Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Use the information given in Exercises \(8-9\) to calculate Spearman's rank correlation coefficient \(r_{s} .\) Do the data present sufficient evidence to indicate an association between variables \(A\) and \(B\) ? Use \(\alpha=.05 .\) $$\begin{array}{l|rrrrrr}\text { A } & 1.2 & .8 & 2.1 & 3.5 & 2.7 & 1.5 \\\\\hline \text { B } & 1.0 & 1.3 & .1 & -.8 & -.2 & .6\end{array}$$

Short Answer

Expert verified
Answer: No, there is not enough evidence to indicate a significant association between variables A and B at the 0.05 significance level.

Step by step solution

01

1. Rank the values of A and B

First, we will rank the values in ascending order for both variables A and B. Assign a rank of \(1\) to the smallest value, \(2\) to the second smallest value, and so on. A ranks: \((2, 1, 3, 6, 5, 4)\) B ranks: \((4, 5, 2, 1, 3, 6)\)
02

2. Calculate the difference between the ranks

Next, we will find the difference between the ranks of each pair of corresponding values for A and B. Let's call the difference \(d_i\) for each pair \((i=1,2,...,6)\). \(d_i = (\text{rank of A}_i - \text{rank of B}_i)\) \(d = (-2, -4, 1, 5, 2, -2)\)
03

3. Square the differences and sum them up

Now, we will square each of the differences and calculate the sum of these squared differences. \(d^2 = (4, 16, 1, 25, 4, 4)\) \(\sum d^2 = 54\)
04

4. Calculate \(r_s\)

We can now calculate Spearman's rank correlation coefficient \(r_s\) using the formula: $$r_s = 1 - \frac{6 \sum d^2}{n(n^2-1)}$$ where \(n\) is the number of pairs. In our case, \(n=6\), so the formula becomes: $$r_s = 1 - \frac{6 \cdot 54}{6(6^2-1)}$$ $$r_s = 1 - \frac{324}{210} \approx 0.453$$
05

5. Test the association

Finally, we will test whether the data presents sufficient evidence to indicate an association between A and B using a \(0.05\) significance level. We can use the critical value for the Spearman's rank correlation. Looking up on a table for Spearman's rank correlation coefficients at \(n=6\) and significance level \(\alpha=0.05\), we find a critical value of approximately \(0.829\). Since the calculated \(r_s\) value of \(0.453\) is less than the critical value of \(0.829\), we conclude that there is not enough evidence to indicate a significant association between variables A and B at the \(.05\) significance level.

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 Vaia!

Key Concepts

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

Correlation Analysis
Correlation analysis is a statistical method used to determine the strength and direction of the linear relationship between two variables. The goal is to find out how closely the changes in one variable are associated with changes in another.

In the given exercise, the Spearman's rank correlation coefficient (denoted as \( r_s \)) was used as a measure of correlation. This particular form of correlation analysis is useful when the data does not necessarily come from a normal distribution or when we deal with ordinal data—data that can be ranked but not necessarily measured on an interval scale.

It operates by assigning ranks to the data and then assessing how well the ranks between two variables correspond to each other. The Spearman's rank correlation coefficient is a nonparametric measure, meaning it does not assume that the data is normally distributed, making it quite versatile in application.
Nonparametric Statistics
Nonparametric statistics are a branch of statistics that is not reliant on data belonging to any particular distribution. This type of statistics is particularly useful when the data does not meet the assumptions of parametric tests, for instance, if the data is not normally distributed or if the sample size is too small to determine the distribution.

Spearman's rank correlation is a notable nonparametric statistic because it is based on the ranks of the data rather than the actual data points, which circumvents the need for the data to fit a normal distribution. Furthermore, nonparametric methods are often more robust to outliers, as they rely on the relative ordering of values rather than their specific quantities.

In practice, nonparametric methods can be advantageous when dealing with real-world data that may not adhere to the strict assumptions required for parametric testing, ensuring more reliable and valid results under a wider range of circumstances.
Statistical Significance
Statistical significance is a fundamental concept in hypothesis testing that helps researchers determine whether the results observed in their data are likely due to chance or if they reflect actual effects or associations. The significance level, denoted as \( \alpha \), is a threshold set by researchers, below which a p-value suggests that the observed results would be very unlikely under the null hypothesis. A common choice for \( \alpha \) is 0.05, meaning there is a 5% risk of concluding that an effect or an association exists when there is none.

In the context of Spearman's rank correlation coefficient, the statistical significance of the observed value of \( r_s \) is determined by comparing it to a critical value from a reference table based on the chosen \( \alpha \) level and the sample size. If the observed \( r_s \) is greater than the critical value, the association is considered to be statistically significant. Otherwise, as in the exercise, if the calculated \( r_s \) is below the critical value, there is insufficient evidence to conclude a significant association.

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

Use the information given in Exercises \(4-7\) to calculate Spearman's rank correlation coefficient, where \(x_{i}\) and \(y_{i}\) are the ranks of the ith pair of observations and \(d_{i}=x_{i}-y_{i} .\) Assume that there are no ties in the ranks. \(\sum d_{i}^{2}=16 ; n=10\)

To compare the effects of three toxic chemicals, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) on the skin of rats, 2 -centimeter-side squares of skin were treated with the chemicals and then scored from 0 to 10 depending on the degree of irritation. Three adjacent 2-centimeter-side squares were marked on the backs of eight rats, and each of the three chemicals was applied to each rat. Thus, the experiment was blocked on rats to eliminate the variation in skin sensitivity from rat to rat. a. Do the data provide sufficient evidence to indicate a difference in the toxic effects of the three chemicals? Test using the Friedman \(F_{r}\) -test with \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test and interpret it.

Property Values In Chapter 10 , you compared the property evaluations of two tax assessors. \(A\) and \(B\), as shown in the table below. $$ \begin{array}{ccc} \hline \text { Property } & \text { Assessor } \mathrm{A} & \text { Assessor B } \\ \hline 1 & 276.3 & 275.1 \\ 2 & 288.4 & 286.8 \\ 3 & 280.2 & 277.3 \\ 4 & 294.7 & 290.6 \\ 5 & 268.7 & 269.1 \\ 6 & 282.8 & 281.0 \\ 7 & 276.1 & 275.3 \\ 8 & 279.0 & 279.1 \\ \hline \end{array} $$ a. Use the sign test to determine whether the data present sufficient evidence to indicate that assessor A tends to give higher assessments than assessor B; that is, \(P\left(x_{\mathrm{A}}\right.\) exceeds \(\left.x_{\mathrm{B}}\right)>1 / 2\). Test by using a value of \(\alpha\) near. \(05 .\) Find the \(p\) -value for the test and interpret its value. b. In Chapter 10 , we used the \(t\) statistic to test the null hypothesis that assessor A tends to give higher assessments than assessor \(\mathrm{B}\), resulting in a \(t\) -value of \(t=2.82\) with \(p\) -value \(=.013 .\) Do these test results agree with the results in part a? Explain why the answers are (or are not) consistent.

Advertising Campaigns The results of an investigation of product recognition following three advertising campaigns were reported in Example \(11.15 .\) The responses were the percentage adults in 15 different groups who were familiar with the newly advertised product. The normal probability plot indicated that the data were not approximately normal and another method of analysis should be used. Is there a significant difference among the three population distributions from which these samples came? Use an appropriate nonparametric method to answer this question. $$ \begin{array}{lcc} \hline & \text { Campaign } & \\ \hline 1 & 2 & 3 \\ \hline 33 & .28 & 21 \\ .29 & .41 & 30 \\ 21 & .34 & 26 \\ 32 & .39 & .33 \\ .25 & .27 & .31 \\ \hline \end{array} $$

Lead Levels in Blood A study reported in the American Journal of Public Health (Science News) followed blood lead levels in handgun hobbyists using indoor firing ranges. \({ }^{6}\) Lead exposure measurements were made on 17 members of a law enforcement trainee class before, during, and after a 3 -month period of firearm instruction at an indoor firing range. No trainee had elevated blood lead levels before the training, but 15 of the 17 ended their training with blood lead levels deemed "elevated." If use of the indoor firing range causes an increase in a person's blood lead levels, then \(p\), the probability that a person's lead level increases will be greater than .5. Use the sign test to determine whether using an indoor firing range has the effect of increasing a person's blood lead level with \(\alpha=.05 .\) (HINT: The normal approximation to binomial probabilities is fairly accurate for \(n=17 .)\)

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.

Sign-up for free