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In Exercises 9.11 to \(9.14,\) test the correlation, as indicated. Show all details of the test. Test for a positive correlation; \(r=0.35 ; n=30\).

Short Answer

Expert verified
To conclude if there's a positive correlation, we need to compare the calculated t-value (using the provided correlation coefficient and sample size) to the critical t-value for a one-tailed test at a significance level of 0.05. We reject the null hypothesis of no positive correlation if the calculated t-value exceeds the critical t-value, indicating a significant positive correlation. Otherwise, we fail to reject the null hypothesis, signifying no significant positive correlation.

Step by step solution

01

State the Hypotheses

Firstly, we identify our null and alternative hypotheses. The null hypothesis ( \( H_0 \) ) posits no positive correlation, so it would state that the correlation coefficient is equal to 0. The alternative hypothesis ( \( H_1 \) ) claims a positive correlation, therefore, it would state that the correlation coefficient is greater than 0. Expressed mathematically, this would be: \( H_0: r = 0 \) and \( H_1: r > 0 \).
02

Calculate the Test Statistic

We use the given correlation coefficient ( \( r = 0.35 \) ) and sample size ( \( n = 30 \) ), along with the equation for the t-test for correlation coefficients to calculate our test statistic. This equation is \( t = r \sqrt{\frac{n-2}{1-r^2}} \). Substituting our given values into this equation yields \( t = 0.35 \sqrt{\frac{30-2}{1-0.35^2}} \).
03

Compare to Critical Value

Next, we find the critical t-value for a one-tailed test (since we are only interested in determining whether the correlation is greater than 0) at a significance level of 0.05 with \( n-2 \) degrees of freedom. Using a t-table, we find the one-tailed critical t-value to be approximately 1.699. Then, we compare our calculated t-value to this critical value.
04

State the Conclusion

If our calculated t-value is greater than the critical t-value, we reject the null hypothesis in favor of the alternative hypothesis. This would indicate a significant positive correlation. If our calculated t-value is less than or equal to the critical t-value, we do not reject the null hypothesis. This would indicate that the correlation is not significantly different from 0.

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Key Concepts

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

Correlation Coefficient
The correlation coefficient, often denoted as \( r \), is a statistic that measures the strength and direction of a linear relationship between two variables. It can range from -1 to 1. An \( r \) value close to 1 indicates a strong positive relationship, whereas a value close to -1 indicates a strong negative relationship. A value near 0 suggests no linear correlation.

In hypothesis testing, when you are provided with an \( r \) value, as in the problem with \( r = 0.35 \), it suggests a moderate positive correlation between the variables in question. However, to determine if this correlation is statistically significant, a t-test for the correlation coefficient is performed.

Key points to remember about correlation:
  • Values closer to -1 or 1 mean a stronger relationship.
  • A value of 0 implies no linear correlation.
  • Correlation does not imply causation.
t-test
A t-test is a statistical test used to compare the means of two groups and determine if they are different from each other or to test if a sample mean differs from a known value. In the context of correlation coefficients, the t-test assesses whether the observed value of \( r \) significantly differs from 0.

For the t-test of correlation coefficient:
  • Use the formula \( t = r \sqrt{\frac{n-2}{1-r^2}} \), where \( r \) is the correlation coefficient and \( n \) is the sample size.
  • Calculate the degrees of freedom as \( n-2 \).
  • The calculated \( t \)-value will determine if the result is significant or not.
In our example, with \( r = 0.35 \) and \( n = 30 \), you substitute these values to find the t-value. This helps you in comparing it against the critical value from the t-distribution table.

Remember that a t-test for correlation seeks to find if the linear relationship observed could be due to chance.
Significance Level
The significance level, commonly denoted as \( \alpha \), is a threshold that determines when to reject the null hypothesis. It reflects the probability of committing a Type I error, which is falsely rejecting the null hypothesis when it is true.

Typical significance levels are 0.05, 0.01, or 0.10. In hypothesis testing, such as in the given exercise with \( \alpha = 0.05 \), it represents a 5% risk of concluding that a correlation exists when there isn’t one.

Key points about significance levels:
  • Lower \( \alpha \) means stricter criteria for rejecting the null hypothesis.
  • Commonly set as 0.05 in many research fields.
  • Used to find the critical t-value from the t-distribution table.
In this exercise, \( \alpha = 0.05 \) guides you to find the critical value against which the calculated t-value is compared to make a decision about rejecting or not rejecting the null hypothesis.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement in statistics that there is no effect or no relationship present. It serves as the default or starting assumption that there is no significant effect or relationship in a statistical hypothesis test.

For correlation testing, \( H_0: r = 0 \) implies there is no actual linear relationship between the variables being studied. The null hypothesis is tested against an alternative hypothesis, which, in this case, is \( H_1: r > 0 \), suggesting a positive correlation.

Key features of the null hypothesis:
  • The hypothesis that researchers aim to test against.
  • Assumes no effect or relationship until evidence suggests otherwise.
  • Serves as the basis for calculating the test statistic and making decisions in hypothesis testing.
In hypothesis testing, if the calculated test statistic exceeds the critical value, the null hypothesis is often rejected, providing evidence in favor of the alternative hypothesis.

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Most popular questions from this chapter

Golf Scores In a professional golf tournament the players participate in four rounds of golf and the player with the lowest score after all four rounds is the champion. How well does a player's performance in the first round of the tournament predict the final score? Table 9.6 shows the first round score and final score for a random sample of 20 golfers who made the cut in a recent Masters tournament. The data are also stored in MastersGolf. Computer output for a regression model to predict the final score from the first-round score is shown. Use values from this output to calculate and interpret the following. Show your work. (a) Find a \(95 \%\) interval to predict the average final score of all golfers whoshoot a 0 on the first round at the Masters. (b) Find a \(95 \%\) interval to predict the final score of a golfer who shoots a -5 in the first round at the Masters. (c) Find a \(95 \%\) interval to predict the average final score of all golfers who shoot a +3 in the first round at the Masters. The regression equation is Final \(=0.162+1.48\) First \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 0.1617 & 0.8173 & 0.20 & 0.845 \\ \text { First } & 1.4758 & 0.2618 & 5.64 & 0.000 \\ S=3.59805 & R-S q=63.8 \% & \text { R-Sq }(a d j) & =61.8 \%\end{array}\) Analysis of Variance Source Regression Residual Error Total \(\begin{array}{rrrrr}\text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ 1 & 411.52 & 411.52 & 31.79 & 0.000 \\ 18 & 233.03 & 12.95 & & \\ 19 & 644.55 & & & \end{array}\)

Test the correlation, as indicated. Show all details of the test. Test for evidence of a linear association; \(r=0.28 ; n=10\)

Test the correlation, as indicated. Show all details of the test. Test for a negative correlation; \(r=-0.41\); \(n=18\).

Use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 77.44 & 14.43 & 5.37 & 0.000 \\ \text { Score } & -15.904 & 5.721 & -2.78 & 0.012 \end{array} $$

Use this information to fill in all values in an analysis of variance for regression table as shown. $$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F-statistic } & \text { p-value } \\ \hline \text { Model } & & & & & \\ \hline \text { Error } & & & & & \\ \hline \text { Total } & & & & & \\ \hline \end{array} $$ SSModel \(=800\) with SSTotal \(=5820\) and a sample size of \(n=40\)

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