AP Statistics
Confidence Interval w/ Means Quiz
Name: _________________________________
___________1.
The correct answer is II, III, and IV. I is not correct since you cannot make a probability statement about a
found interval—the true mean is either in the interval (P = 1) or it isn't (P = 0). II is correct and is just a
restatement of "Intervals constructed by this procedure will capture the true mean 99% of the time." III is true—
it's our standard way of interpreting a confidence interval. IV is true. Since the interval constructed does not
contain 0, it's unlikely that this interval came from a population whose true mean is 0. Since all the values are
positive, the interval does provide statistical evidence (but not proof ) that the program is effective at promoting
weight loss. It does not give evidence that the amount lost is of practical importance.
___________2.
The correct answer is 1.753. n = 16 df = 16 – 1 = 15. Using a table of t distribution critical values, look in
the row for 15 degrees of freedom and the column with 0.05 at the top (or 90% at the bottom). On a TI-84 with
the invT function, the solution is given by invT(0.95,15).
___________3.
The correct answer is 74.5 ± 1.871 . For df = 15 – 1 = 14, t* = 2.264 for a 96% confidence interval (from Table
B; if you have a TI-84 with the invT function, invT(0.98,14)=2.264. The interval is 74.5 ± (2.264)
74.5 ± 1.871.
=
_____________4. A certain type of pen is claimed to operate for a mean of 190 hours. A random sample of 49
pens is tested, and the mean operating time is found to be 188 hours with a standard deviation
of 6 hours. Construct a 95% confidence interval for the true mean operating time of this type
of pen.
t = 2.011 from invT (.975, 48) The interval is 186 ± (2.011)
4
= 188 ± 1.14893….
√49
(184.85, 187.15)
t = 2.011 from invT (.975, 48) The interval is 188 ± (2.011)
(186.56, 189.44)
5
= 188 ± 1.14836….
√49
AP Statistics
Confidence Interval w/ Means Quiz
Name: _________________________________
5. A study was conducted to determine if male and female 10th graders differ in performance in
mathematics. Twenty-three randomly selected males and 26 randomly selected females were each given
a 50-question multiple-choice test as part of the study. The scores were approximately normally
distributed. The results of the study were as follows:
Construct a 99% confidence interval for the true difference between the mean score for males and the
mean score for females. Does the interval suggest that there is a difference between the true means for
males and females?
Girls: (36.13, 42.29)
Guys (36.711, 43.889)
Differences (.581, 1.619)
Does this suggest a different between the true MEANS? NO
Why? There is a minimum amount of variance between the top and bottom numbers which are the barriers of
the 95% confidence interval…not the means. On the bottom end there is a .581 difference which is one
sixteenth of a standard deviation. On the top end, there is a 1.619 difference which is approximately 1/5 of a
standard deviation. On the overall scheme of things, these are very small variations in grade distributions. One
person’s score could ultimately make this difference.
6. Lie detectors are based on measuring changes in the nervous system. The assumption is that lying will
be reflected in physiological changes that are not under the voluntary control of the individual. When a
person is telling the truth, the galvanic skin response scores have a distribution that is normal with a
mean of 48.4/47.4 and a standard deviation of 2.5.
a) What is the probability that a person will have a score less than 45?
Normal cdf (-1E99, 45, 48.4, 2.5) = .0869
Normal cdf (-1E99, 45, 47.4, 2.5) = .169
b) Suppose a SRS of 36 people is taken, what is the distribution of X ?
T-distribution…wider and shorted graph than that of a normal distribution. We use this distribution
because we do not know the standard deviation of the population, only the sample.
c) What is the probability that the average score will be more than 50?
Normal cdf (-1E99, 45, 48.4, 2.5/√36) = .00006153 = 0
Normal cdf (-1E99, 45, 47.4, 2.5/√36) = .0000000002129 = 0
d) Construct a 99% confidence interval.
AP Statistics
Confidence Interval w/ Means Quiz
Name: _________________________________
Assumptions: SRS stated, normal distribution stated, σ known (z distribution)
Calculations: (46.327, 48.473)
(47.327, 49.473)
Conclusion: We are 99% confident that the true mean galvanic skin response scores are between 46.327/47.327 and
48.473/49.473.