Solutions

dependent variable: fee amount

independent variable: time

y = 6x + 8, 4y = 8, and y + 7 = 3x are all linear equations.

The number of flu cases depends on the year. Therefore, year becomes the independent variable and the number of flu cases is the dependent variable.

The y-intercept is 50 (a = 50). At the start of the cleaning, the company charges a one-time fee of $50 (this is when x = 0). The slope is 100 (b = 100). For each session, the company charges $100 for each hour they clean.

12,000 lb of soil

The slope is –1.5 (b = –1.5). This means the stock is losing value at a rate of $1.50 per hour. The y-intercept is $15 (a = 15). This means the price of stock before the trading day was $15.

ŷ = 2.23 + 1.99x

The slope is 1.99 (b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another $1.99 million in pay each year.

It means that there is no correlation between the data sets.

Yes. There are enough data points and the value of r is strong enough to show there is a strong negative correlation between the data sets.

H_a: ρ ≠ 0

$250,120

1326 acres

1125 hours, or when x = 1125

Check student solution.

1. When x = 1985, ŷ = 25,52.
2. When x = 1990, ŷ = 34,275.
3. When x = 1970, ŷ = –725. Why doesn’t this answer make sense? The range of x values was 1981 to 2002; the year 1970 is not in this range. The regression equation does not apply, because predicting for the year 1970 is extrapolation, which requires a different process. Also, a negative number does not make sense in this context, when we are predicting flu cases diagnosed.

Also, the correlation r = 0.4526. If r is compared with the value in the 95 Percent Critical Values of the Sample Correlation Coefficient Table, because r > 0.423, r is significant, and you would think that the line could be used for prediction. But, the scatter plot indicates otherwise.

Check student’ solution.

$ŷ$ = 3,448,225 + 1750x

There was an increase in flu cases diagnosed until 1993. From 1993 through 2002, the number of flu cases diagnosed declined each year. It is not appropriate to use a linear regression line to fit to the data.

Because there is no linear association between year and number of flu cases diagnosed, it is not appropriate to calculate a linear correlation coefficient. When there is a linear association and it is appropriate to calculate a correlation, we cannot say that one variable causes the other variable.

We don’t know if the pre-1981 data were collected from a single year. So, we don’t have an accurate x value for this figure.

Regression equation: ŷ (number of flu cases) = –3,448,225 + 1749.777 (year).

• a = –3,488,225
• b = 1,750
• correlation = 0.4526
• n = 22

No, he is not correct. An outlier is only an influential point if it significantly impacts the slope of the least-squares regression line and the correlation coefficient, r. If omission of this data point from the calculation of the regression line does not show much impact on the slope or r-value, then the outlier is not considered an influential point. For different reasons, it still may be determined that the data point must be omitted from the data set.

Yes. There appears to be an outlier at (6, 58).

The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate as a predictor for the data.

s = 1.75

1. independent variable: age; dependent variable: fatalities
2. independent variable: number of family members; dependent variable: grocery bill
3. independent variable: age of applicant; dependent variable: insurance premium
4. independent variable: power consumption; dependent variable: utility
5. independent variable: higher education (years); dependent variable: crime rates

It means that 72 percent of the variation in the dependent variable (y) can be explained by the variation in the independent variable (x).

b. Check student solution.

c. ŷ = 35.5818045 – 0.19182491x

d. r = –0.57874
For four degrees of freedom and alpha = 0.05, the LinRegTTest gives a p value of 0.2288, so we do not reject the null hypothesis; there is not a significant linear relationship between deaths and age.
Using the table of critical values for the correlation coefficient, with four degrees of freedom, the critical value is 0.811. The correlation coefficient r = –0.57874 is not less than –0.811, so we do not reject the null hypothesis.

f. There is not a linear relationship between the two variables, as evidenced by a p value greater than 0.05.

a. We wonder if the better discounts appear earlier in the book, so we select page as x and discount as y.

b. Check student solution.

c. ŷ = 17.21757 – 0.01412x

d. r = – 0.2752
For seven degrees of freedom and alpha = 0.05, LinRegTTest gives a p value = 0.4736, so we do not reject; there is a not a significant linear relationship between page and discount.
Using the table of critical values for the correlation coefficient, with seven gives degrees of freedom, the critical value is 0.666. The correlation coefficient xi = –0.2752 is not less than 0.666, so we do not reject.

f. There is not a significant linear correlation so it appears there is no relationship between the page and the amount of the discount.

As the page number increases by one page, the discount decreases by $0.01412.

a. Year is the independent or x variable; the number of letters is the dependent or y variable.

b. Check student’s solution.

c. No.

d. ŷ = 47.03 – 0.0216x

e. –0.4280. The r value indicates that there is not a significant correlation between the year the state entered the Union and the number of letters in the name.

g. No. The relationship does not appear to be linear; the correlation is not significant.

Using LinRegTTest, the output for the original least-squares regression line is $y=26.14+0.7539x$ and $r=0.6657\text{.}$

The output for the new least-squares regression line, after omitting the outlier of (56, 95), is $ŷ=6.36+1.0045x$ and $r=0.9757\text{.}$

The slope of the new line is quite a bit different from the slope of the original least-squares regression line, but the larger change is shown in the r-values, such that the new line has an r-value that has increased to a value that is almost equal to one.

Thus, it may be stated that the outlier (56, 95) is also an influential point.

a. and b. Check student solution.

c. The slope of the regression line is –0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32 percent new sparrow hawks, which doesn’t make sense, because if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 30.3 percent.

d. If we examine r₂, we see that only 57.52 percent of the variation in the percentage of new birds is explained by the model and the correlation coefficient, r = –.7584 only indicates a somewhat strong correlation between returning and new percentages.

e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66 percent, our observed new percentage, 6 percent, is almost 6 percent less than the predicted new value of 11.98 percent. If we remove this data pair, we see only an adjusted slope of –0.2789 and an adjusted intercept of 30.9816. In other words, although these data generate the largest residual, it is not an outlier, nor is the data pair an influential point.

f. If there are 70 percent returning birds, we would expect to see y =– 0.2789(70) + 30.9816 = 0.114 or 11.4 percent new birds in the colony.

1. Check student solution.
2. Check student solution.
3. We have a slope of –1.4946 with a y-intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate decreases by 1.5 beats per minute. If the student is not swimming at all, the y-intercept indicates that his heart rate will be 193.88 beats per minute. Although the slope has meaning (the longer it takes to swim 2000 m, the less effort the heart puts out), the y-intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low.
4. Because only 1.5 percent of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship.
5. Point (34.72, 124) generates the largest residual: –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes –2.953, with the y-intercept changing to 247.1616. Although the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y-intercept becomes more meaningful.

If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = –0.0009x + 160, with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set.

c. No. The y-intercept would occur at year 0, which doesn’t exist.

1. Check student's solution.
2. Yes.
3. No, the y-intercept would occur at year 0, which doesn’t exist.
4. ŷ = −266.8863 + 0.1656x.
5. 0.9448, yes.
6. 62.8233, 62.3265.
7. Yes.
8. No, (1987, 62.7).
9. 72.5937, no.
10. Slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656.

1. Size (ounces)	Cost ($)	Cost per ounce
   16	3.99	24.94
   32	4.99	15.59
   64	5.99	9.36
   200	10.99	5.50

   Table 12.42
2. Check student solution.
3. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size.
4. ŷ = 20.2368 – 0.0819x
5. r = –.8086
6. 40-oz: 16.96 cents/oz
7. 90-oz: 12.87 cents/oz
8. The relationship is not linear; the least-squares line is not appropriate.
9. There are no outliers.
10. No. You would be extrapolating. The 300-oz size is outside the range of x.
11. X = –0.08194. For each additional ounce in size, the cost per ounce decreases by 0.082 cents.

1. Size is x, the independent variable, and price is y, the dependent variable.
2. Check student solution.
3. The relationship does not appear to be linear.
4. ŷ = –745.252 + 54.75569x.
5. r = .8944 and yes, it is significant.
6. 32-inch: $1006.93, 50-inch: $1992.53.
7. No, the relationship does not appear to be linear. However, r is significant.
8. No, the 60-inch TV.
9. For each additional inch, the price increases by $54.76.

1. Rank is the independent variable and area is the dependent variable.
2. Check student solution.
3. There appears to be a linear relationship, with one outlier.
4. ŷ (area) = 24177.06 + 1010.478x
5. r = .50047. r is not significant, so there is no relationship between the variables.
6. Alabama: 46,407.576 square miles, Colorado: 62,575.224 square miles.
7. The Alabama estimate is closer than the Colorado estimate.
8. If the outlier is removed, there is a linear relationship.
9. There is one outlier (Hawaii).
10. rank 51: 75,711.4 square miles, no.

11. Alabama	7	1819	22	52,423
    Colorado	8	1876	38	104,100
    Hawaii	6	1959	50	10,932
    Iowa	4	1846	29	56,276
    Maryland	8	1788	7	12,407
    Missouri	8	1821	24	69,709
    New Jersey	9	1787	3	8,722
    Ohio	4	1803	17	44,828
    South Carolina	13	1788	8	32,008
    Utah	4	1896	45	84,904
    Wisconsin	9	1848	30	65,499

    Table 12.43
12. ŷ = –87065.3 + 7828.532x.
13. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier.
14. Yes, with the exception of Hawaii.