chi-Square Test Statistic
The chi-square test statistic is the heart of the Chi-Square Test, a number representing how much our observed frequencies deviate from the expected frequencies. It's calculated by taking each category, finding the difference between observed (O_i) and expected (E_i) frequencies, squaring that difference, dividing by the expected frequency, and summing these values for all categories.

\[\begin{equation}chi^2 = \begin{array}{lc} \text{Day of Week} & \text{Frequency} \table sim \text{Sum} & \text{ } \text{Sunday} & \text { \text{ } } \text{Monday} & \text{ } \text{Tuesday} & \text{ } \text{Wednesday} & \text{ } \text{Thursday} & \text{ } \text{Friday} & \text{ } \text{Saturday} & \text{ } \table sim \text{ } & \frac{(O_i - E_i)^2}{E_i}n\end{array}\end{equation}\]
This gives us a measure of the overall discrepancy between what we observe and what we would expect by mere chance if the null hypothesis were true. A higher chi-square value suggests a greater deviation and thus casts more doubt on the null hypothesis.
Degrees of Freedom
Degrees of freedom is a statistical concept tied heavily to the accuracy and reliability of various test statistics, including the Chi-Square Test. Think of it as the number of independent pieces of information you have in your data, which you're free to vary when estimating statistical parameters.

The degrees of freedom for a Chi-Square Test are determined by the number of categories minus one (number of days - 1). The reason for subtracting one is to account for the constraint that the total observed frequencies must equal the total expected frequencies. In our bicycle accident example, we have seven categories (days of the week) which results in six degrees of freedom (7 - 1 = 6). Knowing the degrees of freedom alongside the significance level (\[\begin{equation}alpha = .05\end{equation}\]) is essential to find the critical value which is a cut-off point determining whether or not to reject the null hypothesis.

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Most popular questions from this chapter

The paper “Overweight Among Low-Income Preschool Children Associated with the Consumption of Sweet Drinks" (Pediatrics [2005]: 223-229) described a study of children who were underweight or normal weight at age 2. Children in the sample were classified according to the number of sweet drinks consumed per day and whether or not the child was overweight one year after the study began. Is there evidence of an association between whether or not children are overweight after one year and the number of sweet drinks consumed? Assume that it is reasonable to regard the sample of children in this study as representative of 2 - to 3 -year-old children and then test the appropriate hypotheses using a \(.05\) significance level. $$ \begin{array}{c|rc} \text { Number of Sweet } & \text { Overweight? } \\ \begin{array}{c} \text { Drinks Consumed } \\ \text { per Day } \end{array} & \text { Yes } & \text { No } \\ \hline 0 & 22 & 930 \\ 1 & 73 & 2074 \\ 2 & 56 & 1681 \\ 3 \text { or More } & 102 & 3390 \\ \hline \end{array} $$

A particular state university system has six campuses. On each campus, a random sample of students will be selected, and each student will be categorized with respect to political philosophy as liberal, moderate, or conservative. The null hypothesis of interest is that the proportion of students falling in these three categories is the same at all six campuses. a. On how many degrees of freedom will the resulting \(X^{2}\) test be based? b. How does your answer in Part (a) change if there are seven campuses rather than six? c. How does your answer in Part (a) change if there are four rather than three categories for political philosophy?

The press release titled "Nap Time" (pewresearch.org, July 2009 ) described results from a nationally representative survey of 1488 adult Americans. The survey asked several demographic questions (such as gender, age, and income) and also included a question asking respondents if they had taken a nap in the past 24 hours. The press release stated that \(38 \%\) of the men surveyed and \(31 \%\) of the women surveyed reported that they had napped in the past 24 hours. For purposes of this exercise, suppose that men and women were equally represented in the sample. a. Use the given information to fill in observed cell counts for the following table: $$ \begin{array}{l|ll} & \text { Napped } & \text { Did Not Nap } & \text { Row Total } \\ \hline \text { Men } & & & 744 \\ \text { Women } & & & 744 \\ \hline \end{array} $$ b. Use the data in the table from Part (a) to carry out a hypothesis test to determine if there is an association between gender and napping. c. The press release states that more men than women nap. Although this is true for the people in the sample, based on the result of your test in Part (b), is it reasonable to conclude that this holds for adult Americans in general? Explain.

Can people tell the difference between a female nose and a male nose? This important (?) research question was examined in the article "You Can Tell by the Nose: Judging Sex from an Isolated Facial Feature" (Perception [1995]: 969-973). Eight Caucasian males and eight Caucasian females posed for nose photos. The article states that none of the volunteers wore nose studs or had prominent nasal hair. Each person placed a black Lycra tube over his or her head in such a way that only the nose protruded through a hole in the material. Photos were then taken from three different angles: front view, three-quarter view, and profile. These photos were shown to a sample of undergraduate students. Each student in the sample was shown one of the nose photos and asked whether it was a photo of a male or a female; and the response was classified as either correct or incorrect. The accompanying table was constructed using summary values reported in the article. Is there evidence that the proportion of correct sex identifications differs for the three different nose views? $$ \begin{array}{l|ccc} & & \text { View } \\ \hline & & & \text { Three- } \\ \text { Sex ID } & \text { Front } & \text { Profile } & \text { Quarter } \\ \hline \text { Correct } & 23 & 26 & 29 \\ \text { Incorrect } & 17 & 14 & 11 \\ \hline \end{array} $$

An article about the California lottery that appeared in the San Luis Obispo Tribune (December 15, 1999) gave the following information on the age distribution of adults in California: \(35 \%\) are between 18 and 34 years old, \(51 \%\) are between 35 and 64 years old, and \(14 \%\) are 65 years old or older. The article also gave information on the age distribution of those who purchase lottery tickets. The following table is consistent with the values given in the article: $$ \begin{array}{lc} \text { Age of Purchaser } & \text { Frequency } \\ \hline 18-34 & 36 \\ 35-64 & 130 \\ 65 \text { and over } & 34 \\ \hline \end{array} $$ Suppose that the data resulted from a random sample of 200 lottery ticket purchasers. Based on these sample data, is it reasonable to conclude that one or more of these three age groups buys a disproportionate share of lottery tickets? Use a chi-square goodness-of-fit test with \(\alpha=.05\).

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