/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q11E In an experiment to compare the ... [FREE SOLUTION] | 91影视

91影视

In an experiment to compare the bond strength of two different adhesives, each adhesive was used in five bondings of two surfaces, and the force necessary to separate the surfaces was determined for each bonding. For adhesive 1, the resulting values were , \(229,286,245\) , \(299\)and \(250\), whereas the adhesive 2 observations were , \(213,179,163,247\)and \(225\) . Let \({\mu _i}\) denote the true average bond strength of adhesive type . Use the Wilcoxon rank-sum test at level \(.05\) to test \({H_0}:{\mu _1} = {\mu _2}\) versus \({H_a}:{\mu _1} > {\mu _2}\).

Short Answer

Expert verified

Therefore,

It Reject \({{\rm{H}}_0}\).

Step by step solution

01

Given.

Given:

\(\begin{array}{l}m = 5\\n = 5\\{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} > {\mu _2}\\\alpha = 0.05\end{array}\)

02

To determine the rank of every data value.

Determine the rank of every data value. The smallest value receives the rank l, the second smallest value receives the rank 2, the third smallest value receives the rank 3, and so on.

If multiple data values have the same value, then their rank is the average of the corresponding ranks.

Sample 1

Rank

Sample 2

Rank

\(229\)

\(5\)

\(213\)

\(3\)

\(286\)

\(9\)

\(179\)

\(2\)

\(245\)

\(6\)

\(163\)

\(1\)

\(299\)

\(10\)

\(247\)

\(7\)

\(250\)

\(8\)

\(225\)

\(4\)

Sum

\(38\)

Sum

\(17\)

03

To determine the each bonding.

\({{\rm{W}}_1}\)is the sum of the ranks in the smaller sample:

\({\rm{w = 38}}\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true.
The P-value is the number (or interval) in the column\({P_0}(W \ge c)\)of the Wilcoxon rank-sum test table in the appendix that contains the w-value in the row\(m = 5,n = 5,c = 38\):

\(0.008 < P < 0.028\)

If the probability (P-value) is less than the significant level\(\alpha \), then reject the null hypothesis:

\(P < 0.05 \Rightarrow {\rm{ Reject }}{H_0}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the large-sample version of the Wilcoxon test at significance level .05 on the data of Exercise 37 in Section 9.3 to decide whether the true mean difference between outdoor and indoor concentrations exceeds .20.

The sign test is a very simple procedure for testing hypotheses about a population median assuming only that the underlying distribution is continuous. To illustrate, consider the following sample of 20 observations on component lifetime (hr):

1.7 3.3 5.1 6.9 12.6 14.4 16.4

24.6 26.0 26.5 32.1 37.4 40.1 40.5

41.5 72.4 80.1 86.4 87.5 100.2

We wish to test \({H_0}:\tilde \mu = 25.0\) versus \({H_0}:\tilde \mu > 25.0\)The test statistic is Y 5 the number of observations that exceed 25

a. Determine the P-value of the test when Y 5 15. (Hint: Think of a 鈥渟uccess鈥 as a lifetime that exceeds 25.0. Then Y is the number of successes in the sample. What kind of a distribution does Y have when\(\tilde \mu = 25.0\)?)

b. For the given data, should H0 be rejected at significance level .05? (Note: The test statistic is the number of differences \({X_i} - 25\)that have positive signs, hence the name sign test.)

The accompanying data is a subset of the data reported in the article 鈥淪ynovial Fluid pH, Lactate, Oxygen and Carbon Dioxide Partial Pressure in Various Joint Diseases鈥 (Arthritis and Rheumatism, 1971: 476鈥477). The observations are pH values of synovial fluid (which lubricates joints and tendons) taken from the knees of individuals suffering from arthritis. Assuming that true average pH for nonarthritic individuals is 7.39, test at level .05 to see whether the data indicates a difference between average pH values for arthritic and nonarthritic individuals.

7.02 7.35 7.34 7.17 7.28 7.77 7.09

7.22 7.45 6.95 7.40 7.10 7.32 7.14

Give as much information as you can about the P-value for the Wilcoxon test in each of the following situations.

a. \(n = 12,\)upper-tailed test, \({s_ + } = 56,\)

b. \(n = 12,\)upper-tailed test, \({s_ + } = 62,\)

c. \(n = 12,\)lower-tailed test, \({s_ + } = 20,\)

d. \(n = 14,\) two-tailed test, \({s_ + } = 21,\)

e. \(n = 25,\) two-tailed test, \({s_ + } = 300,\)

The ranking procedure described in Exercise 35 is somewhat asymmetric, because the smallest observation receives rank 1, whereas the largest receives rank 2, and so on. Suppose both the smallest and the largest receive rank 1, the second smallest and second largest receive rank 2, and so on, and let W鈥欌 be the sum of the X ranks. The null distribution of W鈥欌 is not identical to the null distribution of W, so different tables are needed. Consider the case m = 3, n = 4. List all 35 possible orderings of the three X values among the seven observations (e.g., 1, 3, 7 or 4, 5, 6), assign ranks in the manner described, compute the value of W鈥欌 for each possibility, and the tabulate the null distribution of W0. What is the P-value if w鈥欌 = 9? This is the Ansari-Bradley test; for additional information, see the book by Hollander and Wolfe in the chapter bibliography.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.