Stat 240
Oct. 15
LAB KEY
Activity 1 (Section 9.2): Suppose that, p = .80 (80%) is the proportion of all patients with a certain
disease will be successfully treated with a newly available medication, although this is unknown to
researchers. A clinical trial with n=100 patients will be done, and the sample proportion successfully
treated will be computed in order to estimate the success rate.
Note: Over all possible random samples of n = 100 patients, the mean value of the sample proportion
= p = .80.
a. Calculate the standard deviation of the sampling distribution for potential sample proportions.
b. Use the Empirical Rule to fill in the blanks in the following 
For about 95% of all samples of n = 100 patients, the sample proportion successfully cured will be
between _______ and _________
For about 99.7% % of all samples of n = 100 patients, the sample proportion successfully cured will
be between ______ and _______
c. Suppose that in the clinical trial, 65 of the 100 patients are successfully treated. Explain whether
or not you think the sample result (65 of 100 successfully treated) is “significant” evidence against
the statement that the true population success rate is p = .80. Hint: The results for part (b) might be
useful.
d Use Minitab to find the probability that a sample proportion for 100 patients would be .75 or
smaller if the true success rate for all people is .80. Use Calc>Probability Distribution>Normal.
Enter mean = .80, std. dev. = answer in part a, and .65 as Input Constant.
e. Explain why the answer to part d would be evidence that the true success rate might not be .80.
ANSWERS
p( 1  p )
.8(1  .2)
a.
=
 .04
100
n
b. 95% interval is Mean  2 s.d, which is .80  2(.04), or .72 to .88
99.7% interval is Mean  3 s.d, which is .80  3(.04), or .68 to .92
c. .65 is too low. It’s below the range of values that would be expected if .80 is the true success rate
for all patients.
d. Probability that sample p would be less than .65 = .0001
By hand, z = (.65-.80)/.04 = -3.75. Use the “In the Extreme” portion of the table on page 612 to find
that the probability is .0001.
e. If true p = .80, there is only .0001 chance a sample result could be .65. The observation of .65 is
evidence against .80 because it would be unlikely to occur if .80 is the true population value.
CONFIDENCE INTERVALS CHAPTER 10
Activity 2: For the given sample survey result, determine the 95% confidence interval estimate of the
population percent. Assume the margin of error given is the “95% margin of error.”
In a random sample of 900 Pennsylvania drivers, 600 drivers say they are in favor of harsher penalties for
persons found guilty of driving under the influence. The margin of error is about 3%.
ANSWER. Find sample proportion = 600/900 =.67, which is 67%.
Interval is 67%  3%, or 64% to 70%.
With 95% confidence we can say that 64% and 70% of all Pennsylvania drivers favor harsher penalties
for DUI.
Activity 3 In our class survey done at the beginning of the semester, 37 of 49 students responding said
they believe there is extraterrestrial life.. The following results were created using Minitab, and they
provide a 95% confidence interval for the population proportion that thinks there is extraterrestrial life.
Variable
ET_Life
a.
b.
X
37
N
49
Sample p
0.755
In the output, what value is given under “Sample p?” Using values given in the output, show
how the “Sample p” was computed.
Refer to the previous part. Explain whether the “Sample p” is a statistic or a parameter. What
statistical notation should be used to represent the sample p?
By substituting values into the formula, show how to calculate the approximate “95% margin of
error in this case. The formula is 2 
c.
d.
95.0% CI
(0.632, 0.878)
pˆ (1  pˆ )
, and the answer is about .123.
n
Compute Sample p  Margin of error. Verify that the result equals the 95% confidence interval
given in the output.
What population do you think is represented by the sample of 49 students in our class?
ANSWERS:
a. 37/49
b. Statistic because it is a summary of a sample. The correct notation for a sample proportion is p̂
2
pˆ (1  pˆ )
.755(1  .755)
= 2
 .123
49
n
c. .755  .123 gives the values under “95% CI in the output”
d. This answer will vary from student to student. A conservative answer is that the population is Penn
State students in the majors required to take this class.
We asked the same question in Stat 200, a class taken by Liberal Arts students. The Minitab output for a
9%% confidence interval for the population proportion based on that sample is:
ariable
BelievET
X
153
N Sample p
281 0.544
95.0% CI
(0.486, 0.603)
Notice that the confidence intervals based on the two different samples don’t overlap indicating that
population proportions (believing in ET life) may be different. Liberal Arts students might be less likely
to believe there is extraterrestrial life.