Q29SE High-pressure sales tactics or d... [FREE SOLUTION]

91影视

Probability And Statistics For Engineering And Sciences

Jay L. Devore

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 714 pages

ISBN: 9781305251809

Chapter 15: Q29SE (page 676)

High-pressure sales tactics or door-to-door salespeople can be quite offensive. Many people succumb to such tactics, sign a purchase agreement, and later regret their actions. In the mid-1970s, the Federal Trade Commission implemented regulations clarifying and extending the rights of purchasers to cancel such agreements. The accompanying data is a subset of that given in the article 鈥淓valuating the FTC Cooling-Off Rule鈥� (J. of Consumer Affairs, 1977: 101鈥�106). Individual observations are cancellation rates for each of nine sales people during each of 4 years. Use an appropriate test at level .05 to see whether true average cancellation rate depends on the year.

Short Answer

Expert verified

Reject null hypothesis

Step by step solution

Friedman’s test:

The Friedman鈥檚 test for rendomized block experiment

Test for teststatistic is

$\begin{array}{l}{F_r} = \frac{{12J}}{{I\left( {I + 1} \right)}}\sum\limits_{i = 1}^I {{{\left( {{{\bar R}_i} - \frac{{I + 1}}{2}} \right)}^2}} \\ = \frac{{12J}}{{IJ\left( {I + 1} \right)}}\sum\limits_{i = 1}^I {\bar R_i^2 - 3J} \left( {I + 1} \right)\end{array}$

When the null hypothesis is true test statistic ${F_r}$, has approximately chi-squared distribution with I-1 degrees of freedom.

solving further:

Two table goes together- point${x_{ij}}$in the first table corresponds to the${r_{ij}}$rank in the second table.

I	Blocks
1973	2.8	5.9	3.3	4.4	1.7	3.8	6.6	3.1	0
1974	3.6	1.7	5.1	2.2	2.1	4.1	4.7	2.7	1.3
1975	1.4	0.9	1.1	3.2	0.8	1.5	2.8	1.4	0.5
1976	2	2.2	0.9	1.1	0.5	1.2	1.4	3.5	1.2

I	Ranks																${r_i}$	$r_i^2$
	1	2		3		4		5		6		7		8		9
1973	3		4		3		4		3		3		4		3	1	28	784
1974	4		2		4		2		4		4		3		2	4	29	841
1975	1		1		2		3		2		2		2		1	2	16	256
1976	2		3		1		1		1		1		1		4	3	17	289

testing static value:

The test statistic value is,

$\begin{array}{l}{f_r} = \frac{{12J}}{{I\left( {I + 1} \right)}}{\sum\limits_{i = 1}^I {\left( {{{\bar R}_i} - \frac{{I - 1}}{2}} \right)} ^2}\\ = \frac{{12}}{{IJ\left( {I + 1} \right)}}\sum\limits_{i = 1}^I {\bar R_i^2} - 3J\left( {I + 1} \right)\\ = \frac{{12}}{{9 \cdot 4 \cdot 5}} \cdot \left( {784 + 841 + 256 + 289} \right) - 3 \cdot 9 \cdot 5\\ = 9.62\end{array}$

Degrees of freedom are

$df = I - 1 = 4 - 1 = 3$

Critical value at significant level 0.05 is

$\begin{array}{l}X_{0.05,3}^2 = 7.815\\X_{0.05,3}^2 = 7.815 < 9.62 = {f_r}\end{array}$

reject null hypothesis

Hence, reject null hypothesis.