# 10.3: Inference for Regression and Correlation

• • Contributed by Kathryn Kozak
• Professor (Mathematics) at Coconino Community College

How do you really say you have a correlation? Can you test to see if there really is a correlation? Of course, the answer is yes. The hypothesis test for correlation is as follows:

#### Hypothesis Test for Correlation:

1. State the random variables in words.
x = independent variable
y = dependent variable
2. State the null and alternative hypotheses and the level of significance
$$\begin{array}{l}{H_{o} : \rho=0 \text { (There is no correlation) }} \\ {H_{A} : \rho \neq 0 \text { (There is a correlation) }} \\ {\text { or }} \\ {H_{A} : \rho<0 \text { (There is a negative correlation) }} \\ {\text { or }} \\ {H_{A} : \rho>0 \text { (There is a postive correlation) }}\end{array}$$
Also, state your $$\alpha$$ level here.
3. State and check the assumptions for the hypothesis test
The assumptions for the hypothesis test are the same assumptions for regression and correlation.
4. Find the test statistic and p-value
$$t=\frac{r}{\sqrt{\frac{1-r^{2}}{n-2}}}$$
with degrees of freedom = df = n - 2
p-value: Using the TI-83/84: tcdf(lower limit, upper limit, df)

Note

If $$H_{A} : \rho<0$$, then lower limit is -1E99 and upper limit is your test statistic. If $$H_{A} : \rho>0$$, then lower limit is your test statistic and the upper limit is 1E99. If $$H_{A} : \rho \neq 0$$, then find the p-value for $$H_{A} : \rho<0$$, and multiply by 2.

Using R: pt(t, df)

Note

If $$H_{A} : \rho<0$$, then use pt(t, df), If $$H_{A} : \rho>0$$, then use $$1-\mathrm{pt}(t, d f)$$. If $$H_{A} : \rho \neq 0$$, then find the p-value for $$H_{A} : \rho<0$$, and multiply by 2.

5. Conclusion
This is where you write reject $$H_{o}$$ or fail to reject $$H_{o}$$. The rule is: if the p-value < $$\alpha$$, then reject $$H_{o}$$. If the p-value $$\geq \alpha$$, then fail to reject $$H_{o}$$.
6. Interpretation
This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show $$H_{A}$$ is true, or you do not have enough evidence to show $$H_{A}$$ is true.

Note

The TI-83/84 calculator results give you the test statistic and the p-value. In R, the command for getting the test statistic and p-value is cor.test(independent variable, dependent variable, alternative = "less" or "greater"). Use less for $$H_{A} : \rho<0$$, use greater for $$H_{A} : \rho>0$$, and leave off this command for $$H_{A} : \rho \neq 0$$.

Example $$\PageIndex{1}$$ Testing the claim of a linear correlation

Is there a positive correlation between beer’s alcohol content and calories? To determine if there is a positive linear correlation, a random sample was taken of beer’s alcohol content and calories for several different beers ("Calories in beer," 2011), and the data is in Table 10.2.1. Test at the 5% level.

Solution:

1. State the random variables in words.
x = alcohol content in the beer
y = calories in 12 ounce beer
2. State the null and alternative hypotheses and the level of significance.
Since you are asked if there is a positive correlation, $$\rho> 0$$.
$$\begin{array}{l}{H_{o} : \rho=0} \\ {H_{A} : \rho>0} \\ {\alpha=0.05}\end{array}$$
3. State and check the assumptions for the hypothesis test.
The assumptions for the hypothesis test were already checked in Example 10.1.2.
4. Find the test statistic and p-value.
The results from the TI-83/84 calculator are in Figure 10.3.1. Figure 10.3.1: Results for Linear Regression Test on TI-83/84
Test statistic: t $$\approx$$ 5.938 and p-value: $$p \approx 2.884 \times 10^{-4}$$
The results from R are
cor.test(alcohol, calories, alternative = "greater")
Pearson's product-moment correlation
data: alcohol and calories
t = 5.9384, df = 7, p-value = 0.0002884
alternative hypothesis: true correlation is greater than 0
95 percent confidence interval:
0.7046161 1.0000000
sample estimates:
cor
0.9134414
Test statistic: t $$\approx$$ 5.9384 and p-value: $$p \approx 0.0002884$$
5. Conclusion
Reject $$H_{o}$$ since the p-value is less than 0.05.
6. Interpretation
There is enough evidence to show that there is a positive correlation between alcohol content and number of calories in a 12-ounce bottle of beer.

#### Prediction Interval

Using the regression equation you can predict the number of calories from the alcohol content. However, you only find one value. The problem is that beers vary a bit in calories even if they have the same alcohol content. It would be nice to have a range instead of a single value. The range is called a prediction interval. To find this, you need to figure out how much error is in the estimate from the regression equation. This is known as the standard error of the estimate.

Definition

Standard Error of the Estimate
This is the sum of squares of the residuals

$$s_{e}=\sqrt{\frac{\sum(y-\hat{y})^{2}}{n-2}}$$

This formula is hard to work with, so there is an easier formula. You can also find the value from technology, such as the calculator.

$$s_{e}=\sqrt{\frac{S S_{y}-b^{*} S S_{x y}}{n-2}}$$

Example $$\PageIndex{2}$$ finding the standard error of the estimate

Find the standard error of the estimate for the beer data. To determine if there is a positive linear correlation, a random sample was taken of beer’s alcohol content and calories for several different beers ("Calories in beer," 2011), and the data are in Table 10.2.1.

Solution:

x = alcohol content in the beer

y = calories in 12 ounce beer

Using the TI-83/84, the results are in Figure 10.3.2. Figure 10.3.2: Results for Linear Regression Test on TI-83/84

The s in the results is the standard error of the estimate. So $$s_{e} \approx 15.64$$.

To find the standard error of the estimate in R, the commands are
lm.out = lm(dependent variable ~ independent variable) – this defines the linear model with a name so you can use it later. Then
summary(lm.out) – this will produce most of the information you need for a regression and correlation analysis. In fact, the only thing R doesn’t produce with this command is the correlation coefficient. Otherwise, you can use the command to find the regression equation, coefficient of determination, test statistic, p-value for a two-tailed test, and standard error of the estimate.

The results from R are
lm.out=lm(calories~alcohol)
summary(lm.out)

Call:
lm(formula = calories ~ alcohol)

Residuals:
$$\begin{array} {ccccc} {\text{Min}} & {\text{1Q}} & {\text{Median}} & {\text{3Q}} & {\text{Max}} \\{-30.253}&{-1.624}&{2.744}&{9.271}&{14.271} \end{array}$$

Coefficients:
$$\begin{array}{ccccc} {}&{\text{Estimate Std.}}&{\text{Error}}&{\text{t value}}&{\text{Pr(>|t|)}} \\ {\text{(Intercept)}}&{25.031}&{24.999}&{1.001}&{0.350038}\\{\text{alcohol}}&{26.319}&{4.432}&{5.938}&{0.000577}\end{array}$$
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 15.64 on 7 degrees of freedom
Multiple R-squared: 0.8344, Adjusted R-squared: 0.8107
F-statistic: 35.26 on 1 and 7 DF, p-value: 0.0005768

From this output, you can find the y-intercept is 25.031, the slope is 26.319, the test statistic is t = 5.938, the p-value for the two-tailed test is 0.000577. If you want the p-value for a one-tailed test, divide this number by 2. The standard error of the estimate is the residual standard error and is 15.64. There is some information in this output that you do not need.

If you want to know how to calculate the standard error of the estimate from the formula, refer to Example 10.3.3.

Example $$\PageIndex{3}$$ finding the standard error of the estimate from the formula

Find the standard error of the estimate for the beer data using the formula. To determine if there is a positive linear correlation, a random sample was taken of beer’s alcohol content and calories for several different beers ("Calories in beer," 2011), and the data are in Table 10.2.1.

Solution:

x = alcohol content in the beer

y = calories in 12 ounce beer

From Example 10.1.3 :

$$\begin{array}{l}{S S_{y}=\sum(y-\overline{y})^{2}=10335.56} \\ {S S_{x y}=\sum(x-\overline{x})(y-\overline{y})=327.6666} \\ {n=9} \\ {b=26.3}\end{array}$$

The standard error of the estimate is

\begin{aligned} s_{e} &=\sqrt{\frac{S S_{y}-b^{*} S S_{x y}}{n-2}} \\ &=\sqrt{\frac{10335.56-26.3(327.6666)}{9-2}} \\ &=15.67 \end{aligned}

#### Prediction Interval for an Individual y

Given the fixed value $$x_{0}$$, the prediction interval for an individual y is

$$\hat{y}-E<y<\hat{y}+E$$

where

$$\begin{array}{l}{\hat{y}=a+b x} \\ {E=t_{c} s_{e} \sqrt{1+\frac{1}{n}+\frac{\left(x_{o}-\overline{x}\right)^{2}}{S S_{x}}}} \\ {d f=n-2}\end{array}$$

Note

To find $$S S_{x}=\sum(x-\overline{x})^{2}$$ remember, the standard derivation formula from chapter 3 $$s_{x}=\sqrt{\frac{\sum(x-\overline{x})^{2}}{n-1}}$$

So, $$s_{x}=\sqrt{\frac{S S_{x}}{n-1}}$$

Now solve for $$S S_{x}$$

$$S S_{x}=s_{x}^{2}(n-1)$$

You can get the standard deviation from technology.
R will produce the prediction interval for you. The commands are (Note you probably already did the lm.out command. You do not need to do it again.)
lm.out = lm(dependent variable ~ independent variable) – calculates the linear model
predict(lm.out, newdata=list(independent variable = value), interval="prediction", level=C) – will compute a prediction interval for the independent variable set to a particular value (put that value in place of the word value), at a particular C level (given as a decimal)

Example $$\PageIndex{4}$$ find the prediction interval

Find a 95% prediction interval for the number of calories when the alcohol content is 6.5% using a random sample taken of beer’s alcohol content and calories ("Calories in beer," 2011). The data are in Table 10.2.1.

Solution:

x = alcohol content in the beer

y = calories in 12 ounce beer

Computing the prediction interval using the TI-83/84 calculator:
From Example 10.1.2

$$\begin{array}{l}{\hat{y}=25.0+26.3 x} \\ {x_{o}=6.50} \\ {\hat{y}=25.0+26.3(6.50)=196 \text { calories }}\end{array}$$

From Example #10.3.2

$$s_{e} \approx 15.64$$ Figure 10.3.3: Results of 1-Var Stats on TI-83/84

$$\begin{array}{l}{\overline{x}=5.517} \\ {s_{x}=1.247497495} \\ {n=9}\end{array}$$

Now you can find

\begin{aligned} S S_{x} &=s_{x}^{2}(n-1) \\ &=(1.247497495)^{2}(9-1) \\ &=12.45 \\ d f &=n-2=9-2=7 \end{aligned}

Now look in table A.2. Go down the first column to 7, then over to the column headed with 95%.

$$t_{c}=2.365$$

\begin{aligned} E &=t_{c} s_{e} \sqrt{1+\frac{1}{n}+\frac{\left(x_{o}-\overline{x}\right)^{2}}{S S_{x}}} \\ &=2.365(15.64) \sqrt{1+\frac{1}{9}+\frac{(6.50-5.517)^{2}}{12.45}} \\ &=40.3 \end{aligned}

Prediction interval is

$$\begin{array}{l}{\hat{y}-E<y<\hat{y}+E} \\ {196-40.3<y<196+40.3} \\ {155.7<y<236.3}\end{array}$$

Computing the prediction interval using R:
predict(lm.out, newdata=list(alcohol=6.5), interval = "prediction", level=0.95)
$$\begin{array}{ccc}{}&{\text { fit }} & {\text { lwr }} & {\text { upr }} \\ {1} & {196.1022} & {155.7847}&{236 .4196}\end{array}$$

fit = $$\hat{\mathcal{Y}}$$ when x = 6.5%. lwr = lower limit of prediction interval. upr = upper limit of prediction interval. So the prediction interval is $$155.8<y<236.4$$.

Statistical interpretation: There is a 95% chance that the interval $$155.8<y<236.4$$ contains the true value for the calories when the alcohol content is 6.5%.

Real world interpretation: If a beer has an alcohol content of 6.50% then it has between 156 and 236 calories.

Example $$\PageIndex{5}$$ Doing a correlation and regression analysis using the ti-83/84

Table 10.3.1 contains randomly selected high temperatures at various cities on a single day and the elevation of the city.

 Elevation (in feet) Temperature (°F) 7000 4000 6000 3000 7000 4500 5000 50 60 48 70 55 55 60

Table 10.3.1: Temperatures and Elevation of Cities on a Given Day

1. State the random variables.
2. Find a regression equation for elevation and high temperature on a given day.
3. Find the residuals and create a residual plot.
4. Use the regression equation to estimate the high temperature on that day at an elevation of 5500 ft.
5. Use the regression equation to estimate the high temperature on that day at an elevation of 8000 ft.
6. Between the answers to parts d and e, which estimate is probably more accurate and why?
7. Find the correlation coefficient and coefficient of determination and interpret both.
8. Is there enough evidence to show a negative correlation between elevation and high temperature? Test at the 5% level.
9. Find the standard error of the estimate.
10. Using a 95% prediction interval, find a range for high temperature for an elevation of 6500 feet.

Solution:

a. x = elevation

y = high temperature

b.

1. A random sample was taken as stated in the problem.
2. The distribution for each high temperature value is normally distributed for every value of elevation.
1. Look at the scatter plot of high temperature versus elevation. Figure 10.3.4: Scatter Plot of Temperature Versus Elevation
The scatter plot looks fairly linear.
2. There are no points that appear to be outliers.
3. The residual plot for temperature versus elevation appears to be fairly random. (See Figure 10.3.7.)
It appears that the high temperature is normally distributed.

All calculations computed using the TI-83/84 calculator. Figure 10.3.5: Setup for Linear Regression on TI-83/84 Calculator Figure 10.3.6: Results for Linear Regression on TI-83/84 Calculator

$$\hat{y}=77.4-0.0039 x$$

c.

x y $$\hat{\mathcal{Y}}$$ $$y-\hat{y}$$
7000 50 50.1 -0.1
4000 60 61.8 -1.8
6000 48 54.0 -6.0
3000 70 65.7 4.3
7000 55 50.1 4.9
4500 55 59.85 -4.85
5000 60 57.9 2.1

Table 10.3.2: Residuals for Elevation vs. Temperature Data Figure 10.3.7: Residual Plot for Temperature vs. Election

The residuals appear to be fairly random.

d.

$$\begin{array}{l}{x_{o}=5500} \\ {\hat{y}=77.4-0.0039(5500)=55.95^{\circ} F}\end{array}$$

e.

$$\begin{array}{l}{x_{o}=8000} \\ {\hat{y}=77.4-0.0039(8000)=46.2^{\circ} F}\end{array}$$

f. Part d is more accurate, since it is interpolation and part e is extrapolation.

g. From Figure 10.3.6, the correlation coefficient is r $$\approx$$ -0.814, which is moderate to strong negative correlation.

From Figure 10.3.6, the coefficient of determination is $$r^{2} \approx 0.663$$, which means that 66.3% of the variability in high temperature is explained by the linear model. The other 33.7% is explained by other variables such as local weather conditions.

h.

1. State the random variables in words.
x = elevation
y = high temperature
2. State the null and alternative hypotheses and the level of significance
$$\begin{array}{l}{H_{o} : \rho=0} \\ {H_{A} : \rho<0} \\ {\alpha=0.05}\end{array}$$
3. State and check the assumptions for the hypothesis test The assumptions for the hypothesis test were already checked part b.
4. Find the test statistic and p-value
From Figure 10.3.6,
Test statistic:
$$t \approx-3.139$$
p-value:
$$p \approx 0.0129$$
5. Conclusion
Reject $$H_{o}$$ since the p-value is less than 0.05.
6. Interpretation
There is enough evidence to show that there is a negative correlation between elevation and high temperatures.

i. From Figure 10.3.6,

$$s_{e} \approx 4.677$$

j. $$\hat{y}=77.4-0.0039(6500) \approx 52.1^{\circ} F$$ Figure 10.3.8: Results of 1-Var Stats on TI-83/84

$$\begin{array}{l}{\overline{x}=5214.29} \\ {s_{x}=1523.624} \\ {n=7}\end{array}$$

Now you can find

\begin{aligned} S S_{x} &=s_{x}^{2}(n-1) \\ &=(1523.623501)^{2}(7-1) \\ &=13928571.43 \\ d f &=n-2=7-2=5 \end{aligned}

Now look in table A.2. Go down the first column to 5, then over to the column headed with 95%.

$$t_{c}=2.571$$

So

\begin{aligned} E &=t_{c} s_{e} \sqrt{1+\frac{1}{n}+\frac{\left(x_{o}-\overline{x}\right)^{2}}{S S_{x}}} \\ &=2.571(4.677) \sqrt{1+\frac{1}{7}+\frac{(6500-5214.29)^{2}}{13928571.43}} \\ &=13.5 \end{aligned}

Prediction interval is

$$\begin{array}{l}{\hat{y}-E<y<\hat{y}+E} \\ {52.1-13.5<y<52.1+13.5} \\ {38.6<y<65.6}\end{array}$$

Statistical interpretation: There is a 95% chance that the interval $$38.6<y<65.6$$ contains the true value for the temperature at an elevation of 6500 feet.

Real world interpretation: A city of 6500 feet will have a high temperature between 38.6°F and 65.6°F. Though this interval is fairly wide, at least the interval tells you that the temperature isn’t that warm.

Example $$\PageIndex{6}$$ doing a correlation and regression analysis using r

Table 10.3.1 contains randomly selected high temperatures at various cities on a single day and the elevation of the city.

1. State the random variables.
2. Find a regression equation for elevation and high temperature on a given day.
3. Find the residuals and create a residual plot.
4. Use the regression equation to estimate the high temperature on that day at an elevation of 5500 ft.
5. Use the regression equation to estimate the high temperature on that day at an elevation of 8000 ft.
6. Between the answers to parts d and e, which estimate is probably more accurate and why?
7. Find the correlation coefficient and coefficient of determination and interpret both.
8. Is there enough evidence to show a negative correlation between elevation and high temperature? Test at the 5% level.
9. Find the standard error of the estimate.
10. Using a 95% prediction interval, find a range for high temperature for an elevation of 6500 feet.

Solution:

a. x = elevation

y = high temperature

b.

1. A random sample was taken as stated in the problem.
2. The distribution for each high temperature value is normally distributed for every value of elevation.
1. Look at the scatter plot of high temperature versus elevation.
R command: plot(elevation, temperature, main="Scatter Plot for Temperature vs Elevation", xlab="Elevation (feet)", ylab="Temperature (degrees F)", ylim=c(0,80)) Figure 10.3.9: Scatter Plot of Temperature Versus Elevation
The scatter plot looks fairly linear.
2. The residual plot for temperature versus elevation appears to be fairly random. (See Figure 10.3.10.)
It appears that the high temperature is normally distributed.

Using R:
Commands:
lm.out=lm(temperature ~ elevation)
summary(lm.out)

Output:
Call:
lm(formula = temperature ~ elevation)

Residuals:
$$\begin{array} {ccccccc} {1}&{2}&{3}&{4}&{5}&{6}&{7}\\{ 0.1667}&{-1.6333}&{-5.7667}&{4 .4333}&{5 .1667}&{-4.6667}&{ 2.3000} \end{array}$$

Coefficients:
$$\begin{array}{ccccc} {}&{\text{Estimate Std.}}&{\text{Error}}&{\text{t value}}&{\text{Pr(>|t|)}} \\ {\text{(Intercept)}}&{77.36667}&{6.769182}&{11.429}&{8.98e-05 ***}\\{\text{elevation}}&{-0.003933}&{0.001253}&{-3.139}&{0.0257*}\end{array}$$
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.677 on 5 degrees of freedom
Multiple R-squared: 0.6633, Adjusted R-squared: 0.596
F-statistic: 9.852 on 1 and 5 DF, p-value: 0.0257

From the output you can see the slope = -0.0039 and the y-intercept = 77.4. So the regression equation is:

$$\hat{y}=77.4-0.0039 x$$

c. R command: (notice these are also in the summary(lm.out) output, but if you have too many data points, then R only gives a numerical summary of the residuals.)

residuals(lm.out)

$$\begin{array} {CCCCCCC} {1}&{2}&{3}&{4}&{5}&{6}&{7} \\ {0.1666667}&{-1.63333333}&{-5.766667}&{4 .43333333}&{5 .1666667}&{-4.66666667}&{2.3000000} \end{array}$$

So for the first x of 7000, the residual is approximately 0.1667. This means if you find the $$\hat{y}$$ for when x is 7000 and then subtract this answer from the y value of 50 that was measured, you would obtain 0.1667. Similar process is computed for the other residual values.

To plot the residuals, the R command is

plot(elevation, residuals(lm.out), main="Residual Plot for Temperautre vs Elevation", xlab="Elevation (feet)", ylab="Residuals")
abline(0,0) Figure 10.3.10: Residual Plot for Temperature vs. Elevation

The residuals appear to be fairly random.

d. $$\begin{array}{l}{x_{o}=5500} \\ {\hat{y}=77.4-0.0039(5500)=55.95^{\circ} F}\end{array}$$

e. $$\begin{array}{l}{x_{o}=8000} \\ {\hat{y}=77.4-0.0039(8000)=46.2^{\circ} F}\end{array}$$

f. Part d is more accurate, since it is interpolation and part e is extrapolation.

g. The R command for the correlation coefficient is
cor(elevation, temperature)
 -0.8144564

So, $$r \approx-0.814$$, which is moderate to strong negative correlation.

From summary(lm.out), the coefficient of determination is the Multiple R-squared.

So $$r^{2} \approx 0.663$$, which means that 66.3% of the variability in high temperature is explained by the linear model. The other 33.7% is explained by other variables such as local weather conditions.

h.

1. State the random variables in words.
x = elevation
y = high temperature
2. . State the null and alternative hypotheses and the level of significance
$$\begin{array}{l}{H_{o} : \rho=0} \\ {H_{A} : \rho<0} \\ {\alpha=0.05}\end{array}$$
3. State and check the assumptions for the hypothesis test.
The assumptions for the hypothesis test were already checked part b.
4. Find the test statistic and p-value
The R command is cor.test(elevation, temperature, alternative = "less")
Pearson's product-moment correlation
data: elevation and temperature
t = -3.1387, df = 5, p-value = 0.01285
alternative hypothesis: true correlation is less than 0
95 percent confidence interval:
-1.0000000 -0.3074247
sample estimates:
cor
-0.8144564
Test statistic: $$t \approx-3.1387$$ and p-value: $$p \approx 0.01285$$
5. Conclusion
Reject $$H_{o}$$ since the p-value is less than 0.05.
6. Interpretation
There is enough evidence to show that there is a negative correlation between elevation and high temperatures.

i. From summary(lm.out), Residual standard error: 4.677.

So, $$s_{e} \approx 4.677$$

j. R command is predict(lm.out, newdata=list(elevation = 6500), interval = "prediction", level=0.95)

$$\begin{array}{cccc}{}&{\text { fit }}&{ \text { lwr} } &{ \text { upr }} \\ {1}&{51.8}&{38 .29672}&{65 .30328}\end{array}$$

So when x = 6500 feet, $$\hat{y}=51.8^{\circ} F \text { and } 38.29672<y<65.30328$$.

Statistical interpretation: There is a 95% chance that the interval $$38.3<y<65.3$$ contains the true value for the temperature at an elevation of 6500 feet.

Real world interpretation: A city of 6500 feet will have a high temperature between 38.3°F and 65.3°F. Though this interval is fairly wide, at least the interval tells you that the temperature isn’t that warm.

#### Homework

Exercise $$\PageIndex{1}$$

For each problem, state the random variables. The data sets in this section are in the homework for section 10.1 and were also used in section 10.2. If you removed any data points as outliers in the other sections, remove them in this sections homework too.

1. When an anthropologist finds skeletal remains, they need to figure out the height of the person. The height of a person (in cm) and the length of their metacarpal bone one (in cm) were collected and are in Table 10.1.5 ("Prediction of height," 2013).
1. Test at the 1% level for a positive correlation between length of metacarpal bone one and height of a person.
2. Find the standard error of the estimate.
3. Compute a 99% prediction interval for height of a person with a metacarpal length of 44 cm.
2. Table 10.1.6 contains the value of the house and the amount of rental income in a year that the house brings in ("Capital and rental," 2013).
1. Test at the 5% level for a positive correlation between house value and rental amount.
2. Find the standard error of the estimate.
3. Compute a 95% prediction interval for the rental income on a house worth \$230,000.
3. The World Bank collects information on the life expectancy of a person in each country ("Life expectancy at," 2013) and the fertility rate per woman in the country ("Fertility rate," 2013). The data for 24 randomly selected countries for the year 2011 are in Table 10.1.7.
1. Test at the 1% level for a negative correlation between fertility rate and life expectancy.
2. Find the standard error of the estimate.
3. Compute a 99% prediction interval for the life expectancy for a country that has a fertility rate of 2.7.
4. The World Bank collected data on the percentage of GDP that a country spends on health expenditures ("Health expenditure," 2013) and also the percentage of women receiving prenatal care ("Pregnant woman receiving," 2013). The data for the countries where this information is available for the year 2011 are in Table 10.1.8.
1. Test at the 5% level for a correlation between percentage spent on health expenditure and the percentage of women receiving prenatal care.
2. Find the standard error of the estimate.
3. Compute a 95% prediction interval for the percentage of woman receiving prenatal care for a country that spends 5.0 % of GDP on health expenditure.
5. The height and weight of baseball players are in Table 10.1.9 ("MLB heightsweights," 2013).
1. Test at the 5% level for a positive correlation between height and weight of baseball players.
2. Find the standard error of the estimate.
3. Compute a 95% prediction interval for the weight of a baseball player that is 75 inches tall.
6. Different species have different body weights and brain weights are in Table 10.1.10. ("Brain2bodyweight," 2013).
1. Test at the 1% level for a positive correlation between body weights and brain weights.
2. Find the standard error of the estimate.
3. Compute a 99% prediction interval for the brain weight for a species that has a body weight of 62 kg.
7. A random sample of beef hotdogs was taken and the amount of sodium (in mg) and calories were measured. ("Data hotdogs," 2013) The data are in Table 10.1.11.
1. Test at the 5% level for a correlation between amount of calories and amount of sodium.
2. Find the standard error of the estimate.
3. Compute a 95% prediction interval for the amount of sodium a beef hotdog has if it is 170 calories.
8. Per capita income in 1960 dollars for European countries and the percent of the labor force that works in agriculture in 1960 are in Table 10.1.12 ("OECD economic development," 2013).
1. Test at the 5% level for a negative correlation between percent of labor force in agriculture and per capita income.
2. Find the standard error of the estimate.
3. Compute a 90% prediction interval for the per capita income in a country that has 21 percent of labor in agriculture.
9. Cigarette smoking and cancer have been linked. The number of deaths per one hundred thousand from bladder cancer and the number of cigarettes sold per capita in 1960 are in Table 10.1.13 ("Smoking and cancer," 2013).
1. Test at the 1% level for a positive correlation between cigarette smoking and deaths of bladder cancer.
2. Find the standard error of the estimate.
3. Compute a 99% prediction interval for the number of deaths from bladder cancer when the cigarette sales were 20 per capita.
10. The weight of a car can influence the mileage that the car can obtain. A random sample of cars weights and mileage was collected and are in Table 10.1.14 ("Passenger car mileage," 2013).
1. Test at the 5% level for a negative correlation between the weight of cars and mileage.
2. Find the standard error of the estimate.
3. Compute a 95% prediction interval for the mileage on a car that weighs 3800 pounds.

For hypothesis test just the conclusion is given. See solutions for entire answer.

1. a. Reject Ho,    b. $$s_{e} \approx 4.559$$,    c. $$151.3161 \mathrm{cm}<y<187.3859 \mathrm{cm}$$

3. a. Reject Ho,    b. $$s_{e} \approx 3.204$$,    c. $$62.945 \text { years }<y<81.391 \text{years}$$

5. a. Reject Ho,    b. $$s_{e} \approx 15.33$$,    c. $$176.02 \text { inches }<y<240.92 \text{inches}$$

7. a. Reject Ho,    b. $$s_{e} \approx 48.58$$,    c. $$348.46 \mathrm{mg}<y<559.38 \mathrm{mg}$$

9. a. Reject Ho,    b. $$s_{e} \approx 0.6838$$,    c. $$1.613 \text { hundred thousand }<y<5.432 \text{ hundred thousand}$$

#### Data Source:

Calories in beer, beer alcohol, beer carbohydrates. (2011, October 25). Retrieved from www.beer100.com/beercalories.htm

Capital and rental values of Auckland properties. (2013, September 26). Retrieved from http://www.statsci.org/data/oz/rentcap.html