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Chapter 10: Problem 33

The manufacturer of a hot tub is interested in testing two different heating elements for his product. The element that produces the maximum heat gain after 15 minutes would be preferable. He 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 significant difference; Unit 1 and Unit 2 perform similarly. P-value is 0.714 (normal approximation).

Step by step solution

01

Identify the problem

We need to compare the heat gain performance of two heating units using data collected from samples. We will apply the Wilcoxon rank-sum test to see if there's a statistically significant difference in heat gains at a significance level of \(\alpha = 0.05\).
02

Organize the Data

List all the given heat gain data for Unit 1 and Unit 2, keeping them separate to calculate ranks later. Unit 1: 25, 27, 29, 31, 30, 26, 24, 32, 33, 38. Unit 2: 31, 33, 32, 35, 34, 29, 38, 35, 37, 30.
03

Combine and Rank the Data

Combine all the heat gain data from both units into a single list: 24, 25, 26, 27, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 37, 38, 38. Rank them from smallest to largest; assign average ranks for ties.
04

Calculate Ranks for Each Unit

Calculate the sum of ranks for each unit: - Unit 1 (combined ranks for values): 1, 2, 3, 4, 6.5, 7.5, 9, 10, 13.5, 20 (total rank sum = 77). - Unit 2 (combined ranks for values): 10, 13.5, 9, 16, 14, 6.5, 20, 16, 18, 7.5 (total rank sum = 123).
05

Apply the Wilcoxon Rank-Sum Test

The Wilcoxon Rank-Sum Test is used since the sample sizes are equal and the distribution might not be normal. Calculate the test statistic \( U \): \[ U_1 = n_1 \cdot n_2 + \frac{n_1(n_1+1)}{2} - R_1 = 10 \cdot 10 + \frac{10 \cdot 11}{2} - 77 = 45 \] \[ U_2 = n_1 \cdot n_2 + \frac{n_2(n_2+1)}{2} - R_2 = 10 \cdot 10 + \frac{10 \cdot 11}{2} - 123 = 55 \] Use the smaller \( U \) value to find the p-value, which in this case is 45.
06

Use Normal Approximation (Optional)

For the normal approximation, calculate the mean and standard deviation of the rank-sum under the null hypothesis: Mean: \( \mu_U = \frac{n_1n_2}{2} = 50 \)Standard deviation: \( \sigma_U = \sqrt{\frac{n_1n_2(n_1+n_2+1)}{12}} \approx 13.67 \)Calculate the \( z \)-value using: \[ z = \frac{U - \mu_U}{\sigma_U} = \frac{45 - 50}{13.67} \approx -0.366 \]Find the p-value from the z-table; p-value is approximately 0.714.
07

Conclude the Test

Since the p-value (0.714) is greater than \(\alpha = 0.05\), we do not reject the null hypothesis. There's no statistical evidence to suggest the superiority of one heating unit over the other based on heat gain after 15 minutes.

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Key Concepts

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

Non-parametric Statistics
Non-parametric statistics are a set of statistical methods that do not assume your data comes from a particular distribution, such as the normal distribution. This aspect is particularly useful when dealing with data that does not fit common distribution models, or when the sample size is small.

Non-parametric tests, like the Wilcoxon Rank-Sum Test, are often used when data is ordinal or not sufficiently continuous. They are beneficial because they are more flexible and can be applied more broadly across varying types of data. Instead of relying on parameters like the mean and variance, non-parametric statistics use ranks or medians to make inferences about data.
  • Results are not dependent on strict assumptions.
  • They accommodate outliers and data anomalies better than parametric tests.
  • Non-parametric tests are often simpler to calculate and can be more intuitive to understand.
It's important to note that while non-parametric tests provide flexibility, they can be less powerful than parametric methods, meaning they might not detect a significant effect when there actually is one. They are most effective when the assumptions of parametric tests are violated.
Hypothesis Testing
Hypothesis testing is a statistical method that helps us decide whether to reject a null hypothesis based on sample data. The process typically involves comparing a test statistic calculated from the sample to a theoretical distribution.

In the Wilcoxon Rank-Sum Test, hypothesis testing is used to determine if there is a significant difference between two independent samples. The null hypothesis (\(H_0\)) in this test generally states that there are no differences between the two groups, whereas the alternative hypothesis (\(H_1\)) suggests that there is a difference.
  • First, define a null hypothesis and an alternative hypothesis suitable for your test.
  • Determine a significance level (\(\alpha\)), commonly set at 0.05, which sets your tolerance for a Type I error (false positive).
  • Calculate the test statistic, rank sum, and correspondingly a \(p\)-value to determine the outcome.
The result from the hypothesis testing will help ascertain whether any observed effect is statistically meaningful.
Statistical Significance
Statistical significance is a critical concept in statistical testing. It helps to determine whether the difference or relationship observed in your data is large enough to be unlikely due to randomness or chance alone. This significance is measured by the \(p\)-value in hypothesis testing.

If the \(p\)-value is less than the chosen significance level (\(\alpha\)), the result is deemed statistically significant, leading to the rejection of the null hypothesis. For instance, with an \(\alpha\) of 0.05, a \(p\)-value smaller than 0.05 indicates that the probability of observing the data (or something more extreme) given that the null hypothesis is true, is less than 5%. Therefore, we might infer that our results support the alternative hypothesis.
  • A high \(p\)-value (> \(\alpha\)) indicates weak evidence against the null hypothesis.
  • A low \(p\)-value (鈮� \(\alpha\)) indicates strong evidence against the null hypothesis.
  • Statistical significance does not imply practical significance, and additional context should always be considered.
Understanding statistical significance is crucial as it helps to draw meaningful conclusions from data and inform decision-making.

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Most popular questions from this chapter

An experiment was conducted to compare the filling capability of packaging equipment at two different wineries. Ten bottles of pinot noir from Ridgecrest Vineyards were randomly selected and measured, along with 10 bottles of pinot noir from Valley View Vineyards. The data are as follows (fill volume is in milliliters): (a) What assumptions are necessary to perform a hypothesistesting procedure for equality of means of these data? Check these assumptions. (b) Perform the appropriate hypothesis-testing procedure to determine whether the data support the claim that both wineries will fill bottles to the same mean volume. (c) Suppose that the true difference in mean fill volume is as much as 2 fluid ounces; did the sample sizes of 10 from each vineyard provide good detection capability when \(\alpha=0.05 ?\) Explain your answer.

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The fill volume can be assumed normal, with standard deviation \(\sigma_{1}=0.020\) and \(\sigma_{2}=0.025\) ounces. A member of the quality engineering staff suspects that both machines fill to the same mean net volume, whether or not this volume is 16.0 ounces. A random sample of 10 bottles is taken from the output of each machine. (a) Do you think the engineer is correct? Use \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Calculate a \(95 \%\) confidence interval on the difference in means. Provide a practical interpretation of this interval. (c) What is the power of the test in part (a) for a true difference in means of \(0.04 ?\) (d) Assuming equal sample sizes, what sample size should be used to assure that \(\beta=0.05\) if the true difference in means is 0.04 ? 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 \(\alpha=0.05 .\)

Two different foam expanding agents that can be used in the nozzles of fire- fighting spray equipment. A random sample of five observations with an aqueous film-forming foam (AFFF) had a sample mean of 4.7 and a standard deviation of \(0.6 .\) A random sample of five observations with alcohol-type concentrates (ATC) had a sample mean of 6.9 and a standard deviation 0.8 . (a) Can you draw any conclusions about differences in mean foam expansion? Assume that both populations are well represented by normal distributions with the same standard deviations. (b) Find a \(95 \%\) confidence interval on the difference in mean foam expansion of these two agents.

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?
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