/*! 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 43 The manufacturer of a hot tub is... [FREE SOLUTION] | 91Ó°ÊÓ

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The manufacturer of a hot tub is interested in testing two different heating elements for its product. The element that produces the maximum heat gain after 15 minutes would be preferable. The manufacturer obtains 10 samples of each heating unit and tests each one. The heat gain after 15 minutes (in \({ }^{\circ} \mathrm{F}\) ) follows. $$\begin{array}{l|l}\text { Unit 1: } & 25,27,29,31,30,26,24,32,33,38 \\\\\hline \text { Unit 2: } & 31,33,32,35,34,29,38,35,37,30\end{array}$$ (a) Is there any reason to suspect that one unit is superior to the other? Use \(\alpha=0.05\) and the Wilcoxon rank sum test. (b) Use the normal approximation for the Wilcoxon rank sum test. Assume that \(\alpha=0.05 .\) What is the approximate \(P\) -value for this test statistic?

Short Answer

Expert verified
No, there is no statistical reason to suspect one unit is superior (P-value = 0.65).

Step by step solution

01

Understanding the Wilcoxon Rank Sum Test

The Wilcoxon rank sum test compares two independent samples to determine if they arise from the same distribution. It is a non-parametric method used when the data does not necessarily follow a normal distribution.
02

Assign Ranks to the Combined Data

Combine the samples from Unit 1 and Unit 2: \( 25, 27, 29, 31, 30, 26, 24, 32, 33, 38, 31, 33, 32, 35, 34, 29, 38, 35, 37, 30 \). Sort and rank these values, giving the average rank to ties.
03

Calculate Sum of Ranks

Calculate the sum of the ranks for each unit. After assigning ranks: - Unit 1: Sum of ranks = 8 + 10 + 3.5 + 12.5 + 11 + 6 + 1 + 13 + 15 + 19 = 99 - Unit 2: Sum of ranks = 12.5 + 15 + 13 + 17 + 16 + 8 + 19 + 17 + 18 + 10 = 141
04

Determine Test Statistic

The test statistic for the Wilcoxon rank sum test is the smaller of the two rank sums, which in this case is Unit 1's sum = 99.
05

Compare To Critical Value

Use the Wilcoxon rank sum test table for a 0.05 significance level with sample sizes \( n_1 = 10 \) and \( n_2 = 10 \). Find the critical value. Here, the critical value is 85. Since 99 > 85, we fail to reject the null hypothesis.
06

Normal Approximation

For larger samples, a normal approximation can be used. Calculate the mean:\[ \mu = \frac{n_1(n_1 + n_2 + 1)}{2} = \frac{10(10 + 10 + 1)}{2} = 105 \]Calculate the standard deviation:\[ \sigma = \sqrt{\frac{n_1 n_2(n_1 + n_2 + 1)}{12}} = \sqrt{\frac{10 \times 10 \times 21}{12}} = \sqrt{175} = 13.23 \]
07

Calculate Z-value

Calculate the Z-value using the normal approximation:\[ Z = \frac{W - \mu}{\sigma} = \frac{99 - 105}{13.23} = -0.454 \]
08

Compute P-value and Conclusion

Using the standard normal distribution, find the P-value for \( Z = -0.454 \), which yields a P-value of approximately 0.65. Since 0.65 > 0.05, we fail to reject the null hypothesis.

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Key Concepts

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

Non-parametric Statistical Test
Non-parametric tests are statistical tests that are not based on parameterized families of probability distributions, unlike parametric tests. This is useful in scenarios where data does not follow a normal distribution, or when we have ordinal data and wish to compare two groups. The Wilcoxon Rank Sum Test is a type of non-parametric test that evaluates two independent samples to determine if they have differences in their central tendencies.
One of the main advantages of this test is that it does not assume normal distribution, making it versatile in many experimental conditions.
  • Used when data is not normally distributed
  • Effective for small sample sizes
  • Compares medians between two groups
Ultimately, the Wilcoxon test evaluates rank differences rather than actual data points to make inferences about the populations, enhancing its applicability to a wider range of datasets.
Heat Gain Measurement
In the context of this exercise, heat gain measurement refers to recording temperatures to assess the performance of heating elements. The manufacturer uses this measurement to evaluate which heating unit provides better performance over a set period, resulting in higher temperatures.
From the provided data:
  • Unit 1 sample values indicate heat gains of 25 to 38 degrees Fahrenheit
  • Unit 2 sample values vary from 29 to 38 degrees Fahrenheit
By understanding these measurements, we can accurately analyze the data using statistical methods like the Wilcoxon Rank Sum Test. It helps determine if there's a significant difference in the heat gain contributions of the two units.
Test Statistic Calculation
The calculation of the test statistic is central to implementing the Wilcoxon Rank Sum Test. After combining and ranking data from both samples, the sum of ranks is calculated for each unit.
The test statistic is determined by selecting the smaller sum of ranks. In this exercise, the multitude of ranks includes ties, which are resolved by assigning average ranks. Here, the rank sums are:
  • Unit 1: Sum of ranks = 99
  • Unit 2: Sum of ranks = 141
The test statistic for the Wilcoxon test is the sum of ranks for the smaller sample because it helps evaluate whether this sum is significantly lower than expected if there were no differences between groups.
Normal Approximation Method
The normal approximation method is used when the sample size is large enough, allowing parametric techniques to approximate results of non-parametric tests. In this case, the normal approximation helps to simplify the Wilcoxon Rank Sum Test calculations.
Calculating the normal approximation involves these steps:
  • The mean ( \( \mu \) ) is computed by: \( \mu = \frac{n_1(n_1 + n_2 + 1)}{2} \)
  • The standard deviation ( \( \sigma \) ) is provided by: \( \sigma = \sqrt{\frac{n_1 n_2(n_1 + n_2 + 1)}{12}} \)
  • The Z-value, expressing the deviation from the mean, is determined by: \( Z = \frac{W - \mu}{\sigma} \)
Using this Z-value, one can easily find the P-value from the standard normal distribution, helping decide whether to reject or accept the null hypothesis based on the significance level.

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Most popular questions from this chapter

A random sample of 500 adult residents of Maricopa County indicated that 385 were in favor of increasing the highway speed limit to \(75 \mathrm{mph}\), and another sample of 400 adult residents of Pima County indicated that 267 were in favor of the increased speed limit. (a) Do these data indicate that there is a difference in the support for increasing the speed limit for the residents of the two counties? Use \(\alpha=0.05 .\) What is the \(P\) -value for this test?

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Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst \(1,\) resulting in an average yield of 86 and a sample standard deviation of \(3 .\) Fifteen batches were prepared using catalyst \(2,\) and they resulted in an average yield of 89 with a standard deviation of 2 . Assume that yield measurements are approximately normally distributed with the same standard deviation. (a) Is there evidence to support a claim that catalyst 2 produces a higher mean yield than catalyst \(1 ?\) Use \(\alpha=0.01\). (b) Find a \(99 \%\) confidence interval on the difference in mean yields that can be used to test the claim in part (a).

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