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Chapter 4: Problem 13

A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Do the data confirm the belief that one design is just as likely to be selected as another? Explain. $$\begin{array}{cc} \text { Design } & \text { Number of Times Preferred} \\ 1 & 5 \\ 2 & 15 \\ 3 & 30 \\ 4 & 40 \\ 5 & 10 \end{array}$$

Short Answer

Expert verified
Design preferences are not equally likely; the data does not support equal selection.

Step by step solution

01

Determine the Expected Probability per Design

Since each design is assumed to be equally likely to be chosen, each should have the same probability of being selected. With five designs, the probability for each design is the number of favorable outcomes (1) over the total number of options (5), which is \( \frac{1}{5} = 0.20 \).
02

Define the Hypotheses for Chi-Square Test

We want to test if the observed frequencies match the expected frequencies under the assumption that each design has an equal chance of being selected. - Null Hypothesis (\(H_0\)): Each design is equally likely. - Alternative Hypothesis (\(H_a\)): The designs do not all have the same probability of being chosen.
03

Calculate the Expected Frequencies

Under the null hypothesis, the expected frequency for each design in a sample of 100 consumers is: \[ \text{Expected frequency} = \text{Total consumers} \times \text{Probability per design} = 100 \times 0.20 = 20 \].
04

Compute the Chi-Square Statistic

The chi-square statistic is calculated using the formula:\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency.\[\chi^2 = \frac{(5-20)^2}{20} + \frac{(15-20)^2}{20} + \frac{(30-20)^2}{20} + \frac{(40-20)^2}{20} + \frac{(10-20)^2}{20} \]\[\chi^2 = \frac{225}{20} + \frac{25}{20} + \frac{100}{20} + \frac{400}{20} + \frac{100}{20} = 11.25 + 1.25 + 5 + 20 + 5 = 42.5\]
05

Determine Degrees of Freedom and Critical Value

The degrees of freedom for this test is the number of designs minus one: \(5 - 1 = 4\). Using a chi-square distribution table and a significance level (commonly 0.05), find the critical value for \(df = 4\). The critical value at \(\alpha = 0.05\) is approximately 9.49.
06

Compare the Chi-Square Statistic with Critical Value

The calculated chi-square statistic is 42.5. Since 42.5 is greater than the critical value of 9.49, we reject the null hypothesis. This means the data does not confirm the belief that each design is equally likely to be selected.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a foundational concept in statistical testing. It represents a statement that there is no effect or no difference, serving as a starting point for statistical testing. In the context of the Chi-Square Test, the null hypothesis assumes that the observed distribution of data follows a certain expected distribution.

For the toothpaste packaging design exercise, the null hypothesis (\(H_0\)) states that each design is equally likely to be chosen by consumers. This implies that any deviations observed from this prediction in the experiment might be due to random chance rather than some systematic effect. Formally, the null hypothesis is expressed as: \(H_0: p_1 = p_2 = p_3 = p_4 = p_5\), where each \(p_i\) represents the proportion of selection for a design, and all are presumed equal.

Testing this null hypothesis involves comparing observed data against expected frequencies assuming equality. If the observed deviations are larger than what random chance might produce, the null hypothesis might be rejected, suggesting that not all designs have the same likelihood of being selected.
Expected Frequency
Expected frequency plays a crucial role when employing a Chi-Square Test. It refers to the count of outcomes logically anticipated based on the null hypothesis. These expected values serve as benchmarks against which actual observed data is compared.

In our example with toothpaste designs, if we assume each package is equally preferred, we calculate expected frequencies by spreading the total preference count evenly across all options. With 100 surveyed consumers and 5 designs, the expected frequency would be computed as:\[\text{Expected frequency} = \text{Total participants} \times \text{Probability of each design}\] or \[= 100 \times 0.20 = 20\] for each design.

Discrepancies between these expected frequencies and the actual data are fundamental for evaluating the null hypothesis. Significant differences suggest that consumer preferences might not be uniformly distributed, thus providing crucial evidence on the null hypothesis's validity.
Degrees of Freedom
Degrees of freedom is a statistical term often used in hypothesis testing, including the Chi-Square Test. It refers to the number of values that can vary independently in an analysis, helping determine the appropriate critical value from statistical tables.

In the context of a chi-square test, the degrees of freedom (df) are calculated as one less than the number of categories or groups being examined. For the toothpaste experiment with five package designs, the degrees of freedom would be:\[df = \text{Number of designs} - 1 = 5 - 1 = 4\].

The significance of degrees of freedom is in shaping the chi-square distribution used to determine the critical value. This critical value marks the threshold against which the computed chi-square statistic is compared. If the statistic exceeds this threshold, it suggests enough evidence exists to reject the null hypothesis, indicating non-uniformity in preferences.

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Most popular questions from this chapter

Visa Card USA studied how frequently young consumers, ages 18 to \(24,\) use plastic (debit and credit) cards in making purchases (Associated Press, January 16,2006 ). The results of the study provided the following probabilities. \(\bullet\) The probability that a consumer uses a plastic card when making a purchase is .37. \(\bullet\) Given that the consumer uses a plastic card, there is a .19 probability that the consumer is 18 to 24 years old. \(\bullet\) Given that the consumer uses a plastic card, there is a .81 probability that the consumer is more than 24 years old. U.S. Census Bureau data show that \(14 \%\) of the consumer population is 18 to 24 years old. a. Given the consumer is 18 to 24 years old, what is the probability that the consumer uses a plastic card? b. Given the consumer is over 24 years old, what is the probability that the consumer uses a plastic card? c. What is the interpretation of the probabilities shown in parts (a) and (b)? d. Should companies such as Visa, MasterCard, and Discover make plastic cards available to the 18 to 24 years old age group before these consumers have had time to establish a credit history? If no, why? If yes, what restrictions might the companies place on this age group?

In a BusinessWeek/Harris Poll, 1035 adults were asked about their attitudes toward business (BusinessWeek, September 11, 2000). One question asked: "How would you rate large U.S. companies on making good products and competing in a global environment?" The responses were: excellent- \(18 \%\), pretty good \(-50 \%\), only fair- \(26 \%\), poor \(-5 \%\) and don't know/no answer- \(1 \%\) a. What is the probability that a respondent rated U.S. companies pretty good or excellent? b. How many respondents rated U.S. companies poor? c. How many respondents did not know or did not answer?

The Powerball lottery is played twice each week in 28 states, the Virgin Islands, and the District of Columbia. To play Powerball a participant must purchase a ticket and then select five numbers from the digits 1 through 55 and a Powerball number from the digits 1 through 42\. To determine the winning numbers for each game, lottery officials draw five white balls out of a drum with 55 white balls, and one red ball out of a drum with 42 red balls. To win the jackpot, a participant's numbers must match the numbers on the five white balls in any order and the number on the red Powerball. Eight coworkers at the ConAgra Foods plant in Lincoln, Nebraska, claimed the record \(\$ 365\) million jackpot on February \(18,2006,\) by matching the numbers \(15-17-43-44-49\) and the Powerball number \(29 .\) A variety of other cash prizes are awarded each time the game is played. For instance, a prize of \(\$ 200,000\) is paid if the participant's five numbers match the numbers on the five white balls (http://www.powerball.com, March 19, 2006). a. Compute the number of ways the first five numbers can be selected. b. What is the probability of winning a prize of \(\$ 200,000\) by matching the numbers on the five white balls? c. What is the probability of winning the Powerball jackpot?

The U.S. Census Bureau provides data on the number of young adults, ages \(18-24,\) who are living in their parents' home. ' Let \(M=\) the event a male young adult is living in his parents' home \(F=\) the event a female young adult is living in her parents' home If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude \(P(M)=.56\) and \(P(F)=.42\) (The World Almanac, 2006 ). The probability that both are living in their parents' home is .24 a. What is the probability at least one of the two young adults selected is living in his or her parents' home? b. What is the probability both young adults selected are living on their own (neither is living in their parents' home)?

High school seniors with strong academic records apply to the nation's most selective colleges in greater numbers each year. Because the number of slots remains relatively stable, some colleges reject more early applicants. The University of Pennsylvania received 2851 applications for early admission. Of this group, it admitted 1033 students early, rejected 854 outright, and deferred 964 to the regular admissions pool for further consideration. In the past, Penn has admitted about \(18 \%\) of the deferred early admission applicants during the regular admission process. Counting the additional students who were admitted during the regular admission process, the total class size was 2375 (USA Today, January 24,2001 ). Let \(E, R,\) and \(D\) represent the events that a student who applies for early admission is admitted early, rejected outright, or deferred to the regular admissions pool. a. Use the data to estimate \(P(E), P(R),\) and \(P(D)\) b. Are events \(E\) and \(D\) mutually exclusive? Find \(P(E \cap D)\) c. For the 2375 students admitted to Penn, what is the probability that a randomly selected student was accepted for early admission? d. Suppose a student applies to Penn for early admission. What is the probability the student will be admitted for early admission or be deferred and later admitted during the regular admission process?
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