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Magazine Chapter 15: Q2E (page 660)
Here again is the data on expense ratio (%) for a sample of 20 large cap blended mutual funds introduced in Exercise 1.53:
1.03 1.23 1.10 1.64 1.30 1.27 1.25
.78 1.05 .64 .94 .86 1.05 .75
.09 0.79 1.61 1.26 .93 .84
A normal probability plot shows a distinctly nonlinear pattern, primarily because of the single outlier on each end of the data. But a dotplot and boxplot exhibit a reasonable amount of symmetry. Assuming a symmetric population distribution, does the data provide compelling evidence for concluding that the population mean expense ratio exceeds 1%? Use the Wilcoxon test at significance level .1. (Note: The mean expense ratio for the population of all 825 such funds is actually 1.08.)
Short Answer
Do not reject null hypothesis.
Step by step solution
testing null hypothesis:
When testing null hypothesis
\({H_0}:\mu = {\mu _0}\)
Versus one of the alternative hypothesis, one could use test static value
\({s_ + }\)= the sum of the ranks associated with positive\(\left( {{x_i} = {\mu _0}} \right)'s\)
The P-value depends on the alternative hypothesis
Alternative hypothesis P-values
\(\begin{array}{l}{H_0}:\mu > {\mu _0}\\{H_0}:\mu < {\mu _0}\\{H_0}:\mu \ne {\mu _0}\end{array}\) \(\begin{array}{l}{P_0}\left( {{S_ + } \ge {s_ + }} \right)\\{P_0}\left( {{S_ + } \le {s_ + }} \right) = {P_0}\left( {{S_ + } \ge \frac{{n\left( {n + 1} \right)}}{2} - {s_ + }} \right)\\2{P_0}\left( {{S_ + } \ge \max \left\{ {{s_ + },\frac{{n\left( {n + 1} \right)}}{2} - {s_ + }} \right\}} \right)\end{array}\)
solving further:
The test of interest is
\({H_0}:\mu = 1\)
Versus alternative hypothesis
\({H_a}:\mu > 1\)
The following table represents values required to compute the test statistic value and corresponding P-value. Value 鈥�-1鈥� represents 鈥�-鈥� as a sing (negative difference), and value 1 is 鈥�-+鈥�.
i | \({x_i}\) | \({y_i} = \left| {{x_i} - 1} \right|\) | Rank(\({y_i}\)) | Sign |
1 | 1.03 | 0.03 | 1 | 1 |
2 | 1.23 | 0.23 | 11 | 1 |
3 | 1.1 | 0.1 | 6 | 1 |
4 | 1.64 | 0.64 | 19 | 1 |
5 | 1.3 | 0.3 | 16 | 1 |
6 | 1.27 | 0.27 | 15 | 1 |
7 | 1.25 | 0.25 | 12.5 | 1 |
8 | 0.78 | 0.22 | 10 | -1 |
9 | 1.05 | 0.05 | 2.5 | 1 |
10 | 0.64 | 0.36 | 17 | -1 |
11 | 0.94 | 0.06 | 4 | -1 |
12 | 0.86 | 0.14 | 7 | -1 |
13 | 1.05 | 0.05 | 2.5 | 1 |
14 | 0.75 | 0.25 | 12.5 | -1 |
15 | 0.09 | 0.91 | 20 | -1 |
16 | 0.79 | 0.21 | 9 | -1 |
17 | 1.61 | 0.61 | 18 | 1 |
18 | 1.26 | 0.26 | 14 | 1 |
19 | 0.93 | 0.07 | 5 | -1 |
20 | 0.84 | 0.16 | 8 | -1 |
test static value:
The test statistic value is,
\({s_ + }\)= the sum of the ranks associated with positive\(\left( {x\_i - \mu \_0} \right)'s\)
\(\begin{array}{l} = 1 + 2.5 + 2.5 + ... + 19\\ = 117.5\end{array}\)
In the table in the appendix one could find only particular values ; thus, for n = 20 use table to obtain
\(\begin{array}{l}P = {P_0}\left( {{S_ + } \ge {s_ + }} \right) > {P_0}\left( {{S_ + } \ge 140} \right) = 0.101\\P > 0.1 = \alpha \end{array}\)
Do not reject null hypothesis
Normal probability plot:
Hence, do not reject null hypothesis.
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