Thirty randomly selected seniors at Council High School were asked to report the age (in years) and mileage of their main vehicles. Here is a scatterplot of the data:

We used Minitab to perform a least-squares regression analysis for these data. Part of the computer output from this regression is shown here.

a. Explain what the value of r2tells you about how well the least-squares line fits the data.

b. The mean age of the students’ cars in the sample was x¯=5years. Find the mean mileage of the cars in the sample.

c. Interpret the value of s.

d. Would it be reasonable to use the least-squares line to predict a car’s mileage from its age for a Council High School teacher? Justify your answer.

Short Answer

Expert verified

Part a. We interpret that 77%of the variation between the variables has been explained by the least squares regression line.

Part b. The mean mileage is 105800miles.

Part c. The error made when predicting the mileage using the least square regression line is on average 22723miles.

Part d. The data of students is not representative for the teachers and thus we cannot use the least-squares line to predict the mileage of cars of teachers.

Step by step solution

01

Part a. Step 1. Explanation

It is given in the question the table in which the values are calculated. Thus, from table we have,

The value of r2is given in the output as “R-Sq”:

r2=77%=0.77

From this, we interpret that77% of the variation between the variables has been explained by the least squares regression line.

02

Part b. Step 1. Explanation

As we know that the least square regression line is as:

y^=a+bx

With the predicted mileage and the age.

The constant ais given in the row with “Age” and in the column with “Coef”:

a=-13832

The slope bis given in the row with “Age” and in the column with “Coef”:

b=14954

Then the least square equation then becomes:

y^=-13832+14954x

With y^the predicted mileage and xthe age.

The point (x¯,y¯)lies on the least square regression line thus the mean mileage can be obtained by replacing xby x¯in the least square equation and evaluating the expression:

y¯=y^=-13832+14954x¯=-13832+14954(8)=105800

Thus, the mean mileage islocalid="1664182337396" 105800.

03

Part c. Step 1. Explanation

In the question it is given the table of calculated values.

Thus, from that it is given the value of s, it is as:

s=22723

Thus, from this we interpret that the error made when predicting the mileage using the least square regression line is on average 22723 miles.

04

Part d. Step 1. Explanation

It is not be reasonable to use the least square line to predict a car’s mileage from its age for a Council high school teacher because the least-squares line was determined using data of students. The data of students is not representative for the teachers and thus we cannot use the least-squares line to predict the mileage of cars of teachers.

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Most popular questions from this chapter

The following dot plots show the average high temperatures (in degrees Celsius) for a sample of tourist cities from around the world. Both the January and July average high temperatures are shown. What is one statement that can be made with certainty from an Page Number: 704 analysis of the graphical display?

a. Every city has a larger average high temperature in July than in January.

b. The distribution of temperatures in July is skewed right, while the distribution of temperatures in January is skewed left.

c. The median average high temperature for January is higher than the median average high temperature for July.

d. There appear to be outliers in the average high temperatures for January and July.

e. There is more variability in average high temperatures in January than in July

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select10Variety A and10Variety B tomato plants. Then the researchers divide in half each of10small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The10differences (Variety A − Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is x-=0.343051526=0.200=20%x-A-B=0.34and the standard deviation of the differences is s A-B=0.833051526=0.200=20%=sA-B=0.83.LetμA-B=3051526=0.200=20%μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

The P-value for a test of H0: μA−B=03051526=0.200=20%versus Ha: μA−B≠0 is 0.227. Which of the following is the

correct interpretation of this P-value?

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b. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34is0.227.

c. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34or greater is0.227.

d. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference greater than or equal to0.34or less than or equal to −0.34is0.227.

e. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is not 0, the probability of getting a sample mean difference greater than or equal to 0.34or less than or equal to −0.34is0.227.

Friday the 13thRefer to Exercise 88.

a. Construct and interpret a 90%confidence interval for the true mean difference. If you already defined parameters and checked conditions in Exercise 88, you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise 88.

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male and female young adults.

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