Inference For Regression Complete Test Bank Ch9

Statistics - Unlocking the Power of Data, 3e (Lock)

Chapter 9 Inference for Regression

9.1 Inference for Slope and Correlation

Use the following to answer the questions below:

Computer output from a regression analysis is provided.

The regression equation is Y = 72.9 - 0.519 X

1) What is the sample slope for this model?

A) 72.909

B) 2.037

C) -0.5195

D) 0.1946

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

2) What is the sample intercept for this model?

A) 72.909

B) 2.037

C) -0.5195

D) 0.1946

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

3) What is the standard error of the sample slope?

A) 72.909

B) 2.037

C) -0.519

D) 0.1946

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

4) What is the p-value for testing if the slope in the population is different from zero?

A) 0.5195

B) 0.1946

C) p < 0.001

D) 0.008

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

5) The sample size in this situation is n = 157. What are the degrees of freedom for constructing a confidence interval for, or performing a test about, the population slope?

A) 157

B) 156

C) 155

D) 153

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

6) The sample size in this situation is n = 157. Construct a 95% confidence interval for the population slope. Round the margin of error to four decimal places.

A) -0.9038 to -0.1352

B) -0.9009 to -0.1381

C) -0.8976 to -0.1414

D) -0.9138 to -0.1252

Diff: 2 Type: BI Var: 1

L.O.: 9.1.2

7) Use the p-value for testing if the slope in the population is different from zero (and a 5% significance level) to make a clear conclusion about the effectiveness of the model.

A) p-value = 0.008

There is very strong evidence that the population slope differs from zero, and thus is an effective model for predicting this response variable.

B) p-value = 0.008

There is not enough evidence that the population slope differs from zero, and thus this is not an effective model for predicting this response variable.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.3

Use the following to answer the questions below:

Computer output from a regression analysis is provided.

8) What is the sample slope for this model?

A) 1.6370

B) 0.5453

C) 7.2960

D) 14.5444

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

9) What is the sample intercept for this model?

A) 1.6370

B) 0.5453

C) 7.2960

D) 14.5444

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

10) What is the standard error of the sample slope?

A) 1.6370

B) 0.5453

C) 7.2960

D) 14.5444

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

11) What is the p-value for testing if the slope in the population is different from zero?

A) 0.502

B) 0.622

C) 0.5453

D) 0.00765

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

12) The sample size in this situation is n = 20. What are the degrees of freedom for constructing a confidence interval, or performing a test about, the population slope?

A) 17

B) 18

C) 19

D) 20

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

13) The sample size in this situation is n = 20. Construct a 95% confidence interval for the population slope. Round the margin of error to three decimal places.

A) 0.491 to 2.783

B) 0.568 to 2.706

C) 0.578 to 2.697

D) 0.5327 to 2.741

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

14) Use the p-value for testing if the slope in the population is different from zero (and a 5% significance level) to make a clear conclusion about the effectiveness of the model.

A) p-value = 0.00765

There is very strong evidence that the population slope differs from zero, and thus is an effective model for predicting this response variable.

B) p-value = 0.00765

There is not enough evidence that the population slope differs from zero, and thus is not an effective model for predicting this response variable.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.3

15) The website for the Quantitative Environmental Learning Project (funded by the National Science Foundation) describes data they collected on the lengths and widths of Puget Sound Butter Clams. A scatterplot of the data (with the regression line) is provided.

Use the scatterplot to check each of the conditions for using a linear model with this data. Is using a linear model appropriate for these data?

There are no outliers to be concerned about.

There is a pattern that indicates increasing spread (the widths are more spread out for longer clams [8 or more cm long] than shorter clams). This fanning pattern in the data suggests that we shouldn't use a linear model in this situation.

Diff: 3 Type: ES Var: 1

L.O.: 9.1.6

16) In a random sample of 41 students, the correlation between Math SAT score and college GPA is 0.289. Is there a significant linear association between Math SAT score and college GPA? Use α = 0.05. Include all details of the test. Round the test statistic to two decimal places.

: ρ = 0

: ρ ≠ 0

t = = 1.89

p-value = 0.066 (two-tail in a t distribution with df = 39)

There is no evidence of a significant positive correlation between Math SAT score and college GPA (p-value is less than significance level).

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

17) In a random sample of 41 students, the correlation between Verbal SAT score and college GPA is 0.574. Is there evidence of a positive correlation between Verbal SAT score and college GPA? Use a 5% significance level. Include all details of the test. Round the test statistic to two decimal places.

: ρ = 0

: ρ > 0

t = = 4.38

p-value ≈ 0 (right tail in a t distribution with df = 39)

There is very strong evidence of a positive correlation between Verbal SAT score and college GPA.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

18) In a random sample of 41 college students, the correlation between number of hours of television watched in a typical week and college GPA is -0.125. Is there evidence of a negative correlation between the amount of television watched and college GPA? Use a 5% significance level. Include all details of the test. Round the test statistic to three decimal places.

: ρ = 0

: ρ < 0

t = = -0.787

p-value = 0.218 (left tail in a t distribution with df = 39)

There is no evidence that the amount of television watched in a typical week is negatively correlated with college GPA.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

9.2 ANOVA for Regression

Use the following to answer the questions below:

In a regression analysis based on a sample of size n = 30, SSModel = 750 and SSTotal = 2,500.

1) Use this information to fill in all values in an analysis of variance table as shown.

Diff: 2 Type: ES Var: 1

L.O.: 9.2.1

2) Compute .

A) 30%

B) 83%

C) 1.2%

D) 5.5%

Diff: 2 Type: BI Var: 1

L.O.: 9.2.3

3) Compute the standard deviation of the error term. Use two decimal places in your answer.

A) 7.91

B) 6.25

C) 1.50

D) 7.64

Diff: 2 Type: BI Var: 1

L.O.: 9.2.3

Use the following to answer the questions below:

In a regression analysis with n = 25, SSE = 1,800 and SSTotal = 2,000.

4) Use this information to fill in all values in an analysis of variance table as shown. Round decimal answers to three decimal places.

Diff: 1 Type: ES Var: 1

L.O.: 9.2.1

5) Compute .

A) 10%

B) 0.10%

C) 11%

D) 0.11%

Diff: 2 Type: BI Var: 1

L.O.: 9.2.2

6) Compute the standard deviation of the error term. Use two decimal places in your answer.

A) 8.85

B) 8.49

C) 1.70

D) 1.84

Diff: 2 Type: BI Var: 1

L.O.: 9.2.3

9.3 Confidence and Prediction Intervals

Use the following to answer the questions below:

Students in a small statistics course wanted to investigate if forearm length (in cm) was useful for predicting foot length (in cm). The data they collected are displayed in the provided scatterplot (with regression), and the computer output from the analysis is provided.

Use three decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is Foot (cm) = 9.22 + 0.574 Forearm (cm)

Predicted Values for New Observations

1) Consider the scatterplot. Should we have any significant concerns about the conditions being met for using a linear model with these data?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

2) Use the fitted model to predict the foot length for someone whose arm is 30 cm long. Report your answer with two decimal places.

A) 26.44 cm

B) 25.89 cm

C) 26.12 cm

D) 25.73 cm

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

3) What is the estimated slope in this regression model?

A) 0.574

B) 9.22

C) 0.5735

D) 9.216

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

4) What is the test statistic for a test of the slope? What is the p-value? What is the conclusion of the test, in context?

A) t = 3.63; p-value = 0.004

There is strong evidence that forearm length is a useful predictor of foot length.

B) t = 2.04; p-value = 0.066

There is not enough evidence to conclude that forearm length is a useful predictor of foot length.

C) t = 3.63; p-value = 0.004

There is not enough evidence to conclude that forearm length is a useful predictor of foot length.

D) t = 2.04; p-value = 0.066

There is strong evidence that forearm length is a useful predictor of foot length.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.3

5) Use the ANOVA table to determine the overall sample size.

A) 13

B) 12

C) 11

D) 10

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

6) Construct a 90% confidence interval for the population slope.

A) 0.291 to 0.857

B) 0.265 to 0.883

C) 0.314 to 0.834

D) 0.277 to 0.871

Diff: 2 Type: BI Var: 1

L.O.: 9.1.2

7) Use the ANOVA table to compute and interpret .

A) 0.546

About 55% of the variability in foot lengths in this sample is explained by the person's forearm length.

B) 0.298

About 30% of the variability in foot lengths in this sample is explained by the person's forearm length.

C) 0.454

About 45% of the variability in foot lengths in this sample is explained by the person's forearm length.

D) 0.206

About 21% of the variability in foot lengths in this sample is explained by the person's forearm length.

Diff: 2 Type: BI Var: 1

L.O.: 9.2.2

8) The correlation between foot length and forearm length is 0.7389. Compute and interpret for this regression model.

About 55% of the variability in foot lengths in this sample is explained by the person's forearm length.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.5

9) Use the ANOVA table to find the standard deviation of the error term. Round your answer to three decimal places.

Diff: 2 Type: SA Var: 1

L.O.: 9.2.3

10) Based on the output, provide and interpret a 95% confidence interval for the mean foot length for all individuals with a forearm that is 28 cm long.

We are 95% sure that the mean foot length for all individuals with a forearm that is 28 cm long is between 24.1 cm and 26.4 cm.

Diff: 2 Type: ES Var: 1

L.O.: 9.3.1

11) Based on the output, provide and interpret a 95% prediction interval for the foot length of a specific individual with a forearm that is 28 cm long.

We are 95% sure that the foot length for a single individual with forearm that is 28 cm long is between 21.1 cm and 29.5 cm.

Diff: 2 Type: ES Var: 1

L.O.: 9.3.2

12) When conducting inference for the population slope, it is most common to test if the population slope is different from zero. However, there are other situations where a different test might be more interesting. For instance, it is often said that the length of the forearm is roughly the same as the length of the foot (see, for example, the movie Pretty Woman). What population slope is implied by this statement, and what would the hypotheses for testing the accuracy of this claim look like?

: = 1

: ≠ 1

Diff: 3 Type: ES Var: 1

L.O.: 9.1.0

Use the following to answer the questions below:

Data were collected on GPA and number of Facebook friends for students in a small statistics class. Some summary statistics, partial output from the regression analysis, and a scatterplot of the data (with regression line) are provided. Assume that students in this class are typical of all students at the university.

Use three decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is GPA = 3.830 - 0.000919 FacebookFriends

13) Use the scatterplot to determine whether we should have any strong concerns about the conditions being met for using a linear model with these data.

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

14) Use the equation of the least squares line to predict the GPA for someone with 800 Facebook friends.

A) 3.095

B) 3.830

C) 3.153

D) 3.024

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

15) Use the information in the ANOVA table to determine the number of students included in the dataset.

A) 31

B) 30

C) 29

D) 28

Diff: 2 Type: BI Var: 1

L.O.: 9.2.0

16) Use the information in the ANOVA table to compute and interpret .

A) = 0.470

About 47% of the variability in GPA for students in this sample is explained by number of Facebook friends.

B) = 0.686

About 69% of the variability in GPA for students in this sample is explained by number of Facebook friends.

C) = 0.470

About 53% of the variability in GPA for students in this sample is explained by number of Facebook friends.

D) = 0.686

About 31% of the variability in GPA for students in this sample is explained by number of Facebook friends.

Diff: 2 Type: BI Var: 1

L.O.: 9.2.2

17) Is the linear model effective at predicting GPA? Use the information from the computer output and α = 0.05. Include all details of the test.

: = 1

: ≠ 1

or : The model is ineffective versus : The model is effective

F = 24.85

p-value = 0 (using 1 and 28 degrees of freedom)

There is very strong evidence that the linear model is effective for predicting GPA.

Diff: 2 Type: ES Var: 1

L.O.: 9.2.1

18) Use the information in the computer output to compute the standard deviation of the error term.

A) 0.305

B) 0.093

C) 0.412

D) 0.317

Diff: 2 Type: BI Var: 1

L.O.: 9.2.3

19) Use the provided output to construct a 90% confidence interval for the mean GPA of all students with 800 Facebook friends.

A) 2.993 to 3.197

B) 2.760 to 3.430

C) 3.035 to 3.155

D) 2.916 to 3.274

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

20) Use the provided output to construct a 90% prediction interval for the GPA of a student with 800 Facebook friends.

A) 2.566 to 3.624

B) 1.361 to 4.829

C) 2.784 to 3.406

D) 2.224 to 3.966

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

21) Use the following output to identify and interpret a 95% interval for the mean GPA for all students with 500 Facebook friends.

Predicted Values for New Observations

A) CI: (3.2378, 3.5036)

We are 95% sure that the mean GPA for all students with 500 Facebook friends is between 3.2378 and 3.5036.

B) PI: (2.7315, 4.0089)

We are 95% sure that the GPA of a student with 500 Facebook friends is between 2.7315 and 4.0089.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

22) Use the following output to identify and interpret a 95% interval for the GPA of a single student with 500 Facebook friends.

Predicted Values for New Observations

A) PI: (2.7315, 4.0089)

We are 95% sure that the GPA of a student with 500 Facebook friends is between 2.7315 and 4.0089.

B) CI: (3.2378, 3.5036)

We are 95% sure that the mean GPA for all students with 500 Facebook friends is between 3.2378 and 3.5036.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

23) The correlation between GPA and number of Facebook friends is -0.686. Use the correlation and α = 0.05 to test for a linear association between GPA and number of Facebook friends. Include all details of the test.

: ρ = 0

: ρ ≠ 0

t = = -4.989

p-value ≈ 0 (two-tail probability in t distribution with df = 28)

There is very strong evidence of a linear association between GPA and number of Facebook friends.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

24) The correlation between GPA and number of Facebook friends is -0.686. Use the correlation and α = 0.05 to test for a negative linear association between GPA and number of Facebook friends. Include all details of the test.

: ρ = 0

: ρ < 0

t = = -4.989

p-value ≈ 0 (left tail probability in t distribution with df = 28)

There is very strong evidence of a negative linear association between GPA and number of Facebook friends.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

25) Use the information in the computer output to compute the standard error of the slope, SE. Round the answer to six decimal places.

A) 0.000184

B) 0.000183

C) 0.000992

D) 0.000993

Diff: 2 Type: BI Var: 1

L.O.: 9.2.4

26) Compute the t test statistic for the slope.

A) -4.989

B) -5.022

C) -5.465

D) - 5.479

Diff: 3 Type: BI Var: 1

L.O.: 9.1.0

Use the following to answer the questions below:

Data were collected on the age (in years) and price (in thousands of dollars) of a random sample of 25 used Hyundai Elantras. A scatterplot of the data (with regression line) and computer output from a regression analysis are provided.

Use three decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is Price = 15.3 - 1.71 Age

S = 1.37179 R-Sq = 88.9% R-Sq(adj) = 88.4%

Predicted Values for New Observations

27) Use the scatterplot to determine whether we should have any serious concerns about the conditions being met for using a linear model with these data.

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

28) What is the estimated slope in this regression model? Interpret the slope in context.

A) The estimated slope is -1.71.

For each additional year of age, the predicted price of the car (used Hyundai Elantra) decreases by $1,710.

B) The estimated slope is -1.71.

For each additional year of age, the predicted price of the car (used Hyundai Elantra) decreases by $1.71.

C) The estimated slope is 15.3.

The cost of a new used Hyundai Elantra is approximately $15,300.

D) The estimated slope is 15.3.

For each additional year of age, the predicted price of the car (used Hyundai Elantra) decreases by $1530.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

29) Use the equation of the least squares line to predict the price of a used Hyundai Elantra that is 6 years old.

A) $5,040

B) $13,540

C) $6,750

D) $7,750

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

30) What are the degrees of freedom for constructing a confidence interval for, or performing a test about, the population slope?

A) 25

B) 24

C) 23

D) 22

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

31) Use the computer output to test the slope to determine whether age is an effective predictor of price. Use α = 0.05.

A) There is very strong evidence that age is an effective predictor of price.

B) There is not enough evidence to conclude that age is an effective predictor of price.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.3

32) Construct and interpret a 90% confidence interval for the population slope.

-1.7126 ± 1.714(0.1264)

-1.7126 ± 0.2166

-1.929 to -1.496

We are 90% sure that for each additional year of age, the predicted cost of Hyundai Elantras decreases by between 1.496 and 1.929 thousand dollars ($1,496 and $1,929).

Diff: 2 Type: ES Var: 1

L.O.: 9.1.2

33) What is the for this model? Interpret it in context.

The age of the cars explains about 89% of the variability in the prices of the cars in this sample.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.5

34) Based on the available information, what is the correlation between age and price (in thousands of dollars) of used Hyundai Elantras?

A) 0.943

B) -0.943

C) 9.43

D) -9.43

Diff: 3 Type: BI Var: 1

L.O.: 9.1.5

35) Use the computer output to provide and interpret a 95% interval for the mean price of all 3-year-old used Hyundai Elantras.

We are 95% sure that the mean price of all 3-year-old used Hyundai Elantras is between 9.520 and 10.787 thousand dollars ($9,520 and $10,787).

Diff: 2 Type: ES Var: 1

L.O.: 9.3.1

36) Use the computer output to provide and interpret a 95% interval for the price of a 3-year-old used Hyundai Elantra.

We are 95% sure that the price of a single 3-year-old used Hyundai Elantra is between 7.246 and 13.061 thousand dollars ($7,246 and $13,061).

Diff: 2 Type: ES Var: 1

L.O.: 9.3.2

Use the following to answer the questions below:

Data were collected on the mileage (in thousands of miles) and price (in thousands of dollars) of a random sample of used Hyundai Elantras. A scatterplot of the data (with regression line), some summary statistics, and partial computer output from a regression analysis are provided.

Use three decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is Price = 13.8 - 0.0912 Mileage

37) Use the scatterplot to determine whether we should have any strong concerns about the conditions being met for using a linear model with these data.

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

38) Use the equation of the least squares line to predict the price of a used Hyundai Elantra with 50,000 miles.

A) 9,240

B) $6,900

C) $12,888

D) $13,344

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

39) Use the provided output to compute .

A) 0.793

B) 0.672

C) 0.891

D) 0.736

Diff: 2 Type: BI Var: 1

L.O.: 9.2.2

40) Use the information in the ANOVA table to determine the number of cars in the sample.

A) 25

B) 24

C) 23

D) 22

Diff: 2 Type: BI Var: 1

L.O.: 9.2.0

41) Is the linear model effective at predicting the price of used Hyundai Elantras? Use the information from the computer output and α = 0.05. Include all details of the test.

: = 0

: ≠ 0

(or : The model not effective versus : The model effective)

F = 88.01

p-value = 0 (using 1 and 23 degrees of freedom)

There is very strong evidence that mileage is a useful predictor of the price of used Hyundai Elantras (i.e., the model is effective).

Diff: 2 Type: ES Var: 1

L.O.: 9.2.1

42) Use the provided computer output to compute the standard deviation of the error term.

A) 1.872

B) 3.504

C) 1.832

D) 3.357

Diff: 2 Type: BI Var: 1

L.O.: 9.2.3

43) Use the provided output to construct and interpret a 95% interval for the mean price of all used Hyundai Elantras with 50,000 miles.

9.24 ± 2.069 ∙ 1.872 ∙

9.24 ± 2.069(1.872)(0.2066441)

9.24 ± 0.8

8.440 to 10.040

We are 95% sure that the mean price of all used Hyundai Elantras with 50,000 is between $8,440 and $10,040 (8.440 and 10.040 thousand dollars).

Diff: 2 Type: ES Var: 1

L.O.: 9.3.1

44) Use the provided output to construct and interpret a 95% interval for the price of a single used Hyundai Elantra with 50,000 miles.

9.24 ± 2.069 ∙ 1.872 ∙

9.24 ± 2.069(1.872)(1.021128)

9.24 ± 3.955

5.285 to 13.195

We are 95% sure that the price of a single used Hyundai Elantra with 50,000 miles is between $5,285 and $13,195 (5.285 and 13.195 thousand dollars).

Diff: 2 Type: ES Var: 1

L.O.: 9.3.3

45) Use the following computer output to identify and interpret a 95% interval for the mean price of all used Hyundai Elantras with 30,000 miles.

Predicted Values for New Observations

A) We are 95% sure that the mean price of all used Hyundai Elantras with 30,000 miles is between $10,058 and $12,022.

B) We are 95% sure that the price of a single used Hyundai Elantra with 30,000 miles is between $7046 and $15,034.

C) We are 95% sure that the price of a single used Hyundai Elantras with 30,000 miles is between $10,058 and $12,022.

D) We are 95% sure that the mean price of all used Hyundai Elantras with 30,000 miles is between $7046 and $15,034.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

46) Use the following computer output to identify and interpret a 95% interval for the price of a single used Hyundai Elantra with 70,000 miles.

A) We are 95% sure that the price of a single used Hyundai Elantra with 70,000 miles is between $3,440 and $11,347.

B) We are 95% sure that the mean price of all used Hyundai Elantras with 70,000 miles is between $3,440 and $11,347.

C) We are 95% sure that the price of a single used Hyundai Elantra with 70,000 miles is between $6,593 and $8,193.

D) We are 95% sure that the mean price of all used Hyundai Elantras with 70,000 miles is between $6,593 and $8,193.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

47) Use the information in the computer output to compute the standard error of the slope, SE. Round your answer to four decimal places.

A) 0.0097

B) 0.0099

C) 0.0101

D) 0.0103

Diff: 2 Type: BI Var: 1

L.O.: 9.2.4

48) Compute the t test statistic for the slope.

A) -9.402

B) -9.212

C) -9.030

D) -8.854

Diff: 3 Type: BI Var: 1

L.O.: 9.1.0

Use the following to answer the questions below:

Fast food restaurants are required to publish nutrition information about the foods they serve. Nutrition information about a random sample of 15 McDonald's lunch/dinner menu items (excluding sides and drinks) was obtained from their website. We wish to use the total fat content (in grams) to better understand the number of calories in the lunch/dinner menu items at McDonald's. Computer output from a regression analysis and a scatterplot (with regression line) of the data are provided.

Use two decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is Calories = 137.1 + 15.06 Total Fat (g)

S = 62.7442 R-Sq = 86.5% R-Sq(adj) = 85.5%

Predicted Values for New Observations

49) Using the scatterplot, should we should have any major concerns about the conditions being met for using a linear model with these data?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

50) Use the equation of the least squares line to predict the number of calories in a menu item with 20 grams of fat.

A) 438.30 calories

B) 334.44 calories

C) 289.80 calories

D) 410.60 calories

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

51) What is the estimated slope in this regression model?

A) 15.06

B) 16.7

C) -15.06

D) -16.7

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

52) What are the degrees of freedom for constructing a confidence interval for, or performing a test about, the population slope?

A) 15

B) 14

C) 13

D) 12

Diff: 2 Type: BI Var: 1

L.O.: 9.1.0

53) Use the computer output, and α = 0.05, to test the slope to determine whether total fat content (g) is an effective predictor of the number of calories. Include all details of the test.

: = 0

: ≠ 0

t = 9.13

p-value ≈ 0 (using df = 13)

There is very strong evidence that total fat content (g) is an effective predictor of the number of calories in McDonald's lunch/dinner menu items.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.3

54) Construct and interpret a 99% confidence interval for the population slope.

15.055 ± 3.012(1.649)

15.055 ± 4.967

10.09 to 20.02

We are 99% sure that for each additional gram in the total fat content, the number of calories in McDonald's lunch/dinner menu items is predicted to increase by between 10.09 and 20.02 calories.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.2

55) What is the for this model? Interpret it in context.

A) = 86.5%

86.5% of the variability in the number of calories for lunch/dinner menu items in this sample is explained by the total fat content (g).

B) = 86.5%

86.5% of the variability in the the total fat content in this sample is explained by the number of calories.

C) = 85.5%

85.5% of the variability in the number of calories for lunch/dinner menu items in this sample is explained by the total fat content (g).

D) = 85.5%

85.5% of the variability in the the total fat content in this sample is explained by the number of calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.5

56) Based on the available information, what is the correlation between total fat content (g) and number of calories for McDonald's lunch/dinner menu items in this sample?

A) 0.93

B) -0.93

C) 9.3

D) -9.3

Diff: 2 Type: BI Var: 1

L.O.: 9.1.5

57) Use the computer output to provide and interpret a 95% interval for the mean number of calories in all McDonald's lunch/menu items with 25 total grams of fat.

A) CI: (477.4, 549.5)

We are 95% sure that the mean number of calories in all McDonald's lunch/menu items with 25 total grams of fat is between 477.4 and 549.5 calories.

B) CI: (477.4, 549.5)

We are 95% sure that a single lunch/dinner menu item at McDonald's with 25 total grams of fat will have between 477.4 and 477.5 calories.

C) PI: (373.2, 653.7)

We are 95% sure that the mean number of calories in all McDonald's lunch/menu items with 25 total grams of fat is between 373.2 and 653.7 calories.

D) PI: (373.2, 653.7)

We are 95% sure that a single lunch/dinner menu item at McDonald's with 25 total grams of fat will have between 373.2 and 653.7 calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

58) Use the computer output to provide and interpret a 95% interval for the price of a lunch/dinner menu item with 25 total grams of fat.

A) PI: (373.2, 653.7)

We are 95% sure that a single lunch/dinner menu item at McDonald's with 25 total grams of fat will have between 373.2 and 653.7 calories.

B) PI: (373.2, 653.7)

We are 95% sure that the mean number of calories in all McDonald's lunch/menu items with 25 total grams of fat is between 373.2 and 653.7 calories.

C) CI: (477.4, 549.5)

We are 95% sure that the mean number of calories in all McDonald's lunch/menu items with 25 total grams of fat is between 477.4 and 549.5 calories.

D) CI: (477.4, 549.5)

We are 95% sure that a single lunch/dinner menu item at McDonald's with 25 total grams of fat will have between 477.4 and 477.5 calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

59) The website also provides information about the sugar content in the menu items at McDonald's. For this sample of 15 lunch/dinner menu items, the correlation between number of calories and sugar content (in grams) is 0.35. Test, at the 5% significance level, if there is a significant linear association between number of calories and sugar content for McDonald's lunch/dinner menu items. Include all details of the test. Round the test statistic to three decimal places.

: ρ = 0

: ρ ≠ 0

t = = 1.347

p-value = 0.201

There is no evidence of a significant linear association between the number of calories and sugar content of McDonald's lunch/dinner menu items.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

Use the following to answer the questions below:

Fast food restaurants are been required to publish nutrition information about the foods they serve. Nutrition information about a random sample of McDonald's lunch/dinner menu items (excluding sides and drinks) was obtained from their website. We wish to use the sodium content (in milligrams) to better understand the number of calories in the lunch/dinner menu items at McDonald's. Some summary statistics, partial computer output from a regression analysis, and a scatterplot (with regression line) of the data are provided.

Use two decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is Calories = 99.69 + 0.3698 Sodium (mg)

60) Using the scatterplot, should we should have any serious concerns about the conditions being met for using a linear model with these data.

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

61) Use the equation of the least squares line to predict the number of calories in a lunch/dinner menu item with 1,000 mg of sodium.

A) 469.49 calories.

B) 505.63 calories.

C) 424.89 calories.

D) 448.38 calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

62) What is the estimated slope in this regression model? Interpret the slope in context.

A) 0.3698

For each additional mg of sodium in McDonald's lunch/dinner menu items, we predict the number of calories to increase by 0.3698 calories.

B) 0.3698

For each additional calorie in McDonald's lunch/dinner menu items, we predict the grams of sodium to increase by 0.3698 mg.

C) 99.69

For each additional mg of sodium in McDonald's lunch/dinner menu items, we predict the number of calories to increase by 99.69 calories.

D) 99.69

For each additional calorie in McDonald's lunch/dinner menu items, we predict the grams of sodium to increase by 99.69 mg.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

63) Use the information in the ANOVA table to determine the number of menu items in the sample.

A) 15

B) 14

C) 13

D) 12

Diff: 2 Type: BI Var: 1

L.O.: 9.2.0

64) Use the provided output to compute and interpret .

A) = 0.71

71% of the variability in the number of calories in McDonald's lunch/dinner menu items in the sample is explained by the sodium content (mg).

B) = 0.71

71% of the variability in the sodium content (mg) in McDonald's lunch/dinner menu items in the sample is explained by the number of calories.

C) = 0.42

42% of the variability in the number of calories in McDonald's lunch/dinner menu items in the sample is explained by the sodium content (mg).

D) = 0.42

42% of the variability in the sodium content (mg) in McDonald's lunch/dinner menu items in the sample is explained by the number of calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.2.2

65) Is the linear model effective at predicting the number of calories in lunch/dinner menu items at McDonald's? Use the information from the computer output (and α = 0.05) for this test. Include all details of the test.

: = 0

: ≠ 0

(or : The model is ineffective versus : The model is effective).

F = 31.11

p-value ≈ 0

There is very strong evidence that this model, which uses sodium content as the predictor, is effective at predicting the number of calories in lunch/dinner menu items at McDonald's.

Diff: 2 Type: ES Var: 1

L.O.: 9.2.1

66) Use the provided computer output to compute the standard deviation of the error term. Use two decimal places in your answer.

A) 92.73

B) 89.36

C) 164.6

D) 138.23

Diff: 2 Type: BI Var: 1

L.O.: 9.2.3

67) Use the provided output to construct and interpret a 95% interval for the mean number of calories in all McDonald's lunch/dinner menu items with 1,000 mg of sodium.

A) 417.68 to 521.30

We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 1,000 mg of sodium is between 417.68 and 521.30 calories.

B) 417.68 to 521.30

We are 95% sure that a single McDonald's lunch/dinner menu item with 1,000 mg of sodium has between 417.68 and 521.30 calories.

C) 262.60 to 676.38

We are 95% sure that a single McDonald's lunch/dinner menu item with 1,000 mg of sodium has between 262.60 and 676.38 calories.

D) 262.60 to 676.38

We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 1,000 mg of sodium is between 262.60 and 676.38 calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

68) Use the provided output to construct and interpret a 95% interval for the number of calories in a single McDonald's lunch/dinner menu item with 1,000 mg of sodium.

A) 262.60 to 676.38

We are 95% sure that a single McDonald's lunch/dinner menu item with 1,000 mg of sodium has between 262.60 and 676.38 calories.

B) 262.60 to 676.38

We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 1,000 mg of sodium is between 262.60 and 676.38 calories.

C) 417.68 to 521.30

We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 1,000 mg of sodium is between 417.68 and 521.30 calories.

D) 417.68 to 521.30

We are 95% sure that a single McDonald's lunch/dinner menu item with 1,000 mg of sodium has between 417.68 and 521.30 calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

69) Use the following output to identify and interpret a 95% interval for the mean number of calories in all McDonald's lunch/dinner menu items with 1,200 mg of sodium.

Predicted Values for New Observations

A) We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 1,200 mg of sodium is between 485.7 and 601.1 calories.

B) We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 1,200 mg of sodium is between 334.9 and 751.9 calories.

C) We are 95% sure that the number of calories in a single McDonald's lunch/dinner menu item with 1200 mg of sodium is between 485.7 and 601.1 calories.

D) We are 95% sure that the number of calories in a single McDonald's lunch/dinner menu item with 1200 mg of sodium is between 334.9 and 751.9 calories.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

70) Use the following output to identify and interpret a 95% interval for the number of calories in a single McDonald's lunch/dinner menu item with 800 mg of sodium.

Predicted Values for New Observations

A) We are 95% sure that the number of calories in a single McDonald's lunch/dinner menu item with 800 mg of sodium is between 186.2 and 604.8.

B) We are 95% sure that the number of calories in a single McDonald's lunch/dinner menu item with 800 mg of sodium is between 334.8 and 456.2.

C) We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 800 mg of sodium is between 186.2 and 604.8.

D) We are 95% sure that the mean number of calories in all McDonald's lunch/dinner menu items with 800 mg of sodium is between 334.8 and 456.2.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

71) The website also provides information about the sugar content in the menu items at McDonald's. For this sample of lunch/dinner menu items, the correlation between number of calories and sugar content (in grams) is 0.35. Test, at the 5% significance level, if there is a significant linear association between number of calories and sugar content for McDonald's lunch/dinner menu items. Include all details of the test.

: ρ = 0

: ρ ≠ 0

t = = 1.347

p-value = 0.201

There is no evidence of a significant linear association between the number of calories and sugar content of McDonald's lunch/dinner menu items.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.4

72) Use the information in the computer output to compute the standard error of the slope, SE. Report your answer with four decimal places.

A) 0.0663

B) 0.0639

C) 0.0688

D) 0.0698

Diff: 2 Type: BI Var: 1

L.O.: 9.2.4

73) Compute the t test statistic for the slope.

A) 5.58

B) 6.21

C) 5.74

D) 6.34

Diff: 3 Type: BI Var: 1

L.O.: 9.1.0

74) Two intervals are given for the same value of the explanatory variable. Which interval is the confidence interval for the mean response at this value of the explanatory variable?

A) 32 to 38

B) 29 to 41

Diff: 2 Type: BI Var: 1

L.O.: 9.3.0

75) Two intervals are given for the same value of the explanatory variable. Which interval is the prediction interval for an individual response at this value of the explanatory variable?

A) 115 to 133

B) 120 to 128

Diff: 2 Type: BI Var: 1

L.O.: 9.3.0

Use the following to answer the questions below:

A quantitatively savvy, young couple is interested in purchasing a home in northern New York. They collected data on 48 houses that had recently sold in the area. They want to predict the selling price of homes (in thousands of dollars) based on the size of the home (in square feet).

The regression equation is Price (in thousands) = 17.1 + 0.0643 Size (sq. ft.)

S = 48.5733 R-Sq = 37.5% R-Sq(adj) = 36.1%

Predicted Values for New Observations

76) Using the scatterplot, should we have any serious concerns about the conditions being met for using a linear model with these data.

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

77) Use the equation of the least squares line to predict the selling price of a home that is 1,742 square feet in size.

A) $129,111

B) $299,000

C) $143,773

D) $115,582

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

78) What is the estimated slope in this regression model? Interpret the slope in context.

A) 0.06427

For an increase of 1 square foot in the size of a recently sold house, the predicted selling price increases by $64.27.

B) 0.06427

For an increase of $1,000 in selling price the size of the house increases by 64.27 square feet .

C) 17.1

For an increase of 1 square foot in the size of a recently sold house, the predicted selling price increases by $17.10.

D) 17.1

For an increase of $1,000 in selling price the size of the house increases by 17.1 square feet .

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

79) What are the degrees of freedom for constructing a confidence interval for, or performing a test about, the population slope?

A) 48

B) 47

C) 46

D) 45

Diff: 2 Type: BI Var: 1

L.O.: 10.1.0

80) Use the computer output to test the slope, at the 5% level, to determine whether size (in square feet) is an effective predictor of the selling price of recently sold homes. Include all details of the test.

: = 0 (or Size is not an effective predictor of selling price.)

: ≠ 0 (or Size is an effective predictor of selling price.)

t = 5.25

p-value ≈ 0

There is very strong evidence that size (in square feet) is an effective predictor of the selling price of homes.

Diff: 2 Type: ES Var: 1

L.O.: 9.1.3

81) Construct a 95% confidence interval for the population slope.

A) 0.0397 to 0.0889

B) 0.0395 to 0.0891

C) 0.0392 to 0.0894

D) 0.0389 to 0.0897

Diff: 2 Type: BI Var: 1

L.O.: 9.1.2

82) What is the for this model? Interpret it in context.

A) = 37.5%

37.5% of the variability in the selling prices of the sampled homes is explained by the size (in square feet).

B) = 37.5%

37.5% of the variability in the size (ins square feet) of the sampled homes is explained by the selling price.

C) = 36.1%

36.1% of the variability in the selling prices of the sampled homes is explained by the size (in square feet).

D) = 36.1%

36.1% of the variability in the size (ins square feet) of the sampled homes is explained by the selling price.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.5

83) Based on the available information, what is the correlation between selling price (in thousands) and size (square feet) of the sample of recently sold homes? Use three decimal places in your answer.

Diff: 2 Type: SA Var: 1

L.O.: 9.1.5

84) Use the computer output to provide and interpret a 95% interval for the mean selling price of all 2,000 square foot houses in this portion of northern New York.

A) CI: (131.38, 159.83)

We are 95% sure that the mean selling price of all 2,000 square foot houses in this portion of northern New York is between $131,380 and $159,830.

B) CI: (131.38, 159.83)

We are 95% sure that the selling price of a 2,000 square foot house in this portion of northern New York is between $131,380 and $159,830.

C) PI: (46.80, 244.41)

We are 95% sure that the mean selling price of all 2,000 square foot houses in this portion of northern New York is between $46,800 and $244,410.

D) PI: (46.80, 244.41)

We are 95% sure that the selling price of a 2,000 square foot house in this portion of northern New York is between $46,800 and $244,410.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

85) Use the computer output to provide and interpret a 95% interval for the selling price of a single 2,000 square foot house in this portion of northern New York.

A) PI: (46.80, 244.41)

We are 95% sure that the selling price of a 2,000 square foot house in this portion of northern New York is between $46,800 and $244,410.

B) CI: (131.38, 159.83)

We are 95% sure that the mean selling price of all 2,000 square foot houses in this portion of northern New York is between $131,380 and $159,830.

C) CI: (131.38, 159.83)

We are 95% sure that the selling price of a 2,000 square foot house in this portion of northern New York is between $131,380 and $159,830.

D) PI: (46.80, 244.41)

We are 95% sure that the mean selling price of all 2,000 square foot houses in this portion of northern New York is between $46,800 and $244,410.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

Use the following to answer the questions below:

A quantitatively savvy, young couple is interested in purchasing a home in northern New York. They collected data on houses that had recently sold in the area. They want to predict the selling price of homes (in thousands of dollars) based on the age of the home (in years). Some summary statistics, partial regression output, and a scatterplot of the relationship (with regression line) are provided.

Use two decimal places when reporting the results from any calculations, unless otherwise specified.

The regression equation is Price (in thousands) = 193 - 0.665 Age

Analysis of Variance

86) Using the scatterplot, should we have any major concerns about the conditions being met for using a linear model with these data?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.1.6

87) Use the equation of the least squares to predict the selling price of a 92-year-old home.

A) $131,820

B) $152,754

C) $155,398

D) $124,227

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

88) What is the estimated slope in this regression model? Interpret the slope in context.

A) -0.665

For each additional year of age, the selling price of homes in this portion of northern New York is predicted to decrease by $665.

B) -0.665

For each additional year of age, the selling price of homes in this portion of northern New York is predicted to increase by $665.

C) 193

For each additional year of age, the selling price of homes in this portion of northern New York is predicted to increase by $193.

D) 193

For each additional year of age, the selling price of homes in this portion of northern New York is predicted to decrease by $193.

Diff: 2 Type: BI Var: 1

L.O.: 9.1.1

89) Use the information in the ANOVA table to determine how many homes were used in the sample.

Diff: 2 Type: SA Var: 1

L.O.: 9.2.0

90) Use the provided output to compute .

A) = 0.24

B) = 0.32

C) = 0.43

D) = 0.31

Diff: 2 Type: BI Var: 1

L.O.: 9.2.2

91) Is the linear model effective at predicting the selling price of homes in this portion of northern New York? Use the provided computer output (and α = 0.05) for this test.

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 9.2.1

92) Use the provided computer output to compute the standard deviation of the error term.

Diff: 2 Type: SA Var: 1

L.O.: 9.2.3

93) Construct and interpret a 95% interval for the mean selling price of all 92-year-old homes.

A) 115.57 to 148.07

We are 95% sure that the mean selling price of all 92-year-old homes is between $115,570 and $148,070.

B) 115.57 to 148.07

We are 95% sure that selling price of a 92-year-old home is between $115,570 and $148,070.

C) 22.77 to 240.88

We are 95% sure that selling price of a 92-year-old home is between $22,770 and $240,880.

D) 22.77 to 240.88

We are 95% sure that the mean selling price of all 92-year-old homes is between $22,770 and $240,880.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

94) Construct and interpret a 95% interval for the selling price of a single 92-year-old home.

A) 22.77 to 240.88

We are 95% sure that selling price of a 92-year-old home is between $22,770 and $240,880.

B) 115.57 to 148.07

We are 95% sure that the mean selling price of all 92-year-old homes is between $115,570 and $148,070.

C) 115.57 to 148.07

We are 95% sure that selling price of a 92-year-old home is between $115,570 and $148,070.

D) 22.77 to 240.88

We are 95% sure that the mean selling price of all 92-year-old homes is between $22,770 and $240,880.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

95) Use the following output to identify and interpret a 95% interval for the mean selling price of all 50-year-old homes in this portion of northern New York.

Predicted Values for New Observations

A) CI: (141.39, 178.49)

We are 95% sure that the mean selling price of all 50-year-old homes in this portion of northern New York is between $141,390 and $178,490.

B) CI: (141.39, 178.49)

We are 95% sure that the selling price of a 50-year-old house in this portion of northern New York is between $141,390 and $178,490.

C) PI: (50.52, 269.36)

We are 95% sure that the selling price of a 50-year-old house in this portion of northern New York is between $50,520 and $269,360.

D) We are 95% sure that the mean selling price of all 50-year-old homes in this portion of northern New York is between $50,520 and $269,360.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.1

96) Use the following output to identify and interpret a 95% interval for the selling price of a 50-year-old house in this portion of northern New York.

Predicted Values for New Observations

A) PI: (50.52, 269.36)

We are 95% sure that the selling price of a 50-year-old house in this portion of northern New York is between $50,520 and $269,360.

B) We are 95% sure that the mean selling price of all 50-year-old homes in this portion of northern New York is between $50,520 and $269,360.

C) CI: (141.39, 178.49)

We are 95% sure that the mean selling price of all 50-year-old homes in this portion of northern New York is between $141,390 and $178,490.

D) CI: (141.39, 178.49)

We are 95% sure that the selling price of a 50-year-old house in this portion of northern New York is between $141,390 and $178,490.

Diff: 2 Type: BI Var: 1

L.O.: 9.3.2

97) Use the information in the computer output to compute the standard error of the slope, SE. Report your answer with four decimal places.

Diff: 2 Type: SA Var: 1

L.O.: 9.2.4

98) Compute the t test statistic for the slope.

Diff: 3 Type: SA Var: 1

L.O.: 9.1.0

99) Is there evidence of a negative correlation between the selling price of homes in this portion of northern New York and their age? Use α = 0.05. Include all details of the test.

: ρ = 0

: ρ < 0

t = = -3.81

p-value = 0.00021 (left tail in t with df = 46, using Statkey)

There is very strong evidence of a negative correlation between the selling price of homes in this portion of northern New York and their age.

Diff: 3 Type: ES Var: 1

L.O.: 9.1.4;9.1.5

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