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Chapter 13: Problem 43

Many shoppers have expressed unhappiness because grocery stores have stopped putting prices on individual grocery items. The article "The Impact of Item Price Removal on Grocery Shopping Behavior" (J. Market., 1980: 73-93) reports on a study in which each shopper in a sample was classified by age and by whether he or she felt the need for item pricing. Based on the accompanying data, does the need for item pricing appear to be independent of age? $$ \begin{array}{lccccc} & \multicolumn{5}{c}{\text { Age }} \\ \cline { 2 - 6 } & <\mathbf{3 0} & \mathbf{3 0 - 3 9} & \mathbf{4 0 - 4 9} & \mathbf{5 0 - 5 9} & \geq \mathbf{6 0} \\ \hline \begin{array}{l} \text { Number } \\ \text { in Sample } \end{array} & 150 & 141 & 82 & 63 & 49 \\ \begin{array}{l} \text { Number } \\ \text { Who Want } \\ \text { Item Pricing } \end{array} & 127 & 118 & 77 & 61 & 41 \\ & & & & & \end{array} $$

Short Answer

Expert verified
The need for item pricing is dependent on age.

Step by step solution

01

Establish the Hypotheses

We begin by establishing the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). The null hypothesis \( H_0 \) states that the need for item pricing is independent of age. The alternative hypothesis \( H_a \) is that there is a dependence between the need for item pricing and age.
02

Set Up the Contingency Table

Construct a contingency table using the given data. The table will show the age groups and the number of people who want and do not want item pricing in each group.\[\begin{array}{lcc}\text{Age Group} & \text{Want Item Pricing} & \text{Total in Sample} \\hline<30 & 127 & 150 \30-39 & 118 & 141 \40-49 & 77 & 82 \50-59 & 61 & 63 \\geq 60 & 41 & 49\end{array}\]
03

Calculate Expected Frequencies

Determine the expected frequency for each cell in the table assuming \( H_0 \) is true. The expected frequency for each cell can be calculated using:\[E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}\]Calculate these for all age groups who want item pricing.
04

Perform the Chi-Square Test

Use the chi-square test formula:\[\chi^2 = \sum \frac{(O - E)^2}{E}\]where \( O \) is the observed frequency and \( E \) is the expected frequency. Sum this calculation over all age groups to find the chi-square statistic.
05

Determine the Chi-Square Critical Value

Determine the degrees of freedom for the test, given by:\[\text{Degrees of Freedom} = (\text{Number of Rows} - 1)(\text{Number of Columns} - 1) = (5 - 1)(2 - 1) = 4\]Using the chi-square distribution table, find the critical value for this degrees of freedom at the chosen significance level (typically 0.05).
06

Make a Decision

Compare the calculated chi-square statistic with the critical value. If the chi-square statistic is greater than the critical value, reject \( H_0 \); otherwise, do not reject \( H_0 \).
07

Conclusion

Based on the decision in Step 6, conclude whether the need for item pricing is independent of age or not.

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Key Concepts

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

Chi-Square Test
The chi-square test is a popular statistical method used to determine whether there is a significant association between two categorical variables. In our problem about grocery shoppers, we're interested in assessing if the need for item pricing is independent of the shopper's age.

The chi-square test involves several key components:
  • **Observed Frequencies (O)**: The counts you actually observe in your data for each cell of the table.
  • **Expected Frequencies (E)**: These are calculated based on the assumption that the null hypothesis is true. They represent what the counts would look like if the two variables were independent.
To perform the test, we follow these steps:
  • Establish hypotheses. In our case, the null hypothesis \( H_0 \) is that age and the need for pricing are independent, while the alternative hypothesis \( H_a \) suggests dependence.
  • Calculate expected frequencies for each age group using the formula: \[ E = \frac{\text{(Row Total)} \times \text{(Column Total)}}{\text{Grand Total}} \]
  • Calculate the chi-square statistic using: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \]
  • Compare the chi-square statistic with a critical value derived from a chi-square distribution table.
If the calculated chi-square statistic exceeds this critical value, it suggests a significant association (reject \( H_0 \)); otherwise, no significant association (do not reject \( H_0 \)).
Contingency Table
A contingency table, also known as a cross-tabulation or crosstab, is a type of table that displays the distribution of two or more categorical variables. In the provided exercise, we use a two-dimensional contingency table to illustrate the relationship between age groups and the desire for item pricing.

The table includes:
  • Age Groups: Columns indicate different age ranges, like "<30", "30-39", etc.
  • Response: Rows show whether the individual wants item pricing or not.
  • Counts: Each cell contains the number of individuals that fall into both the specified age group and response category.
To construct a contingency table:
  • First, determine your variables and their possible values.
  • Second, tally the individuals in each category combination based on your data.
  • Finally, calculate row and column totals to facilitate further analysis like the calculation of expected frequencies.
Contingency tables are a crucial tool in hypothesis testing because they neatly organize data to visualize potential associations between variables.
Degrees of Freedom
Degrees of freedom are an essential concept in statistics that represents the number of values in a calculation that can vary independently. In the context of a chi-square test using a contingency table, degrees of freedom help determine the critical value from the chi-square distribution table.

To compute the degrees of freedom for a chi-square test involving a contingency table:
  • Use the formula: \( \text{Degrees of Freedom} = (\text{Number of Rows} - 1)(\text{Number of Columns} - 1) \)
  • This calculation allows for an adjustment for the table's size, facilitating the correct critical value lookup.
In our exercise, we had 5 age categories (rows) and 2 outcome possibilities (columns: want and do not want pricing), resulting in:
\[ \text{Degrees of Freedom} = (5 - 1)(2 - 1) = 4 \]

Understanding degrees of freedom helps you gauge the flexibility of data within your statistical models and ensures the selection of the appropriate chi-square distribution when testing for independence.

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Most popular questions from this chapter

The authors of the article "Predicting Professional Sports Game Outcomes from Intermediate Game Scores" (Chance, 1992: 18-22) used a chisquared test to determine whether there was any merit to the idea that basketball games are not settled until the last quarter, whereas baseball games are over by the seventh inning. They also considered football and hockey. Data was collected for 189 basketball games, 92 baseball games, 80 hockey games, and 93 football games. The games analyzed were sampled randomly from all games played during the 1990 season for baseball and football and for the 1990-1991 season for basketball and hockey. For each game, the late-game leader was determined, and then it was noted whether the late-game leader actually ended up winning the game. The resulting data is summarized in the accompanying table. $$ \begin{array}{lcc} \hline \text { Sport } & \begin{array}{c} \text { Late-Game } \\ \text { Leader Wins } \end{array} & \begin{array}{l} \text { Late-Game } \\ \text { Leader Loses } \end{array} \\ \hline \text { Basketball } & 150 & 39 \\ \text { Baseball } & 86 & 6 \\ \text { Hockey } & 65 & 15 \\ \text { Football } & 72 & 21 \end{array} $$ The authors state, "Late-game leader is defined as the team that is ahead after three quarters in basketball and football, two periods in hockey, and seven innings in baseball. The chi-square value on three degrees of freedom is \(10.52\) \((P<.015) . "\) a. State the relevant hypotheses and reach a conclusion using \(\alpha=.05\). b. Do you think that your conclusion in part (a) can be attributed to a single sport being an anomaly?

The NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16 teams in each region are then ranked (seeded) from 1 to 16. During the 12-year period from 1991 to 2002 , the top-ranked team won its regional tournament 22 times, the second-ranked team won 10 times, the third-ranked team won 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3 . Let \(P_{i j}\) denote the probability that the team ranked \(i\) in its region is victorious in its game against the team ranked \(j\). Once the \(P_{i j}\) 's are available, it is possible to compute the probability that any particular seed wins its regional tournament (a complicated calculation because the number of outcomes in the sample space is quite large). The paper "Probability Models for the NCAA Regional Basketball Tournaments"(Amer. Statist., 1991: 35-38) proposed several different models for the \(P_{i j}\) 's. a. One model postulated \(P_{i j}=.5-\lambda(i-j)\) with \(\lambda=\frac{1}{32}\) (from which \(P_{16,1}=\frac{1}{32}, P_{16,2}=\frac{2}{32}\), etc.). Based on this, \(P\) (seed #1 wins) \(=.27477\), \(P(\) seed \(\\# 2\) wins \()=.20834\), and \(P\) (seed #3 wins \()=.15429\). Does this model appear to provide a good fit to the data? b. A more sophisticated model has \(P_{i j}=.5+\) \(.2813625\left(z_{i}-z_{j}\right)\), where the \(z\) 's are measures of relative strengths related to standard normal percentiles [percentiles for successive highly seeded teams are closer together than is the case for teams seeded lower, and .2813625 ensures that the range of probabilities is the same as for the model in part (a)]. The resulting probabilities of seeds 1,2 , or 3 winning their regional tournaments are \(.45883, .18813\), and \(.11032\), respectively. Assess the fit of this model.

The accompanying two-way frequency table appears in the article "Marijuana Use in College" (Youth and Society, 1979: 323-334). Each of 445 college students was classified according to both frequency of marijuana use and parental use of alcohol and psychoactive drugs. Does the data suggest that parental usage and student usage are independent in the population from which the sample was drawn? Use the \(P\)-value method to reach a conclusion. Standard Level of Marijuana use Never Occasional Regular $$ \begin{array}{ll|r|r|c} \cline { 3 - 5 } \begin{array}{l} \text { Parental } \\ \text { Use of } \end{array} & \text { Neither } & 141 & 54 & 40 \\ \cline { 3 - 5 } \begin{array}{l} \text { Alcohol } \\ \text { and Drugs } \end{array} & \text { One } & 68 & 44 & 51 \\ \cline { 2 - 4 } & \text { Both } & 17 & 11 & 19 \\ \hline \end{array} $$

The article "Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients' Birthdays"' (Psych. Rep., 1992: 944鈥�946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient's birthday. Assuming a 365day year (i.e., excluding leap year), in the absence of any relation, a patient's admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of \(11,24,69\), and 96 for categories 1,2 , 3 , and 4, respectively. State and test the relevant hypotheses using a significance level of \(.01\).

The following data resulted from an experiment to study the effects of leaf removal on the ability of fruit of a certain type to mature ("Fruit Set, Herbivory, Fruit Reproduction, and the Fruiting Strategy of Catalpa speciosa," Ecology, 1980: 57-64). Does the data suggest that the chance of a fruit maturing is affected by the number of leaves removed? State and test the appropriate hypotheses at level \(.01\). $$ \begin{array}{lcc} \hline & \begin{array}{l} \text { Number } \\ \text { of Fruits } \\ \text { Matured } \end{array} & \begin{array}{c} \text { Number } \\ \text { of Fruits } \\ \text { Aborted } \end{array} \\ \hline \text { Control } & 141 & 206 \\ \text { Two leaves removed } & 28 & 69 \\ \text { Four leaves removed } & 25 & 73 \\ \text { Six leaves removed } & 24 & 78 \\ \text { Eight leaves removed } & 20 & 82 \end{array} $$
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