Question

3-The population in the city of Houston from 1900 to 2010 is given below: Year Population

1900 44,633

1910 78,800

1920 138,276

1930 292,352

1940 384,514

1950 596,163

1960 938,219

1970 1,233,505

1980 1,595,138

1990 1,631,766

2000 1,953,631

2010 2,100,263

a. Give a scatter-plot and residual plot of the data.

b. Based on the graphs in part a, propose a model for the data. Show me evidence to support your conclusion. Go through all necessary steps to construct a model of the type you chose

4-In R Studio use the data cars to determine the following. Hint: The data set is already in R studio use the quick reference guide to determine the following. [In R Studio, use command(file$column), such as mean(cars$speed)] Description: The data gives the speed of cars and the distances taken to stop. Note that the data were recorded in the 1920s. Format A data frame with 50 observations on 2 variables. speed numeric Speed (mph) dist numeric Stopping distance (ft) a. Give a scatter plot of the data. Determine the form, direction and strength of the relationship between speed and stopping distance (dist). b. Determine the LSRL for predicting stopping distance based on speed of the car. c. Interpret the slope of this LSRL equation. d. Determine the correlation. Give an interpretation of the correlation. e. Determine the coefficient of determination, R2 . Give an interpretation of R2 . f. One of the cars was going 25 mph and had a stopping distance of 85 feet. Determine the residual of this car.

5- The following two-way table describes the age and marital status of American women in 1991. The table entries are in thousands of women.

age	single	married	widowed	divorced
18-25	9008	3352	8	257
25-39	6658	21769	248	3224

40-64	19	24462	2570	4755
65+	900	7255	8464	925

a. Construct the marginal distributions of this table in counts. b. Draw a bar chart to display the marginal distribution of the marital status for all adult women (use percentages). c. What percent of adult American women under the age of 25 have never married? d. What percent of 25-39 year old American women are divorced? e. Compare the conditional distributions of marital status for women aged 18 to 24 and women aged 40 to 64. Briefly describe the most important differences between the two groups of women, and back up your description with percentages.

For problems 7 – 10 circle the best answer.

7. In the least-squares regression line, the desired sum of the errors (residuals) should be a. positive b. negative c. zero d. maximized

8. Suppose that a least squares regression line equation is ˆy = 1.65 − 2.20x and the actual y value corresponding to x = 10 is −19, what is the residual value corresponding to y = −19? a. 1.35 b. −1.35 c. 2.10 d. −2.10

Answer 1

Done in R

Solution4A:

a. Give a scatter plot of the data. Determine the form, direction and strength of the relationship between speed and stopping distance (dist).

plot(cars$dist,cars$speed,main="scatterplot")

Form:linear

strength:strong

Direction:positive

b. Determine the LSRL for predicting stopping distance based on speed of the car. c. Interpret the slope of this LSRL equation.

LSRL <- lm(cars$dist~cars$speed)
coefficients(LSRL)
Output:

Intercept) cars$speed
-17.579095 3.932409

least squares regression line is

distance= -17.579095 +3.932409 *speed

slope=3.932409

y intercept=-17.579095

d. Determine the correlation. Give an interpretation of the correlation

cor(cars$speed,cars$dist)

output:

0.8068949

r=0.8068949

there exists a strong positive relationship between speed and distance

as distance covered speed is high and viceversa.

e. Determine the coefficient of determination, R2 . Give an interpretation of R2

summary(LSRL)

ANSWER:

Multiple R-squared: 0.6511

R SQ=0.6511

65.11% VARIATION IN DISTANCE IS EXPLAINED BY SPEED

Explained variance=65.11%

unexpalined variance=100-65.11= 34.89%

f. One of the cars was going 25 mph and had a stopping distance of 85 feet. Determine the residual of this car.

we have distance= -17.579095 +3.932409 *speed

=-17.579095 +3.932409 *25

=80.73113

Resiudal=observed distan e-predicted distance

=85-80.73113

= 4.26887