Answers to “Confidence Interval Learning Task” or as we see it “The Trouble With Confidence Intervals:
1. 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.
2. 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)
Confidence Level / z*90% / 1.645
95% / 1.96
99% / 2.576
3. Health Care Concern:
a) Between 38% and 44% of US citizens are extremely concerned about health care.
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 . Solving
this equation, we get n = 1033. For a 99% confidence level, we get:
, and that means n = 1784.
In actuality 1050 people were polled, so the confidence level can be found using:
. This means z* = 1.9765 and the confidence level can be found
using normalcdf(-1.9765, 1.9765) = 95.19% .
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, son more than 1067 people should have been
surveyed. ***Note that which is what we used for the margin
of error last year! Solving would give us n ≥ 1111.
(I don’t know why they started over with a again, but here goes…
a) Adding 500 people increases the sample size needed for a 95% confidence
interval to 1533. The new margin of error is .
Adding 500 people increases the sample size needed for a 99% confidence interval
to 2284. The new margin of error is .
The new margins of error are smaller, by .5% or .3 % respectively.
b) Subtracting 500 people from 1033 creates a new margin of error of:
for the 95% confidence interval or
for the 99% confidence interval. These intervals are 1.24% and .59% larger.
c) Sample sizes of 200 yield margins of error of for the 95%
and for the 99% confidence intervals.
d) 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 = .
e) 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.) 4. Presidential Approval
a) If the margin of error is to be 5%, then we need to solve for 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 a90%
confidence interval can be found using: [.5715, .7885],
and the margin of error is 10.85%.The 95% confidence level means
[.5507, .8093] and that’s a margin of error of 12.93%.
The 99% confidence level means the margin of error is = 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 of the actual proportion. Furthermore, 95% of these sample proportions
would 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 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 S , toggle over toTTESTS, then arrow down to A: 1-PropZInt. The
value of n 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).
f) 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
interval given, (.57743, .78257), find , so our margin of
error is 10.23%.
g) 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.
h) 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.
i) (.34152, .61848), (.57298, .82702) (.48546, .75454) (.5287, .7913)
j) This must be done in class, but I suspect that about 95% of the intervals will
contain 55%.
k) We could reduce the confidence level 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 .
b) The confidence interval should be .
c) is the estimate for . So the approximate confidence interval is .
d) The T distribution approximates 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 must do this in class as well, but have yours ready to share when we get to
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) In class.
h) In class.
i) We must 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.
Solve and get n 18 students.
j) The interval is(57.717, 59.483). I am 99% confident that the resting heart rate
of a high school student is between 57.7 and 59.5.
6. Synthesis of Confidence Intervals - Be ready to share these, too!