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Chapter 12: Problem 29

A survey was conducted in the San Francisco Bay area in which each participating individual was classified according to the type of vehicle used most often and city of residence. A subset of the resulting data are given in the accompanying table (The Relationship of Vehicle Type Choice to Personality. Lifestyle, Attitudinal and Demographic Variables, Technical Report UCD-ITS-RRO2-06, DaimlerCrysler Corp., 2002). $$ \begin{array}{l|crc} & & \text { City } \\ \hline \begin{array}{l} \text { Vehide } \\ \text { Type } \end{array} & \text { Concord } & \begin{array}{c} \text { Pleasant } \\ \text { Hills } \end{array} & \begin{array}{c} \text { North San } \\ \text { Francisco } \end{array} \\ \hline \text { Small } & 68 & 83 & 221 \\ \text { Compact } & 63 & 68 & 106 \\ \text { Midsize } & 88 & 123 & 142 \\ \text { Large } & 24 & 18 & 11 \\ \hline \end{array} $$ Do the data provide convincing evidence of an association between city of residence and vehicle type? Use a significance level of \(.05 .\) You may assume that it is reasonable to regard the sample as a random sample of Bay area residents.

Short Answer

Expert verified
Short answer will depend on the computed test statistic and the critical value for df=6 and alpha=0.05. If the test statistic is larger than the critical value, then the conclusion would be: 'With a significance level of 0.05, we reject the null hypothesis. There is convincing evidence to suggest an association between city of residence and vehicle type.' If the test statistic is not larger than the critical value, the conclusion would be: 'With a significance level of 0.05, we do not reject the null hypothesis. There is not enough evidence to suggest an association between city of residence and vehicle type.'

Step by step solution

01

State the hypotheses

The null hypothesis (H0) is that the vehicle type is independent of the city of residence. The alternative hypothesis (Ha) is that the vehicle type is not independent of the city of residence.
02

Calculate the observed and expected frequencies

First calculate the row, column and total sums of the data. Then, calculate the expected frequency for each cell using the formula: (row total * column total) / grand total . For instance, to calculate the expected frequency for 'Small' in 'Concord', use: (68+83+221) * (68+63+88+24) / (grand total). Repeat this for all cells.
03

Compute chi-square test statistic

The chi-square statistic is calculated by summing the squared difference between observed and expected frequencies divided by expected frequency across all cells. Use the formula: \(\chi^{2} = \sum\left[\frac{(O_{i}-E_{i})^{2}}{E_{i}}\right]\). 'O' denotes observed frequency and 'E' denotes expected frequency.
04

Identify the degrees of freedom

Degrees of freedom (df) for a chi-square test of independence is calculated as: (number of rows - 1) * (number of columns - 1). We have four types of vehicles and three cities, so df = (4-1)*(3-1) = 6.
05

Find the critical value and make a conclusion

Using the chi-square distribution table with df=6 and a significance level of 0.05, get the critical value. If the test statistic calculated in Step 3 is greater than this critical value, reject H0. Otherwise, do not reject H0. Based on this, conclude whether there's evidence to suggest an association between city of residence and vehicle type.

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Key Concepts

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

Hypothesis Testing
In statistics, hypothesis testing is a method used to decide if there is enough evidence to reject a hypothesis. This involves two main types of hypotheses. The first is the **null hypothesis (H0)**, which suggests that there is no relationship or difference between two or more groups. It's often what you want to find evidence against. The second is the **alternative hypothesis (Ha)**, which indicates that there is a relationship or difference, opposite to what the null hypothesis suggests.
For instance, in the case of the survey across San Francisco Bay area concerning vehicle type and city of residence, the null hypothesis is that the type of vehicle someone uses is independent of the city they live in. In contrast, the alternative hypothesis states that the vehicle type is not independent of the city.
This method helps determine the statistical significance, offering insights into whether observed data likely result from chance or reflect true patterns.
Degrees of Freedom
Degrees of freedom (df) can be a bit tricky to grasp but are a key concept in many statistical tests. They refer to the number of independent values or quantities which can be assigned to a statistical distribution. In simpler terms, it's about the number of values in the final calculation of a statistic that are free to vary.
In the context of a chi-square test, degrees of freedom are calculated differently depending on the type of test. For a test of independence, used to assess the relationship between categorical variables in a table format, the formula is: \[(df) = (r - 1) \times (c - 1)\]where \(r\) represents the number of rows (in our case, the vehicle types) and \(c\) the number of columns (the cities).
In our exercise, with four vehicle types and three cities, the degrees of freedom are \((4 - 1) \times (3 - 1) = 6\). Understanding this concept helps in determining the appropriate critical value in hypothesis testing.
Observed and Expected Frequencies
The chi-square test revolves around comparing observed frequencies with expected frequencies to see if there is a significant difference. Observed frequencies are the counts we see in the data 鈥� in our exercise, they represent how many people from each city reported using each vehicle type.
To perform a chi-square test, we need to calculate the expected frequencies for each cell in the table. This is done under the assumption that there's no association between the variables, i.e., the null hypothesis is true. The formula to calculate expected frequency for a cell, say for a small vehicle in Concord, is:\[\text{Expected Frequency} = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}\]Once you have both sets of frequencies, you can compute the chi-square statistic by examining how much the observed frequencies differ from the expected ones. This step is vital to assess whether the differences between observed and expected counts are large enough to reject the null hypothesis.

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Most popular questions from this chapter

In a study to determine if hormone therapy increases risk of venous thrombosis in menopausal women, each person in a sample of 579 women who had been diagnosed with venous thrombosis was classified according to hormone use. Each woman in a sample of 2243 women who had not been diagnosed with venous thrombosis was also classified according to hormone use. Data from the study are given in the accompanying table (Journal of the American Medical Association [2004]: \(1581-1587\) ). The women in each of the two samples were selected at random from patients at a large HMO in the state of Washington. a. Is there convincing evidence that the proportions falling into each of the hormone use categories is not the same for women who have been diagnosed with venous thrombosis and those who have not? b. To what populations would it be reasonable to generalize the conclusions of Part (a)? Explain. $$ \begin{array}{l|rrc} & & \text { Current Hormone Use } \\ \hline & & & \text { Conjugated } \\ & \text { None } & \text { Esterified } & \text { Equine } \\ & & \text { Estrogen } & \text { Estrogen } \\ \hline \text { Venous Thrombosis } & 372 & 86 & 121 \\ \text { No Venous Thrombosis } & 1439 & 515 & 289 \\ \hline \end{array} $$

Drivers born under the astrological sign of Capricorn are the worst drivers in Australia, according to an article that appeared in the Australian newspaper The Mercury (October 26,1998 ). This statement was based on a study of insurance claims that resulted in the following data for male policyholders of a large insurance company. $$ \begin{array}{lc} \text { Astrological Sign } & \begin{array}{c} \text { Number of } \\ \text { Policyholders } \end{array} \\ \hline \text { Aquarius } & 35,666 \\ \text { Aries } & 37,926 \\ \text { Cancer } & 38,126 \\ \text { Capricorn } & 54,906 \\ \text { Gemini } & 37,179 \\ \text { Leo } & 37,354 \\ \text { Libra } & 37,910 \\ \text { Pisces } & 36,677 \\ \text { Sagittarius } & 34,175 \\ \text { Scorpio } & 35,352 \\ \text { Taurus } & 37,179 \\ \text { Virgo } & 37,718 \\ \hline \end{array} $$ a. Assuming that it is reasonable to treat the male policyholders of this particular insurance company as a random sample of male insured drivers in Australia, are the observed data consistent with the hypothesis that the proportion of male insured drivers is the same for each of the 12 astrological signs? b. Why do you think that the proportion of Capricorn policyholders is so much higher than would be expected if the proportions are the same for all astrological signs? c. Suppose that a random sample of 1000 accident claims submitted to this insurance company is selected and each claim classified according to the astrological sign of the driver. (The accompanying table is consistent with accident rates given in the article.) $$ \begin{array}{lc} & \text { Observed Number } \\ \text { Astrological Sign } & \text { in Sample } \\ \hline \text { Aquarius } & 85 \\ \text { Aries } & 83 \\ \text { Cancer } & 82 \\ \text { Capricorn } & 88 \\ \text { Gemini } & 83 \\ \text { Leo } & 83 \\ \text { Libra } & 83 \\ \text { Pisces } & 82 \\ \text { Sagittarius } & 81 \end{array} $$ $$ \begin{array}{lc} & \text { Observed Number } \\ \text { Astrological Sign } & \text { in Sample } \\ \hline \text { Scorpio } & 85 \\ \text { Taurus } & 84 \\ \text { Virgo } & 81 \\ \hline \end{array} $$ Test the null hypothesis that the proportion of accident claims submitted by drivers of each astrological sign is consistent with the proportion of policyholders of each sign. Use the given information on the distribution of policyholders to compute expected frequencies and then carry out an appropriate test.

The 2006 Expedia Vacation Deprivation Survey (Ipsos Insight, May 18,2006 ) described the results of a poll of working adults in Canada. Each person in a random sample was classified according to gender and the number of vacation days he or she usually took each year. The resulting data are summarized in the given table. Is it reasonable to conclude that there is an association between gender and the number of vacation days taken? To what population would it be reasonable to generalize this conclusion? $$ \begin{array}{l|rc} & \text { Gender } \\ \hline \text { Days of Vacation } & \text { Male } & \text { Female } \\ \hline \text { None } & 51 & 42 \\ 1-5 & 21 & 25 \\ 6-10 & 67 & 79 \\ 11-15 & 111 & 94 \\ 16-20 & 71 & 70 \\ 21-25 & 82 & 58 \\ \text { More than } 25 & 118 & 79 \\ \hline \end{array} $$

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