Question

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

We fail to reject the null hypothesis.

Step-by-step solution

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

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.

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.

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.