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Chapter 7: Q58 E (page 418)

Shopping vehicle and judgment. Refer to the Journal of Marketing Research (December 2011) study of grocery store shoppers’ judgments, Exercise 2.85 (p. 112). For one part of the study, 11 consumers were told to put their arm in a flex position (similar to carrying a shopping basket) and then each consumer was offered several choices between a vice product and a virtue product (e.g., a movie ticket vs. a shopping coupon, pay later with a larger amount vs. pay now). Based on these choices, a vice choice score was determined on a scale of 0 to 100 (where higher scores indicate a greater preference for vice options). The data in the next table are (simulated) choice scores for the 11 consumers. Suppose that the average choice score for consumers with an extended arm position (similar to pushing a shopping cart) is known to be \(\mu = 50\) . The researchers theorize that the mean choice score for consumers shopping with a flexed arm will be higher than 43 (reflecting their higher propensity to select a vice product) Test the theory at \(\alpha = 0.05\)

Short Answer

Expert verified

We fail to reject the null hypothesis.

Step by step solution

01

Given Information

The sample size is 11.

The hypothesis are given by

\(\begin{aligned}{l}{H_0}:{\mu _0} = 43\\{H_a}:{\mu _0} > 43\end{aligned}\)

The mean is given by 50

02

Compute standard deviation

The standard deviation is calculated as

\(\begin{aligned}sd &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({X_i} - \bar X)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{36 + 676 + 144 + 49 + 25 + 121 + 144 + 49 + 49 + 121}}{{10}}} \\ &= \sqrt {\frac{{1414}}{{10}}} \\ &= \sqrt {141.4} \\ &= 11.89\end{aligned}\)

Therefore, the standard deviation is 11.89.

03

Test statistic

The test statistic is computed as

\(\begin{aligned}t &= \frac{{\bar x - {\mu _0}}}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{50 - 43}}{{\frac{{11.89}}{{\sqrt {11} }}}}\\ &= \frac{7}{{3.584}}\\ &= 1.95\end{aligned}\)

Therefore, the test statistic is 1.95.

04

Conclusion

For,\(\alpha = 0.05\,\,and\,n - 1 = 10\)

\(\begin{aligned}{t_{\alpha ,n - 1}} &= {t_{0.05,10}}\\ &= 1.812\end{aligned}\)

The calculated value is greater than tabulated value.

Therefore, we fail to reject the null hypothesis.

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Most popular questions from this chapter

In a test of H0:μ=100againstHa:μ>100, the sample data yielded the test statistic z = 2.17. Find and interpret the p-value for the test.

“Streaming” of television programs is trending upward. According to The Harris Poll (August 26, 2013), over one-third of American’s qualify as “subscription streamers,” i.e., those who watch streamed TV programs through a subscription service such as Netflix, Hulu Plus, or Amazon Prime. The poll included 2,242 adult TV viewers, of which 785 are subscription streamers. On the basis of this result, can you conclude that the true fraction of adult TV viewers who are subscription streamers differs from one-third? Carry out the test using a Type I error rate of α=.10. Be sure to give the null and alternative hypotheses tested, test statistic value, rejection region or p-value, and conclusion.

Packaging of a children’s health food. Can packaging of a healthy food product influence children’s desire to consume the product? This was the question of interest in an article published in the Journal of Consumer Behaviour (Vol. 10, 2011). A fictitious brand of a healthy food product—sliced apples—was packaged to appeal to children (a smiling cartoon apple was on the front of the package). The researchers showed the packaging to a sample of 408 school children and asked each whether he or she was willing to eat the product. Willingness to eat was measured on a 5-point scale, with 1 = “not willing at all” and 5 = “very willing.” The data are summarized as follows: \(\bar x = 3.69\) , s = 2.44. Suppose the researchers knew that the mean willingness to eat an actual brand of sliced apples (which is not packaged for children) is \(\mu = 3\).

a. Conduct a test to determine whether the true mean willingness to eat the brand of sliced apples packaged for children exceeded 3. Use\(\alpha = 0.05\)

to make your conclusion.

b. The data (willingness to eat values) are not normally distributed. How does this impact (if at all) the validity of your conclusion in part a? Explain.

If a hypothesis test were conducted using α= 0.05, for which of the following p-values would the null hypothesis be rejected?

a. .06

b. .10

c. .01

d. .001

e. .251

f. .042

Salaries of postgraduates. The Economics of Education Review (Vol. 21, 2002) published a paper on the relationship between education level and earnings. The data for the research was obtained from the National Adult Literacy Survey of more than 25,000 respondents. The survey revealed that males with a postgraduate degree have a mean salary of \(61,340 (with standard error \(Sx\) = \)2,185), while females with a postgraduate degree have a mean of \(32,227 (with standard error \(Sx\) = \)932).

  1. The article reports that a 95% confidence interval for \[{\bf{\mu }}M\] , the population mean salary of all males with post-graduate degrees, is (\(57,050, \)65,631). Based on this interval, is there evidence to say that \[{\bf{\mu }}M\] differs from \(60,000? Explain.
  2. Use the summary information to test the hypothesis that the true mean salary of males with postgraduate degrees differs from \)60,000. Use \(\alpha \) =.05.
  3. Explain why the inferences in parts a and b agree.
  4. The article reports that a 95% confidence interval for \(\mu F\) , the population mean salary of all females with post-graduate degrees, is (\(30,396, \)34,058). Based on this interval, is there evidence to say that \(\mu F\)differs from \(33,000? Explain.
  5. Use the summary information to test the hypothesis that the true mean salary of females with postgraduate degrees differs from \)33,000. Use \(\alpha \) =.05.
  6. Explain why the inferences in parts d and e agree.
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