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Chapter 11: Problem 51

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 \(\alpha=.05\).

Short Answer

Expert verified
From this sample, there is not enough evidence at a \(.05\) significance level to reject the null hypothesis that the proportion of total-bran cereal consumers is the same across all four brands.

Step by step solution

01

State Null and Alternate Hypotheses

Null Hypothesis (\(H_0\)): The proportion of people who prefer each brand is equal. i.e. The proportions are \(0.25\) for Brand A, \(0.25\) for Brand B, \(0.25\) for Brand C, and \(0.25\) for Brand D. Alternate Hypothesis (\(H_a\)): The proportion of people who prefer each brand is not equal.
02

Calculate Expected Frequencies

If there's no preference among the brands, each brand should be preferred by \(0.25\) of people i.e. \(0.25 * 1000 = 250\) for each brand. These are the expected frequencies under the null hypothesis.
03

Perform Chi-Square Goodness of Fit Test

The formula for Chi-Square test statistic is: \[ 蠂^2 = 危 \frac{(O_i-E_i)^2}{E_i} \] where \(E_i\) is the Expected frequency, \(O_i\) is the observed frequency. Plug in the numbers and calculate the Chi-Square statistic \[蠂^2 = (\frac{(212-250)^2}{250}) + (\frac{(284-250)^2}{250}) + (\frac{(254-250)^2}{250}) + (\frac{(250-250)^2}{250}) = 4.32 \] The degree of freedom is given by: \[ df = n - 1 = 4 - 1 = 3 \]
04

Decision Rule and Conclusion

For a Chi-square with 3 degrees of freedom, the critical value at \(\alpha = 0.05\) is \(7.815\). If our calculated chi-square statistic is greater than this critical value, we would reject the null hypothesis. However, in this case, our chi-square statistic \(4.32\) is less than the critical value. Therefore, we fail to reject the null hypothesis.

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Most popular questions from this chapter

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.

Chance Corporation produces beauty products. Two years ago the quality control department at the company conducted a survey of users of one of the company's products. The survey revealed that \(53 \%\) of the users said the produet was excellent, \(31 \%\) said it was satisfactory, \(7 \%\) said it was unsatisfactory, and \(9 \%\) had no opinion. Assume that these percentages were true for the population of all users of this product at that time. After this survey was conducted, the company redesigned this product. A recent survey of 800 users of the redesigned product conducted by the quality control department at the company showed that 495 of the users think the product is excellent, 255 think it is satisfactory, 35 think it is unsatisfactory, and 15 have no opinion. Is the percentage distribution of the opinions of users of the redesigned product different from the percentage distribution of users of this product before it was redesigned? Use \(\alpha=.025\).

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 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 \(2009 .\)

Find the value of \(\chi^{2}\) for 12 degrees of freedom and an area of \(.025\) in the right tail of the chi-square distribution curve.
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