# Lab Activity 13 Correlation And Regression Stat 1350 Part 1 Restaurant Tips Is T

**Lab Activity #13: Correlation and Regression**

**STAT 1350**

**Part 1: Restaurant Tips**

**bill****tip**www.lock5stat.com/statkey**Florida Lakes (Mercury vs Ph)****Restaurant Tips (Tip vs Bill)**

- First, let’s attempt to examine and describe the relationship between
**Bill**and**Tip.**In the space below, comment on the form, direction, and strength of this relationship. If you want to, in StatKey, you can click on the “Show Regression Line” box on the right to see how far the data points fall from the regression line.

· Form:

· Direction:

· Strength:

- Use the summary statistics you are given to construct the regression equation that will allow us to predict
**Tip**based on**Bill**. Be sure to type it in equation below. We will come back to the equation later on in this activity.

- Let’s focus now on the correlation between
**Bill**and**Tip**. You should see this value among the**Summary Statistics**that are automatically computed. - What is the value of the correlation? Please write or type this below.

- What does the correlation coefficient tell you about the strength and direction of the relationship between
**Bill**and**Tip**?

4. Look carefully at the regression equation you wrote down in response to Question 2 above. Please answer the following questions based on this equation.

a. What is the slope? Please write an interpretation of this value.

b. What is the intercept? Again, please write an interpretation of this value.

c. Suppose that George is a customer at the restaurant and his **bill** is $55.75. How much (in dollars) would we predict his **tip** to be?

d. Recall the value for the correlation that you found above. Use this information to find the value of r-squared. Write ths value below and explain how it should be interpreted.

e. What percent of the variability (or variation) in **Tip** can NOT be explained by the regression equation?

f. Suppose you learn of another customer, Lucy, who has a **bill** of $100. Why would we NOT want to use the regression equation to predict theamount of money Lucy will leave for a **tip**?

5. Suppose you are told that one customer’s data was accidentally left out of the data set. Peter had a **bill** of $35, and he left a** tip** of $15. How do you think this one data value would change the correlation between the variables? Why?

**Part 2: More reasoning about scatterplots and correlation**

6. If you switch X and Y, the sign of the correlation changes. (Assume here that X and Y are quantitative variables)

a. True b. False

7. Correlation is not affected by skewness and outliers.

a. True b. False

8. The strength of a correlation depends on its sign, positive or negative

a. True b. False

9. A coefficient of correlation of -0.96 indicates a very strong negative correlation.

a. True b. False

10. If the correlation between total semester score and attendance is 0.85, then ____% of the variation in total semester score is explained by the regression equation.

a. 0.85%

b. 8.5%

c. 72.25%

d. 92.20%

11. A professor examines the relationship between minutes studying and score on the midterm (out of 200 points) for students in his course using data from 320 randomly selected students. The data is presented in the scatterplot below. True or False: The correlation for this data set is in units of “points per minute studying.”

- True
- False

12. In Florida, the pH levels and mercury levels of several lakes are measured. From these measurements, researchers construct a regression equation in order to predict mercury levels based on pH levels. The equation is: Predicted mercury level = 1.531 – 0.152(pH level).

It is found that this regression equation explains approximately 32% of the variability in mercury levels. Based on this information, the correlation between mercury level and pH level must be

a. -0.152

b. -0.32

c. -0.57

d. We cannot answer this based on the given information.

13. Suppose an algebra professor found that the correlation between study time (in hours) and exam score

(out of 100) is r = .80, and the regression equation was found to be: Predicted exam score = 20 + 4(study

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