1. Regression
Year
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Per capita
consumption
of margarine
8.2
7
6.5
5.3
5.2
4
4.6
4.5
4.2
3.7
Divorce rate in
Maine
5
4.7
4.6
4.4
4.3
4.1
4.2
4.2
4.2
3.7
Predicted
Divorce
rate
5.02
4.73
4.62
4.34
4.31
4.03
4.17
4.15
4.08
3.96
Residual
-.017
-.035
-.017
.065
-.012
.070
.029
.053
.123
-.260
Regress the Divorce rate in Maine (Y) on Per capita consumption of margarine (X) (data is real)
a) Calculate the predicted Y for all the observations and enter in the chart above.
b) Calculate the residual for all the observations and enter in the chart above.
c) Write out the regression line:
Divorce = 3.09 + .235 MargarineC
d) Find and interpret the r & r2:
r2 = .917. 91.7% of the variance of Divorces in Maine can be explained by the variance in
consumption of margarine.
r = .957. There is a strong positive correlation between Divorces in Maine and consumption of
margarine.
e) Find and interpret b:
.235. For every additional unit (pound?) consumed of margarine, the divorce rate in Maine will
rise by .235.
f) Find and interpret the Y intercept:
3.09. If people consume no margarine, then the Divorce rate in Maine will be 3.09.
g) Find and interpret the X intercept:
-13.16. People would have to consume -13.16 units (pounds?) of margarine so that there would
be no divorces in Maine.
h) If in 2015, people ate 2.6 pounds of margarine, what is the predicted divorce rate?
3.70 divorce rate.
Conduct a hypothesis test to see if the consumption of margarine has any impact on the Divorce rate in
Maine. Use α=.02 significance level. Note: degrees of freedom = n - 2
i) Using the critical value method, what is/are the critical value(s), and which distribution is being
used?
First, note: Ho: β = 0, H1: β ≠ 0
Two-tailed test. We don’t know standard deviation, so it’s a t-test.
Critical t-values: -2.90 & 2.90
j) Using the critical value method, what is the result (statistically) of the hypothesis test and why?
The test statistic (from running the regression) is: 9.38
Since the test statistics > critical value (that is, 9.38 > 2.90), we reject the null hypothesis and
accept the alternative hypothesis.
k) Draw a diagram to represent the previous test
rejection
rejection
-2.90
crit.
value
0
2.90
crit.
value
9.38
Test
stat
l) Use the p-value method to conduct a hypothesis test
p = .0000137. Since p < α (that is, .00001 < .02), we reject the null and accept the alternative.
m) State in English your results.
There is sufficient, statistically significant evidence to reject the hypothesis that the
consumption of margarine has no effect on the divorce rate in Maine, and we accept the
alternative, that the consumption of margarine does affect the divorce rate in Maine.
2. A sample of beers that were bought at a particular bar were measured for their volume. Test
whether the average beer was less than 16 ounces.
Data Set A: 15.7, 15.8, 16.0, 15.5, 15.7, 15.9, 16.3, 15.1, 15.4, 15.9, 16.0, 15.9,
16.1, 15.8, 15.5, 15.4, 15.8, 15.7, 16.2, 15.6
a. State the null and alternative hypothesis
H0: µ= 16 ounces
H0: µ< 16 ounces
b. Using the critical value method, what is/are the critical value(s), and which distribution
is being used?
Note, α = .01 (stated in class). Also, we MUST assume that the data is normally
distributed since the sample size is 30 or less.
It is a t-test, one-tail (left), 19 degrees of freedom.
Critical value = -2.54
c. Using the critical value method, what is the result (statistically) of the hypothesis test
and why?
Test stat = -3.57
Since the tests stat < critical value (that is -3.57 < -2.54), we reject the null and accept
the alternative hypothesis.
d. Draw a diagram to represent the previous test.
rejection
-3.57 -2.54
Test crit.
stat value
0
e. Using the P-value method, what is the result (statistically) of the hypothesis test and
why?
p = .001. Since p < α (that is .001 < .01), then reject the null, accept the hypothesis
f. State, in English, the result of the hypothesis test
There is sufficient information to conclude that the average beer poured is not 16
ounces, and we accept that the average beer poured is less than 16 ounces.
3. (10 points) 150 randomly selected voters were surveyed. 81 of the voters said they would vote
“yes” on Proposition O and/or P. Use α=.01 significance level. Conduct a hypothesis test to
see if a majority of voters will pass the propositions.
a. What conditions must hold make valid conclusions and are these conditions met?
We must have at least 10 people of each category (10 who would vote yes and 10 who
would vote no). We also much have 20 times as many people in the population as in
the sample. This means the population must be at least 3000 (150 * 20).
b. State the null and alternative hypothesis
H0: p = .5 (50%)
H0: p > .5 (50%)
c. Using the critical value method, what is/are the critical value(s), and which distribution
is being used?
Proportions use the normal distribution (z values). One tail (right) test.
z = 2.33
d. Using the critical value method, what is the result (statistically) of the hypothesis test
and why?
The test statistic = .980
We fail to reject the Null Hypothesis since the test stat is not more extreme than the
critical value (that is .98 < 2.33)
e. Draw a diagram to represent the previous test.
rejection
0 .980 2.33
Test crit.
stat value
f. Using the P-value method, what is the result (statistically) of the hypothesis test and
why?
We fail to reject the null because the p-value is not less than the level of significance,
that is p is not < α. p value = .164, which is not < .01.
g. State, in English, the result of the hypothesis test
There is insufficient evidence to conclude that a majority of voters are in favor
Proposition O & P.