**Given the linear correlation coefficient r and the sample size n, determine the critical values of r_Answer**

**Given the linear correlation coefficient r and the sample size n, determine the critical values of r_Answer**

Given the linear correlation coefficient r and the sample size n, determine the critical values of r_Answer

Question 1

1.

Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05. r = 0.75, n = 9

AnswerCritical values: r = ±0.666, no significant linear correlation

Critical values: r = 0.666, no significant linear correlation

Critical values: r = -0.666, no significant linear correlation

Critical values: r = ±0.666, significant linear correlation

5 points

Question 2

1.

Construct a scatterplot for the given data. Choose A, B, C, or D.

Answer5 points

Question 3

1.

Find the value of the linear correlation coefficient r.

Answer-0.054

0.214

0.109

-0.078

5 points

Question 4

1.

Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha. n = 11, = 0.01

Answerr = 0.765

r = ± 0.602

r = 0.735

r = ± 0.735

5 points

Question 5

1.

Use the given data to find the best predicted value of the response variable. Six pairs of data yield r = 0.789 and the regression equation What is the best predicted value of y for x = 5?

Answer22.0

18.0

18.5

19.0

5 points

Question 6

1.

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary. Choose A, B, C, or D.

Answer5 points

Question 7

1.

Is the data point, P, an outlier, an influential point, both, or neither?

AnswerNeither

Outlier

Both

Influential point

5 points

Question 8

1.

Use the given information to find the coefficient of determination. A regression equation is obtained for a collection of paired data. It is found that the total variation is 24.488, the explained variation is 15.405, and the unexplained variation is 9.083. Find the coefficient of determination.

Answer0.629

0.590

1.590

0.371

5 points

Question 9

1.

Use the computer display to answer the question.

Answer82.7%

17.0%

91.1%

83.0%

5 points

Question 10

1.

Find the explained variation for the paired data. The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the explained variation.

Answer498.103

511.724

87.4757

599.2

5 points

Question 11

1.

Find the unexplained variation for the paired data. The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the unexplained variation.

Answer511.724

87.4757

96.103

599.2

5 points

Question 12

1.

Find the total variation for the paired data. The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the total variation.

Answer599.2

498.103

511.724

87.4757

5 points

Question 13

1.

Find the standard error of estimate for the paired data. The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the the standard error of estimate.

Answer4.1097

7.1720

5.3999

13.060

5 points

Question 14

1.

Construct the indicated prediction interval for an individual y. The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.845 + 3.524x and the standard error of estimate is Se = 5.40. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the test.

Answer

58 < y < 82 62 < y < 78 35 < y < 104 32 < y < 107 5 points Question 15 1. Solve the problem. Answer 1.936 < B1 < 4.874 0.686 < B1 < 6.124 0.322 < B1 < 6.488 0.134 < B1 < 6.676 5 points Question 16 1. Use computer software to find the multiple regression equation. Can the equation be used for prediction? An anti-smoking group used data in the table to relate the carbon monoxide of various brands of cigarettes to their tar and nicotine content. Answer CO = 1.37 - 5.53TAR + 1.33NIC; Yes, because the R2 is high CO = 1.37 + 5.50TAR - 1.38NIC; Yes, because the P-value is high CO = 1.3 + 5.5TAR - 1.3NIC; Yes, because the adjusted R2 is high CO = 1.25 + 1.55TAR - 5.79NIC; Yes, because the P-value is too low 5 points Question 17 1. Use computer software to obtain the multiple regression equation and identify R2, adjusted R2, and the P-value. An anti-smoking group used data in the table to relate the carbon monoxide of various brands of cigarettes to their tar and nicotine content. Answer 0.976, 0.921, 0.002 0.943, 0.934, 0.000 0.931, 0.902, 0.000 0.861, 0.900, 0.015 5 points Question 18 1. Use computer software to obtain the multiple regression equation. Use the estimated equation to find the predicted value. A health specialist gathered the data in the table to see if pulse rates can be explained by exercise and smoking. For exercise, he assigns 1 for yes, 2 for no. For smoking, he assigns 1 for yes, 2 for no. He then used his results to predict the pulse rate of a person whose exercise value was 1 and whose smoking value was 1. Answer 70 beats/min 74 beats/min 81 beats/min 77 beats/min 5 points Question 19 1. Find the indicated multiple regression equation. Answer P-hat = 14.09 + 0.213(Att.) + 0.895(Adapt.) P-hat = 14.09 + 0.895(Att.) + 0.213(Adapt.) P-hat = 14.09 + 0.907(Att.) + 0.014(Adapt.) P-hat = 14.09 + 0.014(Att.) + 0.907(Adapt.) 5 points Question 20 1. Use computer software to find the best multiple regression equation to explain the variation in the dependent variable, Y, in terms of the independent variables, X1, X2, X3. Answer Y-hat = 0.42 + 0.99X2 Y-hat = 1.38 - 5.53X1 + 1.33X2 Y-hat = 1.25 - 1.55X1 + 5.79X2 Y-hat = -0.49 + 14.07X1

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