/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Over the last 3 years, Art's Sup... [FREE SOLUTION] | 91Ó°ÊÓ

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Over the last 3 years, Art's Supermarket has observed the following distribution of modes of payment in the express lines: cash (C) \(41 \%\), check (CK) \(24 \%\), credit or debit card (D) \(26 \%\), and other (N) \(9 \% .\) In an effort to make express checkout more efficient, Art's has just begun offering a \(1 \%\) discount for cash payment in the express checkout line. The following table lists the frequency distribution of the modes of payment for a sample of 500 express-line customers after the discount went into effect. $$ \begin{array}{l|cccc} \hline \text { Mode of payment } & \text { C } & \text { CK } & \text { D } & \text { N } \\ \hline \text { Number of customers } & 240 & 104 & 111 & 45 \\ \hline \end{array} $$ Test at a \(1 \%\) significance level whether the distribution of modes of payment in the express checkout line changed after the discount went into effect.

Short Answer

Expert verified
Yes, the distribution of modes of payment in the express checkout line changed significantly after the discount went into effect.

Step by step solution

01

Calculate the Expected Frequencies

Using the previous distribution percentages and the total sample size: \(N=500\), the expected frequencies are calculated as follows: \(E_C = 0.41*500 = 205\), \(E_{CK} = 0.24*500 = 120\), \(E_D = 0.26*500 = 130\), and \(E_N = 0.09*500 = 45\).
02

Calculate the Chi-Square Test Statistic

The formula used to calculate the Chi-square statistic is: \(\chi^2 = \sum[(O_i - E_i)^2 / E_i]\), where O is the observed frequencies and E is the expected frequencies. Substituting observed and expected values for each payment mode: \(\chi^2 = (240 - 205)^2/205 + (104 - 120)^2/120 + (111 -130)^2 /130 + (45 - 45)^2 /45 = 6.79 + 2.13 + 2.78 + 0 = 11.7.\)
03

Find the Critical Value

The critical value can be obtained from the Chi-square distribution table. With the degrees of freedom being \(df = k - 1 = 4 - 1 = 3\) and the significance level equal to \(0.01\), the critical value is approximately \(11.34\).
04

Make a Decision

Comparing the test statistic with the critical value (i.e., \(\chi^2 > \text{Critical Value}\)), we realize \(11.7 > 11.34\). Hence, the distribution of payments shows a significant change; the null hypothesis is rejected at a \(1 \%\) level of significance.

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Key Concepts

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

Expected Frequencies
Expected Frequencies help predict what the data should ideally look like based on historical or theoretical distributions. They are calculated using the known probabilities or proportions and the total number of observations.
In our exercise, we calculate expected frequencies for different payment modes based on the initial percentages.
These percentages were cash (41%), check (24%), credit/debit card (26%), and other (9%) with a total of 500 customers.
The formula for each expected frequency is:
  • Cash: \( E_C = 0.41 \times 500 = 205 \)
  • Check: \( E_{CK} = 0.24 \times 500 = 120 \)
  • Credit/Debit Card: \( E_D = 0.26 \times 500 = 130 \)
  • Other: \( E_N = 0.09 \times 500 = 45 \)
These values represent what we would expect if nothing had changed after implementing the discount strategy.
Observed Frequencies
Observed Frequencies are the actual data collected from observations or experiments. In this context, they show the real number of customers choosing each payment mode after the initiative.
In our example with Art's Supermarket, once the discount was put into effect, we recorded the following observed frequencies for the 500 customers:
  • Cash: 240 customers
  • Check: 104 customers
  • Credit/Debit Card: 111 customers
  • Other: 45 customers
By comparing these numbers to the expected frequencies, we can determine if there is a significant shift in behavior or if the variations occurred purely by chance.
Significance Level
The significance level is a critical concept in hypothesis testing, determining the probability of rejecting the null hypothesis when it is, in fact, true.
Typically set at 1%, 5%, or 10%, it reflects the level of confidence you require in your results. A smaller significance level means stricter criteria for concluding a significant effect.
For the exercise, we are using a 1% significance level, which suggests high confidence in deciding whether the payment distribution changed post-discount. If the test statistic exceeds the critical value at this level, we infer that the change is unlikely due to random fluctuations alone.
Degrees of Freedom
Degrees of Freedom (df) in statistical tests determine the number of independent values or quantities that can vary.
In a Chi-Square test, degrees of freedom are calculated as the number of categories minus one: \( df = k - 1 \). This accounts for the constraints in the data.
For Arts' Supermarket, there are four categories of payment methods, so the degree of freedom is:
  • \( df = 4 - 1 = 3 \)
This value is essential for determining the test's critical Chi-Square value at the chosen significance level. It helps decide if the observed change is significant or a random occurrence.

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Most popular questions from this chapter

The percentage distribution of birth weights for all children in cases of multiple births (twins, triplets, etc.) in North Carolina during 2009 was as given in the following table: $$ \begin{array}{l|cccc} \hline \text { Weight (grams) } & 0-500 & 501-1500 & 1501-2500 & 2501-8165 \\ \hline \text { Percentage } & 1.45 & 11.02 & 49.23 & 38.30 \\ \hline \end{array} $$ The frequency distribution of birth weights of a sample of 587 children who shared multiple births and were born in North Carolina in 2012 is as shown in the following table? $$ \begin{array}{l|cccc} \hline \text { Weight (grams) } & 0-500 & 501-1500 & 1501-2500 & 2501-8165 \\ \hline \text { Frequency } & 2 & 60 & 305 & 220 \\ \hline \end{array} $$ Test at a \(2.5 \%\) significance level whether the 2012 distribution of birth weights for all children born in North Carolina who shared multiple births is significantly different from the one for \(2009 .\)

One of the products produced by Branco Food Company is Total-Bran Cereal, which competes with three other brands of similar total-bran cereals. The company's research office wants to investigate if the percentage of people who consume total-bran cereal is the same for each of these four brands. Let us denote the four brands of cereal by \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\). A sample of 1000 persons who consume total-bran cereal was taken, and they were asked which brand they most often consume. Of the respondents, 212 said they usually consume Brand A, 284 consume Brand B, 254 consume Brand \(\mathrm{C}\), and 250 consume Brand D. Does the sample provide enough evidence to reject the null hypothesis that the percentage of people who consume total-bran cereal is the same for all four brands? Use \(\alpha=.05\).

A sample of 30 observations selected from a normally distributed population produced a sample variance of \(5.8\). a. Write the null and alternative hypotheses to test whether the population variance is different from \(6.0\) b. Using \(\alpha=.05\), find the critical value of \(\chi^{2} .\) Show the rejection and nonrejection regions on a chi-square distribution curve. c. Find the value of the test statistic \(\chi^{2}\). d. Using a \(5 \%\) significance level, will you reject the null hypothesis stated in part a?

The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at a \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.

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