/*! 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} Problem 31 An electrical engineer must desi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electrical engineer must design a circuit to deliver the maximum amount of current to a display tube to achieve sufficient image brightness. Within her allowable design constraints, she has developed two candidate circuits and tests prototypes of each. The resulting data (in microamperes) are as follows: $$\begin{array}{l|l}\text { Circuit } 1: & 251,255,258,257,250,251,254,250,248 \\\\\hline \text { Circuit } 2: & 250.253 .249 .256 .259 .252 .260 .251\end{array}$$ (a) Use the Wilcoxon rank-sum test to test \(H_{0}: \mu_{1}=\mu_{2}\) against the alternative \(H_{1}: \mu_{1}>\mu_{2} .\) Use \(\alpha=0.025 .\) (b) Use the normal approximation for the Wilcoxon rank-sum test. Assume that \(\alpha=0.05 .\) Find the approximate \(P\) -value for this test statistic.

Short Answer

Expert verified
(a) Do not reject \( H_0 \); (b) \( P \)-value > 0.05, so do not reject \( H_0 \).

Step by step solution

01

Combine and Rank the Data

To begin, list all microampere data from Circuit 1 and Circuit 2 together. Then, rank all data points from smallest to largest, regardless of the circuit they come from. Assign ranks to each observation, with average ranks for ties. For Circuit 1, the values are \( 251, 255, 258, 257, 250, 251, 254, 250, 248 \) and for Circuit 2, the values are \( 250, 253, 249, 256, 259, 252, 260, 251 \).
02

Calculate Rank Sums

After assigning ranks to all data points, determine the sum of ranks for Circuit 1 and the sum of ranks for Circuit 2. These are known as \( R_1 \) for Circuit 1 and \( R_2 \) for Circuit 2, respectively.
03

Determine the Test Statistic

The test statistic for the Wilcoxon rank-sum test is the smaller of \( R_1 \) and \( R_2 \). Calculate \( U_1 \), the rank-sum test statistic for Circuit 1, using the formula:\[ U_1 = n_1 n_2 + \frac{n_1(n_1 + 1)}{2} - R_1 \]where \( n_1 \) is the number of observations in Circuit 1 and \( n_2 \) is the number of observations in Circuit 2.
04

Obtain Critical Value and Conclusion for Part (a)

With \( \alpha = 0.025 \), look up the critical value for the Wilcoxon rank-sum test for the given sample sizes in statistical tables. Compare the test statistic \( U_1 \) against this critical value to decide whether to reject or fail to reject \( H_0 \). If \( U_1 \) is less than or equal to the critical value, reject \( H_0 \) in favor of \( H_1 \).
05

Normal Approximation for the Wilcoxon Test

For part (b), calculate the mean \( \mu_U \) and the standard deviation \( \sigma_U \) of \( U \) using the formulas:\[ \mu_U = \frac{n_1 n_2}{2} \]\[ \sigma_U = \sqrt{\frac{n_1 n_2(n_1 + n_2 + 1)}{12}} \]Then, determine the Z-score as:\[ Z = \frac{U_1 - \mu_U}{\sigma_U} \]
06

Calculate the P-value

Consult standard normal distribution tables to find the \( P \)-value associated with the calculated \( Z \)-score. Compare this \( P \)-value to the significance level \( \alpha = 0.05 \). If the \( P \)-value is less than \( \alpha \), reject the null hypothesis \( H_0 \) that \( \mu_1 = \mu_2 \); otherwise, do not 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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-parametric Statistics
Non-parametric statistics are a branch of statistics that do not assume a specific distribution for the data, such as the normal distribution. This is particularly useful when dealing with real-world data that might not meet the assumptions required by parametric tests—like normality. In the context of the Wilcoxon rank-sum test, non-parametric methods are helpful because they allow testing hypotheses on the medians of two groups without assuming a normal distribution for the data.

The strength of non-parametric tests lies in their versatility and robustness. They are applicable to a wider range of data types and distribution shapes compared to parametric tests. Some benefits include:
  • Handling outliers: Because they use ranks instead of raw data values, non-parametric tests are robust to outliers that would affect the mean in parametric tests.
  • Small sample sizes: Non-parametric tests can be more reliable than parametric tests when dealing with small samples.
  • Ordinal data: Apart from interval data, non-parametric tests are capable of analyzing data measured on an ordinal scale.
These tests play critical roles in hypothesis testing, especially when the data do not satisfy the conditions necessary for parametric tests.
Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics used to determine whether there is enough evidence to reject a null hypothesis, based on sample data. In simple terms, it is a way to test an assumption about a population parameter.

In the case of the Wilcoxon rank-sum test, we aim to test the null hypothesis \( H_0: \mu_1 = \mu_2 \) against the alternative \( H_1: \mu_1 > \mu_2 \). This test is particularly useful when the conditions for a parametric test, such as the t-test, are not met.

The hypothesis testing process involves several steps:
  • State the null and alternative hypotheses.
  • Determine the significance level \( \alpha \), which is the probability of rejecting the null hypothesis when it is actually true.
  • Collect data and calculate the test statistic based on the ranks of sample data.
  • Compare the test statistic with a critical value or calculate the p-value.
  • Make a decision: reject or do not reject the null hypothesis.
The outcome of a hypothesis test helps determine if there is a statistically significant difference between the two groups being compared.
Normal Approximation
Normal approximation is a technique used to approximate the distribution of a non-parametric test statistic with a normal distribution. This is particularly useful when dealing with large sample sizes, where calculating exact probabilities becomes cumbersome.

In the context of the Wilcoxon rank-sum test, when the sample sizes are large, the distribution of the test statistic \( U \) approximately follows a normal distribution. This allows us to use the normal approximation to facilitate the calculation of p-values.

To implement normal approximation:
  • Calculate the mean \( \mu_U \) of the distribution of \( U \) using the formula \( \mu_U = \frac{n_1 n_2}{2} \).
  • Calculate the standard deviation \( \sigma_U \) using the formula \( \sigma_U = \sqrt{\frac{n_1 n_2(n_1 + n_2 + 1)}{12}} \).
  • Determine the z-score from the distribution of \( U \) by \( Z = \frac{U_1 - \mu_U}{\sigma_U} \).
The calculated z-score can then be used to find the p-value from standard normal distribution tables.

Using normal approximation simplifies the hypothesis testing process significantly when assumptions of the Wilcoxon rank-sum test are met for larger samples.

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

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}<\mu_{2} .\) Suppose that sample sizes \(n_{1}=15\) and \(n_{2}=15\) that \(\bar{x}_{1}=6.2\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05 .\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 3 units less than \(\mu_{2} ?\) (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2.5 units less than \(\mu_{2} ?\) Assume that \(\alpha=0.05\)

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2} .\) Suppose that sample sizes are \(n_{1}=15\) and \(n_{2}=15,\) that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) for a true difference in means of \(3 ?\) (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if the true difference in means is \(-2 ?\) Assume that \(\alpha=0.05 .\)

The deflection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 15 pipe specimens are tested, and the deflection temperatures observed are as follows (in \({ }^{\circ} \mathrm{F}\) ): $$\begin{array}{l}\text { Type } 1: 206,188,205,187,194,193,207,185,189,213,192,210,194,178,205 \end{array}$$ $$\begin{array}{l}\text { Type } 2: 177,197,206,201,180,176,185,200,197,192, 198,188,189,203,192 \end{array}$$ (a) Construct box plots and normal probability plots for the two samples. Do these plots provide support of the assumptions of normality and equal variances? Write a practical interpretation for these plots. (b) Do the data support the claim that the deflection temperature under load for type 1 pipe exceeds that of type \(2 ?\) In reaching your conclusions, use \(\alpha=0.05 .\) Calculate a \(P\) -value (c) If the mean deflection temperature for type 1 pipe exceeds that of type 2 by as much as \(5^{\circ} \mathrm{F}\), it is important to detect this difference with probability at least \(0.90 .\) Is the choice of \(n_{1}=n_{2}=15\) adequate? Use \(\alpha=0.05\)

Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but we suspect that the variability in concentration may differ between the two companies. The standard deviation of concentration in a random sample of \(n_{1}=10\) batches produced by company 1 is \(s_{1}=4.7\) grams per liter, while for company \(2,\) a random sample of \(n_{2}=16\) batches yields \(s_{2}=5.8\) grams per liter. Is there sufficient evidence to conclude that the two population variances differ? Use \(\alpha=0.05 .\)

The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process. The standard deviation of active concentration is known to be 3 grams per liter, regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data follow: $$\begin{aligned} \text { Catalyst } 1: & 57.9,66.2,65.4,65.4,65.2,62.6,67.6,63.7, 67.2,71.0\end{aligned}$$ $$\begin{array}{l} \text { Catalyst 2: } 66.4,71.7,70.3,69.3 ,64.8,69.6,68.6,69.4,65.3,68.8\end{array}$$ (a) Find a \(95 \%\) confidence interval on the difference in mean active concentrations for the two catalysts. Find the \(P\) -value. (b) Is there any evidence to indicate that the mean active concentrations depend on the choice of catalyst? Base your answer on the results of part (a). (c) Suppose that the true mean difference in active concentration is 5 grams per liter. What is the power of the test to detect this difference if \(\alpha=0.05 ?\) (d) If this difference of 5 grams per liter is really important, do you consider the sample sizes used by the experimenter to be adequate? Does the assumption of normality seem reasonable for both samples?

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.