Download Answers to Confidence Interval

Answers to “Confidence Interval Learning Task” or as we see it “The Trouble With
Confidence Intervals:
1. Review
a) We can use Z-scores and the normal curve to approximate that, by typing
normalcdf((700-550)/50, 99999999) = .0013499672. And, although I would accept
.1350% on a test, I plan on storing that actual probability into my calculator as P for the
rest of this task.
b) This means 5 or 6 or 7 or 8 or…  C(10, 5)p5(1-p)5 + C(10, 6)p6(1-p)4 + C(10,
7)p7(1-p)3 + C(10, 8)p8(1-p)2 + C(10, 9)p9(1-p)1 + C(10, 10)p10(1-p)0 =
sum(seq((10nCrR)P^R*(1-P)^(10-R), R, 5, 10) = 1.125 or by using 1 – binomcdf(10, P, 4).
2. The Empirical Rule
a) The distribution of the sample mean scores is approximately normal.
b) The mean of the sample mean scores is 500 and the sample standard deviation is
50/10 = 5.
c) 68% of the sample means should fall between 545 and 555.
95% of the sample means should fall between 540 and 560.
99.7% of the sample means should fall between 535 and 565.
The mean score of the sample will fall within 2 standard deviations or 10 points from
the mean.
3. Confidence Intervals
a) The middle 95% of our data should lie within 1.96 standard deviations from the
mean.
b) The middle 90% of our data should lie within 1.645 standard deviations of the
mean.
c)
4.
Confidence Level
z*
90%
1.645
95%
1.96
99%
2.576
President’s Approval Ratings: Part 1
a) Between 43% and 49% of US citizens approved of Obama yesterday.
b) Margin of error = z*∙standard error, so for a 95% confidence interval and a
3% margin of error, .03 = SE ∙ 1.96  the Standard Error is .015306.
c) Using the same logic, .03 = SE ∙ 2.576  the Standard Error is .011646 for a
99% confidence level.
d) The standard deviation for sample proportions is
.
The standard error is an approximation for the SD and is equal to
e) Knowing from part b that the Standard Error for a 95% confidence interval is
.015306, and using the formula from part d, we get
.46(.54)
 .015306 . Solving
n
this equation, we get n = 1060. For a 99% confidence level, we get:
.46(.54)
 .011646 , and that means n = 1831. **Note: We got different answers in
n
class, and it’s because we used .0153 and .0116 for our standard errors. In
that case, we got 1062 or 1068 and 1846 or 1858 – depending on how we
rounded the standard error. It’s important to show your work, so I can tell
where your numbers are coming from!
In actuality 1067 people were polled, so the confidence level can be found using:
z*
.46(.54)
 .03 . This means z* = 1.9661999 and the confidence level can be found
1067
using normalcdf(-1.966199, 1.966199) = 95.07%, which still rounds to a 95%
confidence level.
f) If we maximize
by looking at the graph of y = x(1-x), we see that the
expression has a maximum value when p = .5, so p* = 0.5 .
If
, then n ≥ 1067, so more than 1067 people should have been
surveyed. ***Note that
for the margin of error last year! Solving
which is what you might have used
would give us n ≥ 1111.
g) Adding 500 people increases the sample size needed for a 95% confidence
interval to 1560, 1562 or 1586 (depending on the class). The new margin of error
is 1.96
.5(.5)
 .02481  2.48% - or 2.46%
1560
Adding 500 people increases the sample size needed for a 99% confidence interval
to 2331, 2346 or 2358. The new margin of error is 2.576
.5(.5)
 .026678  2.67% 2331
or 2.66% or 2.65%. The new margins of error are smaller, by .52% or .33 %
respectively. (You can find these other differences.)
h) Subtracting 500 people from 1060 creates a new margin of error of:
1.96
2.576
.5(.5)
 .041413  4.14% (4.13%, 4.05%) for the 95% confidence interval or
560
.5(.5)
 .0353 3.53% (3.51%, 3.50%) for the 99% confidence interval. These
1331
intervals are 1.14% and .53% larger. (Do the math!)
i) Sample sizes of 200 yield margins of error of
and
for the 95%
for the 99% confidence intervals.
j) With p* and the confidence level remaining the same, as n increases, the margin
of error decreases, and vice versa. This is true because if we are only changing n,
is just a transformation of y =
.
k) As the confidence level increases, the margin of error increases, too. This
makes sense because as we get more confident, we must have a larger range to pull
our proportion from. (I’m 10% confident that between 70% to 71% of you play a
musical instrument, but I’m 90% confident that between 40% and 90% of you do.)
5. Presidential Approval
a) If the margin of error is to be  5%, then we need to solve 1.96
.5 2
 .05 for n.
n
Doing so tells us that we must choose 385 people or more to construct a 95%
confidence interval with no more than 5% margin of error.
b) Since 34/50 = 68% approve, we assume that our proportion is 68%. So a 90%
confidence interval can be found using: .068  1.645
.68(.32)
 (.5715, .7885),
50
and the margin of error is 10.85%. The 95% confidence level means
.068  1.96
.68(.32)
 (.5507, .8093) and that’s a margin of error of 12.93%.
50
The 99% confidence level means the margin of error is .068  2.576
.68(.32)
=
50
16.99% so the confidence interval is (.5101, .8499).
c) If we could take all the samples of size 50 in our population and create a
histogram of the sample proportions that we found, that distribution would have a
mean close to the actual proportion. Furthermore, 95% of these sample
proportions should fall between 1.96 standard deviations of that true proportion.
Since we have created an interval by going 1.96 standard errors of our sample
proportion, we believe that our interval - from 55% to 81% - will contain the true
proportion – unless our sample is one of the 5% that doesn’t contain the true
proportion to begin with.
d) “Between 55% and 81% of the student body approves of the job that President
Obama is doing based on recent survey of 50 randomly selected high school
students. We can state this with a 95% confidence level.”
1-PropZint probably means “1 proportion interval based on z scores”. x represents the
number of successes, n represents the number of data entries and C-level represents the
confidence level. This worked when x = 34, n = 50 and C-level was .95
e) Press STAT, toggle over to TESTS, then arrow down to A: 1-PropZInt. The
value of x must be 34, n = 50, and the C-Level = .95. We get (.5507, .8093), which
is exactly what we got earlier…
f) Press STAT, toggle over to TESTS, then arrow down to A: 1-PropZInt. The
value of x must be 58% of 350, so type in .58*350 and 203 will appear. Let n = 350
and change C-level to .90. Press Calculate and you should get (.53661, .62339).
(In order to find the margin of error, just subtract the lower bound from the
upper bound to get the width of the interval. Divide that number by 2. For the
.78257  .57743
 .10257 , so our margin of
interval given, (.57743, .78257), find
2
error is 10.23%).
g) The 88% confidence interval is (.57743, .78257). The margin of error is half of
the difference between these two numbers = .10257. This means another way to
state this confidence interval is to say it is “68% ± 10.257”
h) randBin would randomly return 0 or 1, but on average, a 1 would be returned
55% of the time. This best represents a random sample of our population. The sum
represents the total of students who approve out of a random sample of size 50.
i) My interval was (.38152, .65848), and, yes, 55% lies in this interval. Considering
my proportion was 26/50 = 55%, of course I expected it. But this should happen
95% of the time, no matter what.
j) (.34152, .61848), (.57298, .82702) (.48546, .75454) (.5287, .7913)
k) This must be done in class, but I suspect that about 95% of the intervals will
contain 55%. Make a listing of all your intervals on the chalkboard to see what
proportion really does.
l) We could reduce the margin of error by gathering more than 50 students’
opinions or else accepting a lower level of confidence.
5. Racing Hearts
a) The estimate of the population mean is the same as the mean of the samples.
And the standard deviation of the means is the sample standard deviation divided
by
n . i.e.
 x   and  x 

n
.
b) The confidence interval should be   z *

n
.
c) X is the estimate for . So the approximate confidence interval is x  z *
s
n
.
d) The T distributions approximate the normal distribution as n increases, but at
small values of n, the T distribution is wider, meaning that a 95% confidence
interval would be wider, too.
e) We will find the resting heart rate of a typical high school student, so your
data ready to share in class.
f) Try this using 78, 72, 66, 70, and 62. You should get (62.068, 77.132), with a
mean of 69.6 and a standard deviation of 6.066. (Compare this to a Zinterval.)
g) Use {72, 56, 60, 64, 68} as your other 5 pieces of data.
h) Use {72, 68, 68, 58, 60, 60, 64, 72, 68, 76, 56, 72, 56, 60, 64, 68, 64, 66, 70,
54} as your other 20.
i)
We should use a Z* score instead of a t* score, because t* scores change
depending on the sample size. Z* scores don’t depend on a sample size.
4.9
Solve 2.576 
 3 and get n  18 students.
n
j) The interval is (57.717, 59.483). I am 99% confident that the average resting
heart rate of a high school student is between 57.7 and 59.5.
6. Synthesis of Confidence Intervals