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When \(n \geq 30\), the formula \(r=\frac{\pm z}{\sqrt{n-1}}\) can be used to find the critical values for the rank correlation coefficient. For example, if \(n=40\) and \(\alpha=0.05\) for a two-tailed test, $$ r=\frac{\pm 1.96}{\sqrt{40-1}}=\pm 0.314 $$ Hence, any \(r_{s}\) greater than or equal to \(+0.314\) or less than or equal to \(-0.314\) is significant. For Exercises 29 through \(33,\) find the critical \(r\) value for each (assume that the test is two-tailed). $$ n=50, \alpha=0.05 $$

Short Answer

Expert verified
Critical r-value is \(\pm 0.28\).

Step by step solution

01

Determine the Required Formula

We need to calculate the critical value \(r\) using the given formula: \[r = \frac{\pm z}{\sqrt{n-1}} \]This formula finds the critical values for the rank correlation coefficient based on the sample size \(n\) and the critical z-value corresponding to the given significance level \(\alpha\).
02

Determine the Critical z-value

For a two-tailed test with a significance level of \(\alpha = 0.05\), the critical z-value is 1.96. This z-value corresponds to the cut-off for a standard normal distribution where 2.5% of the probability lies in each tail of the distribution.
03

Calculate the Denominator

Substitute the sample size \(n = 50\) into the denominator of the formula:\[\sqrt{n-1} = \sqrt{50-1} = \sqrt{49} = 7\]This will be used to divide the critical z-value to find the critical rank correlation coefficient \(r\).
04

Compute the Critical r-value

Substitute the z-value and calculated denominator into the formula:\[r = \frac{\pm 1.96}{7} = \pm 0.28\]These are the critical values for the rank correlation coefficient \(r\).
05

Interpret the Results

Since this is a two-tailed test, any sample rank correlation coefficient \(r_s\) greater than or equal to \(+0.28\) or less than or equal to \(-0.28\) is considered statistically significant at the \(\alpha = 0.05\) level.

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Key Concepts

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

Two-tailed Test
In statistics, a two-tailed test is essential when you want to determine if there is a significant difference in both directions—higher or lower. It helps us check if a sample statistic, like our rank correlation coefficient, is significantly different from the hypothesized parameter in either way.

Imagine you are interested in knowing whether a certain factor increases or decreases a result. A two-tailed test will test for both possibilities. When you conduct a two-tailed test, your hypothesis is that there is no effect or difference. The alternative hypothesis, on the other hand, states that there is a definitive effect, either positive or negative.

Why is it two-tailed? Because it doesn't matter where the significant difference lies, it matters that a difference is present. This means when conducting the test, an equal amount of significance level (%] called the alpha level, is distributed across both ends of the probability distribution curve, usually with 2.5% at the lower end and 2.5% at the higher end if the overall alpha is 5%. This ensures that any surprise increase or decrease is given equal importance.
  • A two-tailed test captures the extremes on both sides of the normal distribution.
  • It's versatile, covering both potential directions of effect.
Critical z-value
The critical z-value is a point on the standard normal distribution curve beyond which the observed data is considered statistically significant. During hypothesis testing, the critical z-value is used to decide whether to reject the null hypothesis.

In practical terms, the critical z-value marks the boundaries where 95% of the data falls within for a typical confidence level of 0.05 in a two-tailed test. Simply put, it helps you visualize and set a cutoff where values are too extreme to support the null hypothesis.

For example, a critical z-value of 1.96 in our exercise means that any observed z-value beyond +1.96 or below -1.96 indicates our sample result is unlikely under the null hypothesis. Hence, those values lead us to reject the null hypothesis in favor of the alternative.
  • Determines threshold for significance in hypothesis testing.
  • Impacts decisions on hypothesis testing acceptance or rejection.
Significance Level
The significance level, denoted by 1), is a threshold we set to determine when to reject our null hypothesis. It plays a crucial role in hypothesis testing by indicating the probability of observing an extreme result just by chance. Commonly, a significance level of 0.05 (5%) is used.

When we say our significance level is at 0.05, it means there is a 5% probability of rejecting the null hypothesis when it's actually true, known as a Type I error. It's up to researchers to decide their acceptable risk level.

In any test, the significance level is a balancing act—a smaller alpha means less risk of occasional errors but requires stronger evidence to reject the null hypothesis.
  • Crucial for determining error margin in tests.
  • A higher alpha level increases chances of Type I errors.
Sample Size
Sample size, defined as the number of observations or data points in a study, is foundational in statistical testing. A larger sample size generally provides more reliable test results. It helps in minimizing errors and enhances the power of our tests.

In relation to our rank correlation coefficient, sample size E] is a component in calculating critical values. Larger sample sizes typically lead to more precise estimates, allowing us to detect more subtle effects.

In our exercise, for example, with a sample size of 50, we significantly reduce variance and improve confidence in our results' accuracy. However, practical constraints and research goals often dictate the feasibility of larger sample sizes.
  • Impacts accuracy and reliability of statistical results.
  • Larger samples reduce variance and increase consistency.

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Most popular questions from this chapter

For Exercises 5 through \(20,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Television Viewers A researcher read that the median age for viewers of the Jimmy Fallon show is 52.7 years. To test the claim, 75 randomly selected viewers were surveyed, and 27 were under the age of \(52.7 .\) At \(\alpha=0.05\) test the claim. Give one reason why an advertiser might like to know the results of this study. Use the \(P\) -value method.

For Exercises \(9-14,\) use the Wilcoxon signed-rank test to test each hypothesis. Bowling Scores Eight randomly selected volunteers at a bowling alley were asked to bowl three games and pick their best score. They were then given a bowling ball made of a new composite material and were allowed to practice with the ball as much as they wanted. The next day they each bowled three games with the new ball and picked their best score. At the 0.05 level of significance, did scores improve? $$ \begin{array}{l|ccccccc}{\text { Bowler }} & {\mathrm{A}} & {\mathrm{B}} & {\mathrm{C}} & {\mathrm{D}} & {\mathrm{E}} & {\mathrm{F}} & {\mathrm{G}} & {\mathrm{H}} \\ \hline \text { Day 1} & {141} & {176} & {178} & {174} & {135} & {190} & {182} & {141} \\ \hline \text { Day 2} & {158} & {144} & {135} & {153} & {195} & {151} & {151} & {183}\end{array} $$

For Exercises 3 through \(12,\) use the Wilcoxon rank sum test. Assume that the samples are independent. Also perform each of these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Hunting Accidents A game commissioner wishes to see if the number of hunting accidents in counties in western Pennsylvania is different from the number of hunting accidents in counties in eastern Pennsylvania. Random samples of counties from the two regions are selected, and the numbers of hunting accidents are shown. At \(\alpha=0.05,\) is there a difference in the number of accidents in the two areas? If so, give a possible reason for the difference. $$ \begin{array}{l|lllllllll}{\text { Western Pa. }} & {10} & {21} & {11} & {11} & {9} & {17} & {13} & {8} & {15} & {17} \\ \hline \text { Eastern Pa. } & {14} & {3} & {7} & {13} & {11} & {2} & {8} & {5} & {5} & {6}\end{array} $$

For Exercises 3 through \(12,\) use the Wilcoxon rank sum test. Assume that the samples are independent. Also perform each of these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Winning Baseball Games For the years \(1970-1993\) the National League (NL) and the American League (AL) (major league baseball) were each divided into two divisions: East and West. Below are random samples of the number of games won by each league's Eastern Division. At \(\alpha=0.05,\) is there sufficient evidence to conclude a difference in the number of wins? $$ \begin{array}{l|c|ccccccccccc}{\mathrm{NL}} & {89} & {96} & {88} & {101} & {90} & {91} & {92} & {96} & {108} & {100} & {95} \\ \hline \text { AL } & {108} & {86} & {91} & {97} & {100} & {102} & {95} & {104} & {95} & {89} & {88} & {101}\end{array} $$

For Exercises 5 through \(20,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Routine Maintenance and Defective Parts A manufacturer believes that if routine maintenance (cleaning and oiling of machines) is increased to once a day rather than once a week, the number of defective parts produced by the machines will decrease. Nine machines are randomly selected, and the number of defective parts produced over a 24 -hour operating period is counted. Maintenance is then increased to once a day for a week, and the number of defective parts each machine produces is again counted over a 24 -hour operating period. The data are shown. At \(\alpha=0.01,\) can the manufacturer conclude that increased maintenance reduces the number of defective parts manufactured by the machines? $$ \begin{array}{l|cccccccc}{\text { Machine }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} \\ \hline \text { Before } & {6} & {18} & {5} & {4} & {16} & {13} & {20} & {9} & {3} \\ \hline \text { After } & {5} & {16} & {7} & {4} & {18} & {12} & {14} & {7} & {1}\end{array} $$

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