Elementary Statistics
Ch 12 Practice Test
Daphne Skipper
Chapter 12
1. Suppose we are interested in estimating a bear’s weight based on its age. Figure ??
is a scatterplot of bear ages to bear weights. The sample correlation coefficient for
this paired data is r = 0.749. The critical r value is rc = ±0.268. The p-value in a
hypothesis test for linear correlation is 0.000. The regression equation for the line of
best fit is ŷ = 65.2 + 2.7x. Can we use the regression equation to estimate the weight of
a bear given its age in months?
Figure 1: Bear age in months (x) versus Bear weight (y)
Solution: The correlation coefficient and p-value WOULD indicate that there is a
linear correlation in the underlying population if the scatterplot showed a somewhat
linear relationship between bear ages and weights. However, although the bear weight
seems to generally increase with age, there is not a reliable linear pattern across the
range of bear ages. The weights are clustered for young bears, then fan out as bears
get older. Linear correlation and regression analysis is only appropriate for data that
does not display a distinctly nonlinear pattern, including fanning, in the scatterplot.
2. In the scatterplot below, x = car weight in pounds and y = highway gas mileage (miles
per gallon, or mpg).
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Daphne Skipper
(a) Which is the most likely value of the sample correlation coefficient, r:
0.7927, −0.7927, 0.2314, or −0.2314?
Solution: −0.7927; There is a reasonably strong negative linear correlation.
As car weight increases, cars get fewer miles per gallon.
(b) Using the regression line, estimate the gas mileage of a car that weighs 3100 pounds.
Solution: ŷ = 31.5mpg
(c) Approximate the residual associated with the car in the data set that weighs 3100
pounds.
Solution: The car that weighs 3100 pounds looks like it gets about 26 mpg.
So the residual is about y − ŷ = 26 − 31.5 = −5.5.
(d) Circle any potential outlier(s) (if any) on scatterplot. What is special about this
(these) cars?
Solution: The only potential outlier on graph is represented by the point at
the very top of the graph, because it has a much greater residual than any other
point. This car gets very good gas mileage, even though it is not extremely
light. Maybe it is a hybrid or something. (A common misconception is that any
graph not on the regression line is an “outlier”. That is not the case. Outliers
are located by the size of their residuals (vertical distance from the regression
line) compared to the residuals of the other points in the plot.)
3. Is it possible to tell the temperature by counting cricket chirps? The following data is
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Daphne Skipper
the number of chirps in 15 seconds counted at various temperatures.
Number of chirps Degrees Farenheit
20
89
18
82
20
93
18
84
17
81
16
75
15
70
16
74
14
69
16
83
15
80
17
83
16
81
17
84
14
76
19
86
(a) Decide which variable should be independent (x) variable and which should be the
dependent (y) variable.
Solution: independent (x) = number of chirps; dependent (y) = temperature
(b) Make a scatterplot of the data. Does it appear to follow a linear trend?
Solution: Use STATPLOT. Yes, it appears to follow a linear trend.
(c) Write the null and alternative hypotheses for a hypothesis test for linear correlation
of the underlying population of paired data.
Solution: H0 : ρ = 0 (There is no linear correlation in the population.)
Ha : ρ 6= 0 (There is linear correlation in the population.)
(d) Find the sample correlation coefficient r and the r critical values. Plot all on a
numberline from -1 to 1.
Solution: r = 0.865 and rc = ±0.497. The numberline extends from -1 to 1.
The “linear correlation” intervals are from -1 to -0.497 and 0.497 to 1. The “no
evidence of linear correlation” region is between -0.497 and 0.497. r falls in the
(positive) linear correlation interval.
(e) Find the p-value.
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Daphne Skipper
Solution: p-value = 0.00002
(f) Is there evidence of linear correlation in the underlying population?
Solution: Yes. (Based on the r critical value and on the p-value.)
(g) Find the regression equation of the line of best fit. Use one decimal place for a and
b.
Solution: ŷ = 30.5 + 3.0x
(h) Estimate the temperature when there are 19 chirps in 15 seconds.
Solution: ŷ = 30.5 + 3.0(19) = 87.5◦
(i) Estimate the temperature when there are 30 chirps in 15 seconds.
Solution: ŷ = 30.5 + 3.0(30) = 120.5◦
(j) Does your answer in the previous part make sense? Explain.
Solution: No. 120.5◦ is REALLY hot. 30 chirps is outside of the data range,
so we can’t rely on this estimate.
Many of these problems are from Barbara Illowsky & Susan Dean. “Introductory Statistics.” OpenStax College, 2013. iBooks.
(Chapter 2) https://itun.es/us/kFeL1.l