/*! 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 7 Acid Rain In \(2002,\) the media... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 7

Acid Rain In \(2002,\) the median pH level of the rain in Glacier National Park, Montana, was \(5.25 .\) A biologist thinks that the acidity of rain has decreased since then, which would suggest the pH has increased. She obtains a random sample of 15 rain dates in 2015 and obtains the following data: $$ \begin{array}{lllll} \hline 5.31 & 5.19 & 5.55 & 5.38 & 5.37 \\ \hline 5.19 & 5.26 & 5.29 & 5.27 & 5.19 \\ \hline 5.27 & 5.36 & 5.22 & 5.28 & 5.24 \\ \hline \end{array} $$ Test the hypothesis that the median pH level has increased from 5.25 at the \(\alpha=0.05\) level of significance.

Short Answer

Expert verified
Do not reject the null hypothesis; no evidence that the median pH has increased.

Step by step solution

01

State the Hypotheses

Formulate the null and alternative hypotheses. The null hypothesis is that the median pH level has not changed from 5.25, while the alternative hypothesis is that the median pH level has increased. Null Hypothesis (\(\text{H}_0\)): The median pH level is 5.25. Alternative Hypothesis (\(\text{H}_a\)): The median pH level is greater than 5.25.
02

Select the Test

Since the sample size is small, use the Wilcoxon Signed-Rank Test to compare the median pH level to 5.25.
03

Calculate Differences

Calculate the differences between each sample value and 5.25. The differences are: \(5.31-5.25=0.06,\) \(5.19-5.25=-0.06,\) \(5.55-5.25=0.30,\) \(5.38-5.25=0.13,\) \(5.37-5.25=0.12,\) \(5.19-5.25=-0.06,\) \(5.26-5.25=0.01,\) \(5.29-5.25=0.04,\) \(5.27-5.25=0.02,\) \(5.19-5.25=-0.06,\) \(5.27-5.25=0.02,\) \(5.36-5.25=0.11,\) \(5.22-5.25=-0.03,\) \(5.28-5.25=0.03,\) \(5.24-5.25=-0.01\).
04

Rank the Absolute Differences

Rank the absolute values of the differences, ignoring signs, and assign ranks starting from 1. If there are ties, assign the average of the ranks to each tied value.
05

Assign Signs and Sum Ranks

Assign ranks to the differences while keeping their signs. Then, sum the ranks of positive differences (let this sum be \(W^+\)) and the ranks of negative differences (let this sum be \(W^-\)). Ranked absolute differences: \(0.30 (1),\) \(0.13 (2),\) \(0.12 (3),\) \(0.11 (4),\) \(0.06 (5),\) \(0.06 (5),\) \(0.04 (7),\) \(0.03 (8),\) \(0.03 (8),\) \(0.02 (10),\) \(0.02 (10),\) \(0.01 (12),\) \(0.01 (12),\) \(0.01 (12),\) \(0.01 (12).$$W^+=9+2+3+4+5+8+8+12+10+12=83.\) \(W^- =5+5+12 =22.\)
06

Determine Critical Value and Decision Rule

\(W^+\) = 83 Critical value = 95
07

Conclusion

Since \(W^+\) is not greater than the critical value, do not reject the null hypothesis. Hence, at the \(0.05\) significance level, there is not enough evidence to conclude that the median pH level has increased.

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!

Key Concepts

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

null hypothesis
In hypothesis testing, the null hypothesis is a statement that indicates no effect or no difference. It represents a status quo belief where any observed difference is presumed to be due to sampling or measurement error.
For instance, in the Acid Rain exercise, the null hypothesis is that the median pH level of the rain in Glacier National Park has not changed from 5.25 since 2002.
We denote it as \(H_0\): The median pH level is 5.25. This means that we assume the median pH level remains consistent unless proven otherwise by the sample data.
The null hypothesis serves as a starting point for statistical analysis and is essential for determining if the observed data deviates significantly from what was expected.
alternative hypothesis
The alternative hypothesis contrasts the null hypothesis. It indicates the presence of an effect or a difference. In statistical tests, it suggests that the observed data does indeed show a significant change from the status quo.
In the given exercise, the alternative hypothesis is that the median pH level has increased since 2002. This is represented as \(H_a\): The median pH level is greater than 5.25.
The alternative hypothesis can be one-sided or two-sided. Here, it is one-sided because we are specifically testing whether the pH has increased, not just changed.
If the test statistics provide sufficient evidence to support the alternative hypothesis, we reject the null hypothesis, indicating a significant change in the measured variable.
Wilcoxon Signed-Rank Test
The Wilcoxon Signed-Rank Test is a non-parametric test used to compare paired samples to determine if there is a significant difference in their median values. It is useful when the sample size is small and does not assume a normal distribution of the data.
Here is how the test was applied in the Acid Rain exercise:
  • Step 1: Calculating the differences between each sample value and the median (5.25).
  • Step 2: Ranking the absolute values of these differences and ignoring the signs.
  • Step 3: Assigning signs to the ranks and summing the ranks of positive and negative differences separately (denoted as W+ and W-).
  • Step 4: Comparing the sum of the ranks to a critical value to make a decision.

In this exercise, we followed these steps meticulously. We calculated the differences, ranked the absolute values, added up the ranks of positive differences (W+ = 83) and compared it to the critical value. Since W+ was not greater than the critical value of 95, we did not reject the null hypothesis. This indicated that, at the 0.05 significance level, there isn't enough evidence to conclude that the median pH level had increased.

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

Effects of Exercise To determine the effectiveness of an exercise regimen, a physical therapist randomly selects 10 women to participate in a study. She measures their waistlines (in inches) both before and after a rigorous 8 -week exercise program and obtains the data shown. Is the median waistline before the exercise program more than the median waistline after the exercise program? Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{ccc|ccc} \text { Woman } & \text { Before } & \text { After } & \text { Woman } & \text { Before } & \text { After } \\ \hline 1 & 23.5 & 19.75 & 6 & 19.75 & 19.5 \\ \hline 2 & 18.5 & 19.25 & 7 & 35 & 34.25 \\ \hline 3 & 21.5 & 21.75 & 8 & 36.5 & 35 \\ \hline 4 & 24 & 22.5 & 9 & 52 & 51.5 \\ \hline 5 & 25 & 25 & 10 & 30 & 31 \\ \hline \end{array} $$

Describe the difference between parametric statistical procedures and nonparametric statistical procedures.

Outpatient Treatment The median length of stay for male substance-abuse outpatient treatment completers is 105 days. The following data represent the length of stays for a random sample of female substance-abuse outpatient treatment completers. $$ \begin{array}{rrrrrr} \hline 117 & 111 & 81 & 110 & 103 & 94 \\ \hline 92 & 91 & 116 & 108 & 98 & 84 \\ \hline \end{array} $$ Using the Wilcoxon signed-ranks test at the \(\alpha=0.05\) level of significance, does the median length of stay seem different for males and females?

Problems 17 and 18 illustrate the use of the sign test to test hypotheses regarding a population proportion. The only requirement for the sign test is that our sample be obtained randomly. When dealing with nominal data, we can identify a characteristic of interest and then determine whether each individual in the sample possesses this characteristic. Under the null hypothesis in the sign test, we expect that half of the data will result in minus signs and half in plus signs. If we let a plus sign indicate the presence of the characteristic (and a minus sign indicate the absence), we expect half of our sample to possess the characteristic while the other half will not. Letting \(p\) represent the proportion of the population that possesses the characteristic, our null hypothesis will be \(H_{0}: p=0.5 .\) Use the sign test for Problems 17 and 18 , following the sign convention indicated previously. Trusting the Press In a study of 2302 U.S. adults surveyed online by Harris Interactive 1243 respondents indicated that they tend to not trust the press. Using an \(\alpha=0.05\) level of significance, does this indicate that more than half of U.S. adults tend to not trust the press?

Zillow.com allows people to research real estate listings. One feature is the "Zestimate," which represents the price Zillow believes the property should sell for. If the Zestimate is accurate, the median difference between the actual selling price and the Zestimate should be zero. The following data represent the difference between the actual selling price and Zestimate for a random sample of 10 real estate transactions. Do the data suggest the Zestimate is inaccurate? Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{rrrrr} \hline 2500 & -500 & -79,500 & -13,500 & -500 \\ \hline-1000 & 14,500 & 3,500 & -12,500 & -6000 \\ \hline \end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation
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.