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87E

Expert-verifiedFound in: Page 501

Book edition
13th

Author(s)
James T. McClave, P. George Benson, Terry Sincich

Pages
888 pages

ISBN
9780134506593

** Shopping vehicle and judgment. Refer to the Journal of**** Marketing Research (December 2011) study of shopping cart design, Exercise 2.85 (p. 112). Recall that design engineers want to know whether the mean choice of the vice-over-virtue score is higher when a consumer’s arm is flexed (as when carrying a shopping basket) than when the consumer’s arm is extended (as when pushing a shopping cart). The average choice score for the n1 = 11 consumers with a flexed arm was $\overline{x}{1}$= 59, while the average for the n2 = 11 Consumers with an extended arm was $\overline{x}{2}$ = 43. In which scenario is the assumption required for a t-test to compare means more likely to be violated, ${S}{1}$ = 4 and ${{S}}_{{2}}$ = 2, or, ${{S}}_{{1}}$ = 10 and ${{S}}_{{2}}$ = 15? Explain.**

The answer can be reduced from the following steps.

Referring to Exercises 8.8 and 8.9, there is no difference between the n1 and n2 consumers' average scores. The sample mean of consumers with a flexed arm and extended arm was different, that is ** $\overline{x}1$**= 59,** $\overline{x}2$ **= 43. The standard deviation in consumers with a flexed arm and consumers with an extended are dissimilar.

A statistical test called a t-test is employed to contrast the means of two clusters. It is frequently employed in hypothesis testing to establish whether one procedure or treatment affects the target group or even if two groups vary.

The two different kinds of t-tests are as follows.

- Use a two-tailed t-test if the only thing that matters is how the two populations vary from each other.
- Use a one-tailed t-test to determine if one population mean is higher or lower compared to the other.

t-test hypotheses are as follows

$H0:\mu 1=\mu 2\phantom{\rule{0ex}{0ex}}H1:\mu 1\ne \mu 2$

The application of test statistics is as follows.

$t=\frac{\left(\overline{x}1-\overline{x}2\right)-\left(\mu {1}^{}-\mu 2\right)}{\sqrt{\frac{S{1}^{2}}{n1}+\frac{S{2}^{2}}{n2}}}$

Following is test statistics with the degree of freedom.

$df=n1-1$and $df={n}_{2}-1$ are similar because of the mean of consumers with a flexed arm and extended arm are similar. As ${H}_{0}:\mu 1=\mu 2$and ${n}_{1}={n}_{2}$, reject the claim that the value of t is higher. Therefore, the hypothesis will most likely be rejected if ${S}_{1}$ and ${S}_{2}$ are lower.

**Case-1: **If ${S}_{1}=4$and ${S}_{2}=2$

The test statistics is as follows.

$t=\frac{59-43}{\sqrt{\frac{{4}^{2}}{11}+\frac{{2}^{2}}{11}}}\phantom{\rule{0ex}{0ex}}t=\frac{16}{\sqrt{\frac{20}{11}}}\phantom{\rule{0ex}{0ex}}t=\sqrt[8]{\frac{11}{5}}\phantom{\rule{0ex}{0ex}}t=11.866$

**Case-2: **If ${S}_{1}=10$ and ${S}_{2}=15$

$t=\frac{59-43}{\sqrt{\frac{{10}^{2}}{11}+\frac{{15}^{2}}{11}}}\phantom{\rule{0ex}{0ex}}t=\sqrt[16]{\frac{11}{325}}\phantom{\rule{0ex}{0ex}}t=2.946$

**The p-value for case 1 will be higher than case 2 as t increases. In case 1, it is more likely that the hypotheses will be rejected.**

** **

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