Stat 250.3
November 10, 2003
SOLUTIONS
Activity 7: Confidence Intervals and Hypothesis Tests
Situation 1:
A sociologist wants to know how much more money on average a husband makes than his wife. The sociologist records the salaries
for 40 random couples (assume both husband and wife are employed).
A) What type of data is this? Paired
B) Suppose that the sample mean salary for husbands was $52,000 and the sample mean salary for women was $41,000. Differences
are taken for each pair and the husband’s salary is $11,000 more on average with a standard deviation of the differences equal to
$8000. Construct a 95% confidence interval for the mean difference in salaries between husbands and wives.
Parameter of Interest: μd: average difference of salaries in a couple (husband minus wife).
Statistic of Interest:
d
=$11,000
Conditions: n=40>30
Confidence interval calculations: df = 40-1=39. Looking in the Table A.2, for 30 df (next lowest for 39df) and Confidence level 95%,
we have that the multiplier t* = 2.04.
sd 8000
=
 1264.9 .
n
40
Thus, the 95% CI for μd is d ± t* s.e.( d
The s.e.(
d
)=
) = 11 000 ± (2.04) (1264.9) = ($8,414.58, $13,580.42)
Interpretation of the CI: We are 95% confident that on average the salary of a husband is from $8,414.58 to $13,580.42 higher
than the salary of his wife.
C) Test the hypothesis that on average the salaries of the husbands are higher than the ones of their wives.
H0: μd =0 vs Ha: μd≠0
t
d  0 11000

 8.69 , and the p-value=P(T> 8.9) > 0.05 (where T is a t random variable with 39 df.).
sd
1264.9
n
So the conclusion is that we reject the H0, and conclude the on average, the husband salary is higher than the one of his wife.
Using the CI above, without any further calculations we could have rejected H0 at significance level alpha=0.025. However,
after we got the p-value, we can see that H0 is rejected for alpha = 0.05 as well.
Situation 2:
A scientist wants to know how effective a new drug is at lowering blood pressure. He randomly assigns 70 patients to receive either the
blood pressure lowering drug, or a placebo (half get each). The average decrease in blood pressure for the drug group is 8 with a
standard deviation of 10. The average decrease for placebo was 2 with a standard deviation of 4.
A) Construct an appropriate 99% confidence interval for the difference in the mean decrease in blood pressure between the drug and
placebo.
Parameter of Interest: µ1-µ2 , where µ1 is the average decrease in blood pressure for people using the drug, and µ 2 is the
average decrease in blood pressure for people using the placebo.
Statistic of Interest:
x1  x2  8  4  4
Conditions: n1=n2=35 > 30, and the two samples are random and independent.
Confidence interval calculations: df = min(35-1, 35,1)= 34. Looking in the Table A.2, for 30 df (next lowest for 34df) and
Confidence level 99%, we have that the multiplier t* = 2.75.
s.e.( x1  x ) 
s1 2 s 2 2


n1 n2
10
2
35

4
2
35
 1.82
Thus the 99% CI for µ1-µ2 is 4 ± (2.75) (1.82) = (-1.005, 9.005).
B) Test the hypothesis that the drug is effective. (i.e. works better than the placebo) at alpha=0.01 significance level.
H0: µ1-µ2 = 0 vs Ha: µ1-µ2 ≠0
Since zero falls in the 99%CI for µ1-µ2 , we fail to reject the null hypothesis at alpha=0.01 significance level, and conclude that
there is no significant difference between the two means. Therefore, the drug is not significantly effective testing at
alpha=0.01 significance level.
Stat 250.3
November 10, 2003
Situation 3: 12.7 From the Text
A study was done by randomly assigning 200 volunteers with sore throats to either drink a cup of herbal tea or use a sore throat
lozenge to ease their pain. The percent reporting relief for the two methods were compared. What are the research question, the
population parameter, and the sample statistic for this study?
Research question: How much difference is there between the percent getting relief from sore throat symptoms if herbal tea
is used versus if throat lozenges are used?
Population parameter: p1  p 2 = difference in percents reporting relief for population that would use herbal tea and population
that would use throat lozenge.
Sample estimate:
volunteers.
pˆ 1  pˆ 2
= difference in observed percents reporting relief for the two different methods in the study of 200
Situation 4:
A random sample of students was surveyed on whether they smoked pot, and whether they ever tried a “harder” drug (cocaine, heroin,
etc.) The summary statistics for the two variables were as follows:
X = Number who Used a Harder Drug
Test and CI for Two Proportions
Sample
Pot
No Pot
X
N Sample p
34 140 0.242857
40 248 0.161290
Estimate for p(1) - p(2): 0.0815668
95% lower bound for p(1) - p(2): 0.0106494
Test for p(1) - p(2) = 0 (vs > 0): Z = 1.89 P-Value = 0.029
A) Create a 95% for the difference in proportion of students that smoked pot and that did not smoked pot that tried a “harder” drug.
Let p1 be the proportion of students that smoked pot, that tried a harder drug, and p2 be the proportion of students that
haven’t smoked pot, that tried a harder drug,
pˆ 1  pˆ 2  34 / 140  40 / 248  0.08156,
se(
pˆ 1  pˆ 2 ) 
pˆ 1(1  pˆ 1) pˆ 2(1  pˆ 2)

 0.04311
n1
n2
So, the 95% CI for p1 -p2 is 0.08156 ± 2(0.04311) = (-0.0046, 0.16779)
B) Is there enough evidence to conclude that a higher proportion of students who smoked pot than students who did not try pot have
tried a “harder” drug?
Since zero falls in the interval calculated in part (A), we fail to reject the null hypothesis that the two proportions are equal at
alpha=0.025 significance level.
However, if you go through the steps to test the hypothesis H0: p1 -p2 =0 vs Ha p1 -p2 >0, you will get z-stat = 1.89 p-value =
0.029. Using the p-value, we can conclude that we REJECT H0 at alpha=0.05.