Q32SE The study reported in 鈥淕ait Pa... [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: Q32SE (page 676)

The study reported in 鈥淕ait Patterns During Free Choice Ladder Ascents鈥� (Human Movement Sci., 1983: 187鈥�195) was motivated by publicity concerning the increased accident rate for individuals climbing ladders. A number of different gait patterns were used by subjects climbing a portable straight ladder according to specified instructions. The ascent times for seven subjects who used a lateral gait and six subjects who used a four-beat diagonal gait are given.
Lateral 0.86 1.31 1.64 1.51 1.53 1.39 1.09
Diagonal 1.27 1.82 1.66 0.85 1.45 1.24
a. Carry out a test using a 5 .05 to see whether the data suggests any difference in the true average ascent times for the two gaits.
b. Compute a 95% CI for the difference between the true average gait times.

Short Answer

Expert verified

Solution for

a)Do not reject null hypothesis

b) $\left( {{d_{ij\left( 8 \right)}},{d_{ij\left( {35} \right)}}} \right)$=(-0.41, -0.29)

Step by step solution

testing null hypothesis for (a):

When testing null hypothesis

Versus one of the alternative hypothesis, one could use test static value

$w = \sum\limits_{i = 1}^m {{r_i}} $

The P-value depends on the alternative hypothesis

Alternative hypothesis P-values

$\begin{array}{l}{P_0}\left( {W \ge \omega } \right)\\{P_0}\left( {W \le \omega } \right) = {P_0}\left( {W \ge m\left( {m + n + 1} \right) - \omega } \right)\\2{P_0}\left( {{W_ + } \ge \max \left\{ {\omega ,m\left( {m + n + 1} \right) - \omega } \right\}} \right)\end{array}$

solving further:

The test of interest is

${H_0}:{\mu _1} - {\mu _2} = 0$

Versus alternative hypothesis

${H_a}:{\mu _1} - {\mu _2} \ne 0$

The following table represents values required to compute the test statistic value

i	${x_i}/{y_i}$	${r_i}$	$\sum {{r_i}} $
1	0.86	2	48
2	1.31	6
3	1.64	11
4	1.51	9
5	1.53	10
6	1.39	7
n = 7	1.09	3
1	1.27	5	$\omega $=43
2	1.82	13
3	1.66	12
4	0.85	1
5	1.45	8
m=6	1.24	4

test static value:

The alternative ofP-value is,

$2{P_0}\left( {W \ge \left\{ {\omega ,m\left( {m + n + 1} \right) - \omega } \right\}} \right) = 2{P_0}\left( {W \ge 43} \right)$

In the table in the appendix one could find only particular values ; thus, for$\alpha = 0.05$use table to obtain

$m\left( {m + n + 1} \right) - 56 = 28 < \omega = 43 < {W_{m,n.\alpha /2}} = 56$

Do not reject null hypothesis

solving for (b):

Let,

${d_{ij}} = {x_i} - {y_i}$

Where${x_1},{x_2},....,{x_m}$and${y_1},{y_2},...{y_n}$are observed values of continuous distribution that only in the location but not in shap the general form of a$100(1 - \alpha )$confidence interval for${\mu _1} - {\mu _2}$

Is,

$\left( {{d_{ij\left( {mn - c + 1} \right)}},{d_{ij\left( c \right)}}} \right)$

Where${d_{ij\left( 1 \right)}},{d_{ij\left( 2 \right)}}...,{d_{ij\left( {mn} \right)}}$are the ordered difference, use appendix values of c.

For m = 6, n = 7 and for 94.5% confidence interval, c = 35, the confidence interval is the form of

$\left( {{d_{ij\left( {mn - c + 1} \right)}},{d_{ij\left( c \right)}}} \right) = \left( {{d_{ij\left( {6.7 - 35 + 1} \right)}},{d_{ij\left( {35} \right)}}} \right) = \left( {{d_{ij\left( 8 \right)}},{d_{ij\left( {35} \right)}}} \right)$

Compute all difference, find 8^th and 35th ordered difference values. The four smallest ordered difference are

The 8^th and 35th ordered difference are -0.41 and 2.9, respectively; a 94.5% confidence interval is

$\left( {{d_{ij\left( 8 \right)}},{d_{ij\left( {35} \right)}}} \right)$=(-0.41, -0.29)