In a car rental company 's fleet, 70% of the cars are American brands, 20% are Japanese, and the rest are German. The company notes that manufacturers' recalls seem to affect 2% of the American cars, but only 1% of the others.

What's the probability that a recalled car is American?

Are good grades in high school associated with family togetherness? A Simple Random Sample of 142 highschool students was asked how many meals per week their families ate together. Their responses produced a mean of 3.78 meals per week, with a standard deviation of 2.2. Researchers then matched these responses against the students' grade point averages. The scatterplot appeared to be reasonably linear, so they went ahead with the regression analysis, as seen below. No apparent pattern emerged in the residuals plot.

Dependent variable is GPA

A-squared = 11.0%

s = 0.6682 with 142 - 2 = 140 df

$$\begin{array}{lcc} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } \\ \text { Intercept } & 2.7288 & 0.1148 \\ \text { Meals/wk } & 0.1093 & 0.0263 \end{array}$$

Is there evidence of an association? Test an appropriate hypothesis and state your conclusion.

Are good grades in high school associated with family togetherness? A Simple Random Sample of 142 highschool students was asked how many meals per week their families ate together. Their responses produced a mean of 3.78 meals per week, with a standard deviation of 2.2. Researchers then matched these responses against the students' grade point averages. The scatterplot appeared to be reasonably linear, so they went ahead with the regression analysis, as seen below. No apparent pattern emerged in the residuals plot.

Dependent variable is GPA

A-squared = 11.0%

s = 0.6682 with 142 - 2 = 140 df

$$\begin{array}{lcc} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } \\ \text { Intercept } & 2.7288 & 0.1148 \\ \text { Meals/wk } & 0.1093 & 0.0263 \end{array}$$

Are your answers to parts a and b contradictory? Explain.

Are good grades in high school associated with family togetherness? A random sample of 142 high school students was asked how many meals per week their families ate together. Their responses produced a mean of 3.78 meals per week, with a standard deviation of 2.2. Researchers then matched these responses against the students' grade point averages (GPAs). The scatterplot appeared to be reasonably linear, so they created a line of regression. No apparent pattern emerged in the residuals plot. The equation of the line was $\widehat{GPA}$ = 2.73 + 0.11 Meals.

Interpret the slope in this context.

Are good grades in high school associated with family togetherness? A random sample of 142 high school students was asked how many meals per week their families ate together. Their responses produced a mean of 3.78 meals per week, with a standard deviation of 2.2. Researchers then matched these responses against the students' grade point averages (GPAs). The scatterplot appeared to be reasonably linear, so they created a line of regression. No apparent pattern emerged in the residuals plot. The equation of the line was $\widehat{GPA}$ = 2.73 + 0.11 Meals.

If a student in this study had a negative residual, what did that mean?

Are good grades in high school associated with family togetherness? A random sample of 142 high school students was asked how many meals per week their families ate together. Their responses produced a mean of 3.78 meals per week, with a standard deviation of 2.2. Researchers then matched these responses against the students' grade point averages (GPAs). The scatterplot appeared to be reasonably linear, so they created a line of regression. No apparent pattern emerged in the residuals plot. The equation of the line was $\widehat{GPA}$ = 2.73 + 0.11 Meals.

What was the mean GPA for these students?

Are good grades in high school associated with family togetherness? A random sample of 142 high school students was asked how many meals per week their families ate together. Their responses produced a mean of 3.78 meals per week, with a standard deviation of 2.2. Researchers then matched these responses against the students' grade point averages (GPAs). The scatterplot appeared to be reasonably linear, so they created a line of regression. No apparent pattern emerged in the residuals plot. The equation of the line was $\widehat{GPA}$ = 2.73 + 0.11 Meals.

Interpret they-intercept in this context.

At a barbershop music singing competition, choruses are judged on three scales: Music (quality of the arrangement, etc.), Performance, and Singing. The scales are supposed to be independent of each other, and each is scored by a different judge, but a friend claims that be can predict a chorus's singing score from the other two scores. He offers the following regression based on the scores of all 34 choruses in a recent competition:

Dependent variable is: Singing

R-squared = 90.9% R-squared (adjusted) = 90.3%

s = 6.483 with 34 - 3 = 31 degrees of freedom

$$\begin{array}{lcccr} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 2.08926 & 7.973 & 0.262 & 0.7950 \\ \text { Performance } & 0.793407 & 0.0976 & 8.13 & \leq 0.0001 \\ \text { Music } & 0.219100 & 0.1196 & 1.83 & 0.0766 \end{array}$$

Complete the analysis. Check assumptions and conditions to the extent you can with the information provided

At a barbershop music singing competition, choruses are judged on three scales: Music (quality of the arrangement, etc.), Performance, and Singing. The scales are supposed to be independent of each other, and each is scored by a different judge, but a friend claims that be can predict a chorus's Singing score from the other two scores. He offers the following regression based on the scores of all 34 choruses in a recent competition:

Dependent variable is: Singing

R-squared = 90.9% R-squared (adjusted) = 90.3%

s = 6.483 with 34 - 3 = 31 degrees of freedom

$$\begin{array}{lcccr} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 2.08926 & 7.973 & 0.262 & 0.7950 \\ \text { Performance } & 0.793407 & 0.0976 & 8.13 & \leq 0.0001 \\ \text { Music } & 0.219100 & 0.1196 & 1.83 & 0.0766 \end{array}$$

State the standard null hypothesis for the coefficient of Performance and complete the /-test at the 5% level. State your conclusion.

At a barbershop music singing competition, choruses are judged on three scales: Music (quality of the arrangement, etc.), Performance, and Singing. The scales are supposed to be independent of each other, and each is scored by a different judge, but a friend claims that be can predict a chorus's Singing score from the other two scores. He offers the following regression based on the scores of all 34 choruses in a recent competition:

Dependent variable is: Singing

R-squared = 90.9% R-squared (adjusted) = 90.3%

s = 6.483 with 34 - 3 = 31 degrees of freedom

$$\begin{array}{lcccr} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 2.08926 & 7.973 & 0.262 & 0.7950 \\ \text { Performance } & 0.793407 & 0.0976 & 8.13 & \leq 0.0001 \\ \text { Music } & 0.219100 & 0.1196 & 1.83 & 0.0766 \end{array}$$

Complete the analysis. Check assumptions and conditions to the extent you can with the information provided.

At a barbershop music singing competition, choruses are judged on three scales: Music (quality of the arrangement, etc.), Performance, and Singing. The scales are supposed to be independent of each other, and each is scored by a different judge, but a friend claims that be can predict a chorus's Singing score from the other two scores. He offers the following regression based on the scores of all 34 choruses in a recent competition:

Dependent variable is: Singing

R-squared = 90.9% R-squared (adjusted) = 90.3%

s = 6.483 with 34 - 3 = 31 degrees of freedom

$$\begin{array}{lcccr} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 2.08926 & 7.973 & 0.262 & 0.7950 \\ \text { Performance } & 0.793407 & 0.0976 & 8.13 & \leq 0.0001 \\ \text { Music } & 0.219100 & 0.1196 & 1.83 & 0.0766 \end{array}$$

What do you think of your friend's claim? Can he predict Singing scores? Explain.

At a barbershop music singing competition, choruses are judged on three scales: Music (quality of the arrangement, etc.), Performance, and Singing. The scales are supposed to be independent of each other, and each is scored by a different judge, but a friend claims that be can predict a chorus's singing score from the other two scores. He offers the following regression based on the scores of all 34 choruses in a recent competition:

Dependent variable is: Singing

R-squared = 90.9% R-squared (adjusted) = 90.3%

s = 6.483 with 34 - 3 = 31 degrees of freedom

$$\begin{array}{lcccr} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 2.08926 & 7.973 & 0.262 & 0.7950 \\ \text { Performance } & 0.793407 & 0.0976 & 8.13 & \leq 0.0001 \\ \text { Music } & 0.219100 & 0.1196 & 1.83 & 0.0766 \end{array}$$

What do you think of your friend's claim? Can be predict singing scores? Explain.

Are twin births becoming more common? In the earlier exercise looked at the number of twin births by year from 1980 to 2014. Now we can include some inference:

Response variable is: TwinBirths/1000

R squared = 97.1 %

s = 0.9184 with 35 - 2 = 33 degrees of freedom

$$\begin{array}{lcccc} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-Ratio } & \text { P-Value } \\ \text { Intercept } & 18.2127 & 0.3039 & 59.9 & <0.0001 \\ \text { Year since 1980 } & 0.508261 & 0.0154 & 33.1 & <0.0001 \end{array}$$

Do you think the growth in twin births is due to the amount of Carbon Dioxide in the atmosphere? Explain your reasoning.

Are twin births becoming more common? In the earlier exercise looked at the number of twin births by year from 1980 to 2014. Now we can include some inference:

Response variable is: TwinBirths/1000

R squared = 97.1 %

s = 0.9184 with 35 - 2 = 33 degrees of freedom

$$\begin{array}{lcccc} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-Ratio } & \text { P-Value } \\ \text { Intercept } & 18.2127 & 0.3039 & 59.9 & <0.0001 \\ \text { Year since 1980 } & 0.508261 & 0.0154 & 33.1 & <0.0001 \end{array}$$

State and test the standard null hypothesis for the slope.

An Internet blogger claims that the increase in twin births can be explained by the increase in Carbon Dioxide levels in the atmosphere. He posts this regression for the same time period:

Response variable is: TwinBirths/1000

R squared = 97.4

s = 0.8751 with 35 - 3 = 32 degrees of freedom

$$\begin{array}{lccrr} \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-Ratio } & \text { P-Value } \\ \text { Intercept } & 92.3382 & 35.58 & 2.60 & 0.0142 \\ \text { Year since 1980 } & 0.891661 & 0.1846 & 4.83 & <0.0001 \\ \text { CO2 (ppm) } & -0.220668 & 0.1059 & -2.08 & 0.0453 \end{array}$$