CI for a Proportion

7/23/2019 CI for a Proportion

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Confidence Interval for a Proportion 1

Confidence Interval for a Proportion

Example

74% of a company’s customers would like to see new product packaging. A random sample of

50 customers is taken. X is the number of customers in the sample who would like to see the new

packaging; then the sample proportion is n X p ˆ . The mean and standard deviation for p̂ are

Mean of p̂ : p p ˆ = 0.74

Standard deviation of p̂ :

0620350

26.074.01ˆ

n

p p p

Since np = 50(0.74) = 37 and n(1 – p) = 50(0.26) = 13 are both at least 10, the distribution of p̂

is approximately Normal. Using the Normal approximation we know that the probability is

approximately 95% that p̂ falls within 1.96 standard deviations of the mean.

95% of all samples have a sample proportion p̂ between 0.74 – 1.96(0.06203) = 0.618

and 0.74 + 1.96(0.06203) = 0.862.

As 1.96(0.06203) = 0.1216 we could equivalently state “ p̂ is within 0.122 of 0.74.”

For this situation approximately 95% of all possible samples yield a proportion p̂ within

0.122 of 0.740 (within 12.2% of 74.0%).

In general, provided a Normal approximation can be used, 95% of all possible samples yield a

proportion p̂ within n p p 196.1 of p.

An approximate 95% Confidence Interval

The section begins with a description of steps that lead to a usable result. (A rigorous treatment

of the issue requires a good deal of mathematical statistics.) The explanations provided below are

simplified.

We’ve observed the following:

For approximately 95% of all samples p̂ is within n p p 196.1 of p.

Flip-flopping p and p̂ yields the following statement (which is true):

For approximately 95% of all samples p is within n p p ˆ1ˆ96.1 of p̂ .

In this second statement the interval is random. Different samples yield different values for p̂ ,

which result in different intervals.




The Result – a 95% Confidence Interval for p

For one sample yielding a result p̂ , the interval

E p ˆ where n p p E ˆ1ˆ96.1

forms an approximate 95% confidence interval for p. p̂ is the point estimate of p; E is the error

margin associated with the estimate.

That is: (approximately) 95% of all random samples (of size n) produce an interval including p

within the bounds. When we obtain the random sample and compute the interval from it, we no

longer have anything random. At this point we state that we are (approximately) 95% confident

that p is within the interval bounds.

A couple restrictions are on this formula.

1. It should be applied to situations where units are randomly selected.

2.

If sampling is without replacement, check the 20 Times Rule – the population must be at

least 20 times the sample size to use this result. (If not, and you know the population size,

you can use the adjustment described a bit later in this section. However: small

populations are uncommon, and the adjustment is rarely needed. You should recognize

that when the population is not at least 20 times the sample size, then our recipe above for

error margin does not work.1)

3. The actual counts of Successes X and Failures (n – X ) must both be at least 10.2 (If not,

either you have too small a sample, or the Success probability is likely too close to 0 or 1,

for the Normal to be a decent approximation.) If this is not the case, you must seek

alternative strategies. (Minitab, and most other statistical software, can obtain theconfidence interval by an “exact” method that doesn’t require the Normal distribution.

When the counts X and (n – X ) both at least 10, the exact method and the approximate

interval given here will be quite similar.)

We state such an interval in one of four equivalent styles:

E p p E p ˆˆ E p E p ˆ,ˆ E p E p ˆtoˆ E p ˆ

The first is preferred, as it indicates what is being estimated: p. In the first through third versions,

the lower value is always stated first. Every confidence interval should be accompanied by an

interpretation that states the confidence level (here 95%).

1 In fact our formula gives too large an error margin. So in essence you can be more than 95% confident in our result

when its misapplied to situations where the population is small. Probably not the worse error in the world.2 Some sources use 5 in place of 10. 10 is better. 5 is somewhat OK, but if either of these values is between 5 and 9,

you’d be better off getting some help from a statistician, rather than using our methods.




Example

A simple random sample of 1000 adults finds that 343 approve of the President. Obtain a 95%

confidence interval for the proportion of all adults who approve of the President.

This is a random sample where p = the proportion of all adults who approve of the President ( p

is an unknown, but fixed and unvarying, quantity). The population size is huge – much larger

than 20(1,000) = 20,000. The number of Successes and Failures are 343 and 657 respectively –

both are well above 10). We can obtain and then interpret a confidence interval with the

presented method.

x = 343 out of n = 1000 trials. So p̂ = 0.3430. Then

0294.001501.096.11000657.0343.096.1ˆ1ˆ96.1 n p p E .

This 95% confidence interval for p (unknown) can be written any of four ways:

0.3136 < p < 0.3724 (0.3136, 0.3724) 0.3136 to 0.3724 0.3430 ± 0.0294The interval is properly interpreted as follows.

We are (approximately) 95% confident that the proportion of all adults who approve of

the President is…

…within 0.0294 of 0.3430 (2.94% of 34.30%).

…between 0.3136 and 0.3724 (31.36% and 37.24%).

Approximately 95% of all possible samples yield a proportion p̂ such that the confidence

interval includes p. We have one such sample, chosen at random. We are (approximately) 95%

confident that the interval obtained from our sample, 0.3136 to 0.3724, includes p.

Notice! 1000657.0343.096.1 is computed. This involves both the prevalence,3 in the sample,

of Successes/approvers = 0.343 and the prevalence of Failures/disapprovers = 0.657.

3 Define “prevalence” as pretty much the same thing as “proportion.” In disease and addiction, the word prevalence

is often used: “The proportion of individuals that has some disease…” For instance: “The prevalence of smoking in

US adults is approximately 25%.”

These two are equivalent.




Confidence interval for a (population) proportion p

This section summarizes our results. We extend the treatment to include a confidence interval for

a population proportion p with any confidence level.

The typical textbook treatment of the issue takes an overblown approach to notation. In addition

to the confidence C % is the “error rate” (lack of confidence?) for the procedure, which is

generally denoted with , where almost always is rather small. C % = 1 - .

Then take z /2 to be the Z score with /2 area in the right tail of the Normal distribution; its

opposite – z /2 is the Z score with /2 area in the left tail. Between – z /2 and z /2 is area

(1 – ) = C % – the confidence.

The 1 Sample Z Conf idence I nterval for a Proporti on

An approximate C % confidence interval for p is

E p ˆ

where E is the error margin

n

p p z E

ˆ1ˆ2

,

with z /2 from the Standard Normal distribution, taking into consideration the confidence C %.

When should you use this formula for a confidence interval?

You have a random sample drawn from a population with unknown population

proportion p.

If sampling is without replacement, the population must be at least 20 times the size of

the sample (in most practical cases this is almost trivially true).

The sample result has at least 10 Successes and 10 Failures.

What if the population is too small or the sample is too large?

The “at least 20 times rule” is violated. There is a relatively simple fix, but to employ it

you must know – at least reasonably accurately – the size N of the population. The error

margin becomes

1

ˆ1

ˆ

2

N n N

n p p z E

If you examine the adjustment (a multiplication) you can see the logic in the 20 times

rule: When N is more than 20 times n this multiplier will be quite close to, and just

below, 1. In ignoring the adjustment, we end up with a slightly bigger E than is required –

so if anything we will be understating the confidence.




What if the “both at least 10” rule isn’t met?

There is a method that works whether or not the “both at least 5” rule is met. We won’t

cover it here. (If you like, learn the “Wald Interval” descr ibed in textbooks. This is also

an approximate method – but it pretty well for counts below 10.)

What if the sampling isn’t random?

No statistical method can guarantee results with any particular reliability (confidence)

when sampling is not random. The situation may well be hopeless.

Example

A company’s human resources department investigates the application materials submitted by 84

applicants for an entry level position over a six month period. One finding is that 15 of the

applicants falsified information in the application materials.

Assume that the 84 applicants are a random sample from a larger pool of similar applicants. Give

a 99% confidence interval for the proportion of all applicants who falsify information inapplication materials.

Solution

First check conditions. We take it for granted that the complete pool of applications is at least 20

times larger than the 84 in the sample; we are told to assume the sample is random. (In reality,

random sampling in such a circumstance might be hard to accomplish. Still, we might reasonably

assume that the sample applicants are representative of the population of all applicants.) Both the

number of falsifiers (15) and nonfalsifiers (69) exceed 10. We may use the 1 Sample Z

Confidence Interval.

For 99% confidence, 005.02 z z = 2.576. The estimated/observed proportion of falsifiers is p̂

= 15/84 = 0.1786. Then the error margin is

1076.00418.0576.284

8214.01786.0576.2

84

1786.011786.0576.2

ˆ1ˆ

2

n

p p z E

Then 0.179 0.108 is the 99% confidence interval: 0.071 < p < 0.287. We are 99% confident

that between 7.1% and 28.7% of all applicants falsify information.

Why should you be able use this formula for a confidence interval?

Of course a computer can do this, so why should you know how to it? Two reasons: 1)

The formulas are relatively straightforward4, and requires mostly that you be able to

identify and understand the relevant quantities. These same quantities are the inputs into

4 Granted: The formula for error margin is not trivial. If you have trouble getting the proper error margin, you need

to practice using the formula until doing so correctly becomes second nature.




statistical software. 2) The formula – along with trial and error and some examples – can

assist with your understanding of properties of confidence intervals.

Fact worth knowing

95% is the standard confidence level for scientific polls published in the media and online. If a

poll does not publish an error margin, you may assume that sampling is not random – the poll isnot a scientific one. Keep in mind also that many polls with stated error margins are not done

properly. You should have less than 95% confidence in results from such polls.

Nomenclature

There’s some terminology that goes with each of these quantities. (The terminology is useful

because it is generalized to other situations.)

The confidence level (or just confidence) is C . Usually C = 0.90, 0.95 or 0.99 – generally

we prefer to have a high amount of confidence in our statements. However: There is

nothing illegal or necessarily wrong about a 50% CI. (It’s just that 50% CIs miss the

target quantity half of the time.)

Don’t confuse the confidence level (or just plain confidence) with the confidence

interval. The confidence interval is the interval of values you obtain.

p̂ is the (point) estimate of p. It’s a single “good” estimate for p from the sample of data.

It is the prevalance of Successes in the sample.

2 z is the critical value (from the Standard Normal) that goes with C % confidence.

(Some reference materials use simpler notation like z * - and leave it to common sense

that this z is the one that goes with the confidence C .)

The two endpoints of the interval are the bounds: lower bound and upper bound. The

width W of a confidence interval is the distance from the lower to the upper bound. The

part in total is referred to as the “error margin” E . 2 E = W .

n

p p ˆ1ˆ is often called the “(estimated) standard error of p̂ .” A standard error is

essentially a standard deviation.5 Recall: Standard deviation measures “typical deviation

from mean.” The deviation of p̂ from its mean p is a (sampling) error. It’s common to see

the abbreviation SE for standard error. In this case:

n

p p

pSE

ˆ1ˆˆ

.

The error margin E for this interval can be expressed pSE z E ˆ2

.

5 This quantity really is an estimated standard deviation, as the standard deviation of p̂ is n p p 1 .




Mind your p 's and q’ s

Many textbooks make the formulas look shorter by using a second letter q to stand for Failure

rate. So q = (1 – p) and n xn pq ˆ1ˆ . When this is done

n

q p

n

p p ˆˆˆ1ˆ

.

More on interpreting the interval

If you followed the development above, you can deduce the proper interpretation of a confidence

interval. It's also possible to take the justification for granted, and come to an interpretive

understanding.

Different samples give different results. Consider all possible samples. Obtain, for each sample,

a 95% confidence interval. Some of these intervals include p, some do not. Most do. In fact: 95%

of all of them do.

In a statistical study, a single sample is drawn randomly. The data are collected and summarized,

and a 95% confidence interval is computed. We have one sample - selected at random from the

collection of all samples. Because 95% of all samples lead to an interval that covers p, we are

95% confident that the particular interval we have covers p.

We use the word confidence, rather than probability. In statistical applications where parameters

are estimated, those parameters are thought of as fixed values describing populations. They do

not vary. The parameter p is either in the interval or not. There is no probability involved.

Where did the probability go?

There was probability - before the sample was selected. This is similar to tossing a coin. Beforeit's tossed the probability of a Head is 1/2. But once the toss is completed, the probability - for

that toss - is either 0 or 1, depending on the outcome. In this application, the probability is either

0 (the interval covers p) or 1 (the interval doesn't), depending on whether in fact the interval does

or does not cover p. Not knowing p, we cannot tell. All we know is that 95% of all samples yield

an interval covering p. So we are 95% confident that ours does. (Similarly, after the coin is

tossed, if you're unable to see the result, you can be 50% confident it's a Head. The word

probability doesn't apply here.)

In short: Use the word "probability" for random things that haven't yet taken place. Once they've

taken place, even if there are unknowns, use the word confidence. The unknowns merely reflect

human ignorance about events.




Properties of Confidence Intervals

Three values impact the error margin of a confidence interval.

1. the prevalence of Success ( p̂ )

2.

the sample size (n)

3. the confidence level (C )

Undertake an investigation: How do changes in each of these impact the error margin? These

issues are addressed through the exercises.

In some respects the properties you discover will convince you that statistics makes sense: The

numbers work out in ways that common sense would anticipate in advance (Common sense

would never anticipate the precise results.6 But certain procedural properties do make sense.

That's what you want to discover.)

What a Confidence Interval Cannot Do

Notice that the 95% in a 95% confidence interval refers to the percent of all samples that yield an

interval that covers p. If we choose one such sample at random, we’re 95% confident in that

result. The error margin in a confidence interval addresses errors due to random sampling.

The error margin in a confidence interval does not include the effects of other errors. Poorly

recorded data is one source of error. Or perhaps the study didn’t really sample randomly. In these

cases, quantifying sampling error is not enough.

All these other factors will lead to additional estimation error – error that is not captured by our

formula. So while you can still use the formula when other types of error are present, it doesn’t

give a 95% confidence interval. The actual confidence is unknown. For analyses involving

nonrandom data, the actual confidence will be considerably lower than 95%. That’s a real issue

in many studies.

Polling Refusals

Suppose (to oversimplify) that 88 million people approve of the President and 72 million

disapprove. So the President’s approval rating is p = 88/160 = 0.55.

A telephone poll is taken. But: The people that approve of the President are crankier than those

that do not. They are less likely to put up with an intruding phone call. In fact, 40% of the

approvers will not respond (that’s 35.2 million people). The disapprovers are more willing totake the call: only 10% of them will refuse (that’s 7.2 million people). Here’s the breakdown

6 In fact, given the randomness involved in selecting the samples, and the various other attributes that change from

problem to problem (n, p, x, as well as the size of the population), it is remarkable that the formula we have is so

simple.




Approve Disapprove Total

Respond 52.8 64.8 117.6

Refuse 35.2 7.2 42.4

Total 88.0 72.0 160.0The problem here is that our sample is going to reflect the views of only the responders. Of the

117.6 million responders, 52.8 million approve, for a rating of 52.8/117.6 = 0.45.

While people will be randomly called, those who refuse to respond will not be included in the

results. So: Our poll will be estimating 0.45. With a sample size of 1000, the error margin will be

around 0.03. While some samples will give results higher than 0.45, it is highly unlikely that

we’ll get a sample that produces a confidence interval including 0.55. After all: The interval is

designed to include 0.45.

This is an example of a biased estimation. A result is biased if it systematically [on average]

produces the wrong result. Yes: We could get a random sample that has unusually high amountsof Approvers, and luckily gives an interval including 0.55. But we are unlikely to do so, because

on average our estimate is 0.45 not 0.55. That’s what bias is: The average result from the

sampling procedure is not equal to the intended result.

If we know that nonresponse occurred with 40% probability among Approvers, and 10% among

Disapprovers, we could adjust the survey results accordingly, and produce an unbiased estimate.

But generally nonresponse rates are unknown, and the rate changes from survey to survey,

depending on what the issue is. It is difficult to adjust results to compensate for the nonresponse

issue.

Poll results are even harder to interpret when sampling is not done randomly. Internet pollschoose subjects by convenience and interest. Only people who care enough to vote will vote.

These people may be significantly different in their views than the population of interest. No

error margin can fix up such polls. (Hopefully results are stated without an error margin.)

Statistical Software

Statistical software will compute the confidence interval for p. All you need to do is input three

values: n, x, and the confidence C , along with specifying the method the software should use.

The interval you have learned is the (approximate) 1 sample Z interval for a proportion.

Good software has other choices – they use a different "formula" than that above. The formula

you have is approximate, and requires at least ten Successes and Failures to allow using the

Normal. For cases where this condition is not met, you may have statistical software compute the

interval using a different method/formula.7 In fact, where there are at least ten Successes and

7 Usually called the “Exact Binomial” interval or method.




Failures, you may use the alternative method in place of the 1 sample Z interval; you’ll get

slightly different results.8 For really large samples, these differences will be quite small.

One quirk about one method software may use: The intervals may not balanced: The estimate p̂

is not exactly in the middle of the interval. This is particularly noticeable for results with small

precents of either Successes or Failures. (If you have, say, 2 Successes in only 10 – 20 trials, thenultimately the value of p is quite small. So the distribution is clamped against the left edge of the

range of values, and has right skew. Skewness is an expression of imbance; so it’s no surprise

that the interval is not balanced. This is a case where you could not use the formula stated above

– 2 is too few Successes. The “both at least 10” restriction prevents use of a Normal

approximation when things aren’t at all close to Normal.) If both x and in n – x are large, the

interval – not matter which method is used – is nearly balanced about p̂ , and in fact, the exact

interval and the interval from your formula will give very similar results.

Sample Size Determination

The error margin for our confidence interval is n

p p z pSE z E

ˆ1ˆˆ

22

. This is

equivalent to p p E

z n ˆ1ˆ

2

2

. Suppose, prior to the study, we desire an error margin of E . If

we can produce a reasonable educated guess for the prevalence of Successes ( p̂ ), then an

appropriate minimum sample size for the study is p p E

z n ˆ1ˆ

2

2

.

Example 1Suppose you want to estimate the proportion of students at a large university who are

nearsighted. The prevalence for the general population is around 0.45. Use this as a guess to

determine how many students would need to be included in a random sample if you wanted the

error margin for a 95% confidence interval to be less than or equal to 2%.

Recall that the error margin quantifies the maximum reasonable difference between the observed

value p̂ and the population value p.

8 For really large samples, these differences will be quite small. One quirk about the Exact Binomial method: The

intervals may not balanced: The estimate is generally not exactly in the middle of the interval. This is particularly

noticeable for results with small prevalence of either Successes or Failures. (Any p other than 0.5 implies some

asymmetry, and these intervals reflect this.) If both x and n – x are large, the Exact Binomial interval is nearly

symmetric about the sample propotion, and in fact, the exact interval and the interval from your formula will give

very similar results.




Solution: The desired error margin is E = 0.02. Our guess is p̂ = 0.45. The required sample size

is 99.237655.045.002.0

96.1 2

n . Of course we cannot sample 0.99 of a student, so we move

up to 2377.

The actual study was run and it turned out that 951 of 2377 randomly sampled students were

nearsighted. This yields a 95% confidence interval of 0.4001 0.0197.

Remark 1

If our guess is closer to 0.5 than prevalence p̂ observed in the data, then the actual error margin

will be greater than desired. If the guess is further from 0.5, then the actual error margin will be

less than the desired E d . (If the two are equal then someone is a very lucky guesser.)

Example 2

In a study of a new drug, the researchers assume that the cure rate for the drug is the same, 0.60,

as for the established drug. What sample size is required to obtain a 99% confidence interval

with error margin no greater than 0.05?

Solution: The desired error margin is E = 0.05. Our guess is 0.60. The required sample size is

03.63740.060.005.0

576.2 2

n . Of course one cannot sample 0.03 of a patient. To ensure a

large enough sample size, round up to 638.

When the data are collected, it’s found that the drug is much more effective than the established

drug: 573 of the 638 patients (that’s 89.8%) are cured. The 99% confidence interval is

0.898 0.021.

Notice that the error margin is far below 0.05. But this came at a cost. If they had known that the

prevalence would be around 0.90, the researchers could have used 0.90 for a guess, and

determined a sample size of 239 (because 89.2381.09.005.0

576.2 2

n ). Notice that

239

1.09.0576.2

= 0.0500. They ended up sampling 399 more patients than necessary. This cost

them time and money – if they’d had any clue the cure rate would be far higher, they could have

taken advantage of it.

Remark 2

If the actual error margin is less than the desired value, then, while the error margin is bettered,

the expense of conducting the study was larger than necessary. A smaller sample size would

have sufficed to obtain the desired error margin.

There is no way to choose exactly the right sample size. In some cases, we may have no idea

what the prevalence is in advance. If no guess is possible – we are completely in the dark as to




30

40

50

60

70

80

90

100

0.02 0.03 0.04 0.05 0.06

proportion p

s a m p l e s i z e n

the prevalence – we can use 0.5. This guarantees that actual error margin to be no larger than

what is desired. On the other hand, it also pretty much guarantees that we will take a larger

sample than is necessary (only if the prevalence turns out to be 50% will the sample be “just

large enough”).

Remark 3Whenever the range of plausible guesses includes 0.5, use 0.5 as the guess. This rule works

when one has no idea what the prevalence is: the range of plausible guesses is from 0 to 1,

which certainly includes 0.5.

Most two-candidate political races are reasonably close. Pollsters9 generally use 0.50 to

determine the sample size. Using a guess of 0.50 tends not to lead to dramatic oversampling

unless the result falls below 1/3 or above 2/3.

Example 3

Production line defects occur infrequently at an industrial plant. In the past the rate has generally been between 2% and 6% (this value would change over time as the production line, and the

employees working on it, change). What sample size is required to estimate the current rate at

90% confidence with error margin no larger than 4%?

If we assume 2%, then the required size

is 33; if 6% is assumed, the required size

is 96. Here’s a plot of the relationship.

(The relationship is not exactly linear.

However: Linear interpolation would

work well here. In general, as long as the

proportion is confined to a small range

of values to one side of 0.50,

interpolation does work fine.)

You can see that 6% requires the largest n (it’s closest to 0.5). To cover all historical

possibilities, use n = 96. If the rate is actually less than 0.06, you will have oversampled. What if

you sample less than 96? Perhaps a good idea, but if the rate is near 0.06 you won’t get the

desired error margin. And, of course, if production falls seriously out of control, you might see a

result much higher than 6% – leading to an error margin considerably larger than 0.04.

Remark 4

A good idea is to produce a range of plausible guesses, and find the sample size for a number of

values within that range. Graph this relationship. If the final decision isn’t yours, you can place

your graph in front of the decision maker.

9 People or organizations who are paid to conduct polls.




0

200

400

600

800

1000

0 0.2 0.4 0.6 0.8 1

proportion p

s a m p l e s i z

e n

This point is illustrated by Example 3. The decision maker on sample size needs to see that

graph.

At right is a plot of the required sample

size for 95% confidence intervals having

error margin 3%. The curve has the sameshape for other confidence levels and

error margins. You can see that a

prevalence of 0.50 requires the largest

sample size.

Appendix

A general format for confi dence in tervals

The confidence interval we just studied is

pˆ

z /2 SE ( pˆ

)

Where

z /2 is the critical value, found from the Normal, taking into consideration the desired

level of confidence, and

the standard error of the estimate is SE ( p̂ ) =

n

p p ˆ1ˆ , yielding

the error margin

n

p p z pSE z E

ˆ1ˆ*ˆ*

.

More generally, provided the conditions are right, a confidence interval is determined with

Estimate Error margin

Where

Error margin = critical value SE (estimate)

This formula is broadly applicable to all sorts of data analyses. In almost all circumstances

SE ( Estimate) has the square root of the sample size(s) in the denominator.




Exercises

1. There are 8640 students enrolled at SUNY Oswego this semester; 5146 live more than 50

miles from campus. A professor (unaware of these figures) samples 92 students and finds

that 60 of them live more than 50 miles from campus.

a)

Identify the following: i) The population proportion p; ii) The sample count X ; iii) The

sample proportion p̂ .

b) Which of p and p̂ is a parameter? Which is a statistic?

A student (also unaware of the whole-campus figures) is about to randomly select 142

students to estimate the proportion who live more than 50 miles away.

c) For the student: What are the mean and standard deviation for p̂ ? Interpret this mean.

(Be sure to include the phrase “all possible samples” in your statement.)

2.

The saturation rate for a particular kind of marketing via a newspaper ad is 15%. That is:15% of all newspaper buyers will read the ad. For a new ad, marketers randomly sample 30

buyers and determines that 2 have read the ad.

a) Identify values for p, X , and p̂ .

b) Which of p and p̂ is a parameter? Which is a statistic?

For the following exercises, when you interpret results, use the word “all” or “population.”

3. A random sample of 212 adoptive parents finds that 85 of them stated “No Preference” for

their child’s gender. Use this sample data to construct a 95% confidence interval estimate for

the proportion of adoptive parents who state “No Preference.” Explicitly identify thefollowing:

a) The point estimate.

b) The critical value ( Z /2).

c) The error margin.

d) Write the interval bounds in this format: p̂ E .

e) Express the interval in this format: ________ < ________ < ________ .

f)

What confidence do you have in this result?

g) Explain what p represents in this situation. Is its value known?

h) p̂ : Parameter or Statistic? p: Parameter or Statistic?

i) Interpret your interval in words. We are 95% confident that…




4. The Genetics and IVF Institute conducted a clinical trial of the XSORT method designed to

increase the probability of conceiving a girl. 325 babies were born to parents using XSORT,

and 295 of them were girls. Use this data to construct a 99% confidence interval for the

proportion of girls born to parents using XSORT. Interpret your result.

5.

Do individuals have the ability to temporarily postpone death to survive a major holiday?(The hypothesis would be that these holidays are family affairs that give a dying person

incentive to live a bit longer.) In one study, 12000 deaths, over the period from one week

before to one week after Thanksgiving, were examined. Of these, 6062 occurred in the week

before Thanksgiving. Give a 95% confidence interval for the proportion of deaths in this two

week period that occur in the earlier week. Interpret your result. Does your data conclusively

support the “postpone death” theory? (Hint: Check where 0.5 lands relative to your interval.)

6. Complete the small table indicating which critical value from the Standard Normal table goes

with the given levels of confidence.

C 50% 75% 90% 95% 98% 99% 99.9% 99.99%

Z

/2 1.645 1.960 3.891

7. Over a period of 11 years in Hidalgo County, Texas, 870 people were selected for grand jury

duty, and 39% of them were Mexican-American. Notice that you are told the value of p̂ -

you don’t have to compute it: p̂ = 0.39. From this you can deduce that the number X of

Mexican-Americans in the sample. Since 0.39(870) = 339.3, the number must be 339 (it

can’t be 339.3 – you can’t select 3-10ths of a Mexican-American). The given value is

rounded for convenience: 339/870 = 0.3899 to four significant digits, and 0.390 to three,

which is sufficient for computing purposes.)

a)

Assume these data represent a random sample of jury-duty-eligible county citizens.

Obtain a 99% confidence for the percent of all county citizens that are Mexican-

American. Interpret your result.

b) It was determined that 79.1% of all county citizens were Mexican-American. What does

your confidence interval suggest about selection for jury duty?

8. Perform an investigation of the relationship between confidence and error margin. Here’s

how.

a) Take exercise 7, where n = 870 and p̂ = 0.390. You’ve already obtained a 99%

confidence interval: 0.390 0.043. The error margin is 0.043. Now obtain a 95%confidence interval; determine the error margin.

b) Compute intervals for each of the confidence levels specified in the table. Fill in the table

below with the error margin for the various levels of confidence.




C 50% 75% 90% 95% 99% 99.9%

E 0.032 0.043

c) Write a sentence describing how the error margin changes as the confidence is increased

(decreased).

9. A recent survey of 4276 randomly selected households showed that 94.0% of them had

telephones.

a) How many of the 4276 households have telephones? Answer with a whole number.

b) What is the value of p̂ to the nearest 0.0001?

c) Using these results, construct a 99% confidence interval for the proportion of households

with telephones. Interpret your result. What is the error margin for this interval?

d) Give a 99% confidence interval for the proportion of households without telephones.

How does the error margin compare to that in part c?10. Gregor Mendel was responsible for famous genetics experiments with peas. In one

experiment he crossed lines of peas, and the results included 428 green peas and 152 yellow

peas.

a) Find a 95% confidence interval for the proportion of all peas that are green. Interpret your

result.

b) Mendel’s theory of genetic propagation of inherited traits predicted that 75% of all peas

would be green. Is the theory refuted by his data?

11. Perform an investigation of the relationship between sample size and error margin. Here’s

how. Take exercise 7 where p̂ = 428/580 = 0.7393. You’ve already obtained a 95%

confidence interval: 0.7393 0.0358 (keep figures to the nearest 0.0001 for this). The error

margin is 0.0358.

a) Suppose hypothetically the study had investigated four times fewer peas, but the percent

that are green is the same: 107 green and 38 yellow. Determine a 95% confidence interval

for this outcome. Place the error margin in the table below. The sample size is 4 times

smaller: How many times larger is this error margin?

b) Suppose hypothetically the study had investigated twenty-five times more peas than in

the actual study, with 10700 green and 3800 yellow. Determine a 95% confidence

interval for this outcome. Place the error margin in the table below. The sample size is 25

times larger: How many times smaller is this error margin?

n 145 580 14500

E 0.0358




c) Write a sentence describing how the error margin changes when the sample size is k

times larger. Check the solution to make sure you have the right result in mind as you go

forward.

12. Jack conducts a student opinion poll and gets an error margin of 10% for his result. He is

not happy. He wants

3%. How must he adjust his sample size? (Assume the confidence andsample proportion remain the same.)

13. A poll of 4000 people gives an error margin of 0.01. What would the error margin be for a

similar poll of 800 people? (Assume the confidence and sample proportion remain the same.)

Summary

Here is the formula for the error margin: n

p p z E

ˆ1ˆ2

. At this point you ought to be able

to look at the formula and deduce that10

:

The error margin increases when the confidence is increased. (This happens through the

value z /2.)

The error margin decreases when the sample size is increased. (The sample size appears in

the denominator of the formula.) In particular, the relationship is that increasing (decreasing)

the sample size by a factor of k decreases (increases) the error margin by a factor of k .

(That’s because the sample size appears in the square root.)

The error margin does not depend on what is called a Success and what is called a Failure.

(Exercise 9 parts c and d explicitly address this.)

14. Perform an investigation of the relationship between the sample prevalence of Success and

error margin. Here’s an exercise that will help you do this.

What is the relation between the socio-economic status of parents and college graduation of

their children? Different groups are sampled in order to make comparisons. For each socio-

economic status, n = 400 children are sampled and tracked through adulthood. The number of

the 400 who graduate from college is recorded.

a) Obtain a 95% CI for each socio-economic status. Determine the error margin for each

interval. Place your results in the table below.

b) How do error margins compare for the cases p̂ = 0.10 and p̂ = 0.90? How about p̂ =

0.20 and p̂ = 0.80? p̂ = 0.30 and p̂ = 0.70? Why does this make sense?

10 Assuming that all other factors stay the same.




c) Write a single sentence describing the relationship between the proportion p̂ and the

error margin of the confidence interval.

Parents’ Status # of grads p ˆ 95% CI Error Margin

Welfare 40

Poor 80

Low Income 120

Middle Income 200

High Income 280

Wealthy 320

Super rich 360

15.

Go back to problem 10. The 95% confidence interval is 0.3575 < p < 0.4223. The pointestimate for the proportion of green peas is p̂ = 0.7379 with error margin is 0.0358. State the

95% confidence interval for the proportion of peas that are yellow. You should be able to do

so using the results shown here and addition and subtraction.

16. A poll reveals that candidate D has 44.2% of sampled voters leaning towards D (error margin

3.5%). Remember: All media polls are done at 95% confidence unless stated otherwise.

a) Interpret this result.

b) Suppose there is only one other candidate, R. Give the 95% confidence interval for

candidate R.

c) Suppose there are instead three candidates, R, D and U. Is it possible to give an estimate

and error margin for candidate R?

d) In a two candidate race, is it possible that D is actually ahead? (Hint: Suppose U doesn’t

stand for a candidate, but indicates “Undecided.”)

17. A recent newspaper opinion poll found that 81% of Americans are in favor of a military

drawdown in Iraq (error margin 4%). Interpret this statement. Include a confidence level.

Summary

In #12 above you confirmed that (assuming the sample size and confidence stay the same) the

error margin is largest when p̂ = 0.5 and gets smaller as the proportion falls away from 0.5. This

makes intuitive sense: When the prevalence is near 50% there is more uncertainty; when the

prevalence is near 0 (or 1) there is more certainty.




18. At SUNY Oswego n = 125 students are randomly selected; 100 of them are opposed to a

proposal that calls for the college to jam cell phone signals in classrooms. (This would

prevent texting in class.) You can confirm that a 90% confidence interval is

0.741 < p < 0.859.

a)

Give the values of the point estimate and error margin for the interval. b) This survey was also conducted at Penn State University. The results of the survey were

exactly the same: 100 of 125 sampled students opposed the proposal. What is the 90%

confidence interval for Penn State?

c) It should be noted that Penn State has about 5 times more students than does SUNY

Oswego. What impact does the population size have on the error margin for a confidence

interval? Consult the formula: Where does the population size play in to matters?

19. Jake, John and Jaspar are conducting a study: What proportion of SUNY Oswego students

stay in Oswego over the Halloween weekend.

a)

Describe in words what p represents. Is p a parameter or a statistic? Would the value of p

be easy to obtain?

b) They sample students randomly: 56 of 80 sampled students stay in Oswego. This gives

p̂ = 0.70. Is this value a statistic or a parameter?

c) Jake determines correctly that the error margin for a 95% confidence interval is 0.100.

The confidence interval is 0.600 < p < 0.800. John says “I think we should use a 90%

confidence level.” If John’s directive is followed, will the error margin increase or

decrease?

d)

Jaspar too is frustrated by these results. He’s OK with the 90% confidence. But for him,the error margin of 10% is too large. Jaspar says “I want an error margin of about 2%.”

Assuming the result falls at 70%, is a larger or smaller sample required to achieve an

error margin of 0.02? What sample size is required to achieve an error margin of 0.02?

20. In one study of college students, 83.0% admitted to having cheated on a test, with an error

margin of 7.0% (using 95% confidence). A second similar study found the same result of

83.0%; however, the sample size was three times as large. Is the error margin for the second

study (again at 95% confidence) larger or smaller than 7.0%? What is the error margin for

the second study?

21.

Surveys of people were taken in three countries: Mexico, the United States, and Canada. Thesame number of people were surveyed within each country. In the U.S., 40% of people

agreed with the statement “There is an urgent need to take action on global warming.” In

Mexico the result was 25%; in Canada 80%.

a) Does the size of the country’s population have anything to do with the error margin?




b) Convince yourself that the answer to part a is “No.” For which of these countries is the

error margin for a confidence interval the largest? The smallest? (Assume the same

confidence level is used for all three results.)

22. You want a 95% confidence interval estimate with error margin 4% for the proportion of

science majors who are left handed. How many science majors do you sample?a) Describe in words the parameter you are estimating. What symbol is it given?

b) Assume you have no idea what the prevalence of lefties is for this population. Use a

guess of 0.5 to determine the required sample size.

c) In the general population, 10% of people are lefties. Use this value to determine the

sample size.

d) Which of the answers from c or d is the better choice?

e) It turns out that 24 of 217 sampled science majors are lefties. Obtain the confidence

interval. How does the error margin compare to 4%?

23. Suppose you undertook a study of the day of the week that babies are born. You are

interested in the proportion of babies born on a weekend (Saturday or Sunday). Your goal is

a 90% confidence interval with error margin no greater than 3.5%.

a) Explain why a guess of 0.50 is unreasonable.

b) What is a better value for this guess?

c) In fact, 25% is probably an adequate value for the guess. If “all days are equally likely,”

then 2/7 = 28.6% should be born on weekends. However, in recent years there is more of

a trend for doctors to induce pregnancy, which usually happens on a weekday! Use 0.25to obtain a sample size for this study.

d) If the actual prevalence is p̂ = 0.25, determine the confidence interval when the sample

size from c is used. Identify the error margin – does it meet the goal of 0.035? What

would such a result say about the 2/7 hypothesis?

e) If the actual prevalence is 0.20 and the sample size from c is used, how will the error

margin compare to 0.035? Explain.

24. Consider a large city’s mayoral race where there are two candidates.

a)

Determine the required sample size for a media poll to estimate the percent of peoplewho favor the Republican candidate with error margin 3%.

b) Does the required sample size depend on the population of the city?

25. What proportion of people die during summer (as officially defined)? You decide to

investigate this issue by collecting data. How many obituaries would you examine in order to

obtain a 98% confidence interval estimate with error margin of 1%?




Solutions

1. a) i) p = 5146/8640 = 0.5956; ii) X = 60; iii) p̂ = 60/92 = 0.6522. p = 0.5956 is a parameter;

p̂ = 0.6522 is a statistic, c) The mean is p = 0.5956. If we examined all possible samples of 142

students, determining – for each sample – the proportion who live more than 50 miles from

home, the mean of these proportions is 0.5956. The standard deviation of these proporitons is

1424044.05956.0 = 0.0412.

2. a) p = 0.15. This is a parameter. X = 2. p̂ = 2/30 = 0.06667. p = 0.15 is a parameter; p̂ is are

statistic.

3. a) p̂ = 85/212 = 0.4009. b) 1.96. c) E = 0.0660

3. 0660.0

212

5991.04009.096.1

212

212

127

212

85

96.1

E

d) 0.4009 0.0660. e) 0.3335 < p < 0.467. f) 95%. g) p is the proportion of all adoptive parents

who state “No Preference” for their child’s gender. h) p̂ : Statistic; p: Parameter. i) We are 95%

confident that between 33.5% and 46.7% of all adoptive parents state “No Preference” for their

child’s gender.

4. 0.908 0.041 or (0.866, 0.949) or 0.866 < p < 0.949. These are equivalent and either is

correct. You are 99% confident that between 86.6% and 94.9% of all babies conceived using

XSORT will be girls.

5. (0.496, 0.514) or 0.496 < p < 0.514. I am 95% confident that between 49.6% and 51.4% of all

deaths occurring in the two week period surrounding Thanksgiving take place in the week before

it. There is very little evidence supporting the theory. 0.5 is within this interval, indicated that the

“postpone death” theory is not conclusively supported by the data.

6. Results are shown to the nearest 0.001. Typically it’s ok to report Z scores to the nearest 0.01.

However, when you compute with Z scores, you should use as much accuracy as is possible.

Since accuracy is automatically preserved by the spreadsheet function NORMSDIST (even if

you round the cell display), you should be actually using all decimal places of accuracy when

you compute with a Z score.

C 50% 75% 90% 95% 98% 99% 99.9% 99.99%

Z /2 0.675 1.150 1.645 1.960 2.326 2.576 3.291 3.891

7. a) (0.347, 0.432) or 0.347 < p <0.432. I am 99% confident that between 34.7% and 43.2% of

all jurors are Mexican-American. b) Since this interval estimates the population proportion, and

0.791 is the actual value for this proportion, something is wrong. It must be the case that jury

selection is not random, and that, in fact, there is systematic bias keeping Mexican-Americans

from jury duty.

8. For all these, the estimated proportion is 0.390.




a) The 95% confidence interval is 0.3575 < p < 0.4223. The error margin is 0.0324.

b) Error margins are stated in the table.

C 50% 75% 90% 95% 99% 99.9%

E 0.011 0.019 0.027 0.032 0.043 0.054c) As the confidence increases, the error margin increases.

9. a) 4276(0.940) = 4019.44 – so 4019 have telephones. b) p̂ = 4019/4276 = 0.9399.

c) The error margin is 0.009363 – about 0.0094. (0.9305, 0.9493) or 0.9305 < p < 0.9493. I am

99% confident that the proportion of all households (the population of households) having a

telephone is between 0.9305 and 0.9493. The error margin is a little short of 1%. d) 0.0601

0.0094 or 0.0507 < p < 0.0695. The error margins are the same. The intervals are essentially the

same: One is expressed in terms of a proportion of people having a telephone; the other in terms

of those not having a telephone.

10. a) 0.702 < p < 0.774 or 0.7379 0.0358. b) This does not refute the theory, as 0.75 is among

the plausible values for p given in the interval (in which we have fairly high confidence).

11. a) For n = 145 the interval is 0.6664 < p < 0.8095 or 0.7379 0.0716. 0.0716 is the error

margin. This is twice as large as 0.0358. b) The interval is 0.7379 0.0072. Error margin E =

0.0072 (0.00718 to be more precise). This is 5 times smaller than in the actual study.

c) If the sample size is increased to k times larger, the error margin decreases by k times. Or,

if the error margin is to be k times smaller, the sample size must be k 2 times larger.

12. Jack wants an error margin that is 10/3 = 3.33 times smaller. His sample size must be 3.332

times larger: 3.332

= 11.11 times larger.

13. Since the sample size is 5 times smaller, the error margin will be 5 = 2.24 times smaller.

0.01/2.24 = 0.0045.

14. a)

Parents’ Status # of grads p ˆ 95% CI Error Margin

Welfare 40 0.100 0.0706 < p < 0.1294 0.0294

Poor 80 0.200 0.1608 < p < 0.2392 0.0392

Low Income 120 0.300 0.2551 < p < 0.3449 0.0449

Middle Income 200 0.500 0.4510 < p < 0.5490 0.0490

High Income 280 0.700 0.6551 < p < 0.7449 0.0449

Wealthy 320 0.800 0.7608 < p < 0.8392 0.0392

Super rich 360 0.900 0.8706 < p < 0.9294 0.0294




b) They are the same for each pair. A Success prevalence of 90% is equivalent to a Failure

prevalence of 10%, so the error margins must be the same.

c) Error margin is largest for prevalence p̂ = 0.5 and drops (symmetrically) as the prevalence

gets further from 0.5 – on either side of 0.5.

15. Yellow has prevalence 1 – 0.7379 = 0.2621. So the interval is 0.2621 0.0358. Or take

1 – 0.702 = 0.298 and 1 – 0.774 = 0.226 to get (0.226, 0.298).

16. a) I am 95% confident that between 40.7% and 47.7% of all voters lean towards D.

b) 55.8% 3.5%. c) No. We don’t know how the 55.8% is split up. d) Yes. If there are, for

instance, 20% undecided, then the result for R is around 35.8%.

17. This is a media poll – the confidence is 95%. I am 95% confident that between 77% and 85%

of all Americans favor a drawdown.

18. a) The point estimate is 0.800, the error margin is 0.059. b) The interval at Penn State is

exactly the same. c) The population size does not play into this. The formula for error margindepends only upon the prevalence p̂ and the sample size n. This is an underappreciated fact

about sampling and statistical analysis: As long as a population is “large” (at least 20 times

bigger than the sample), its size is pretty much immaterial. What matters in most practical

situations is the sample size.

19. a) p is the proportion of all SUNY Oswego students who stay in Oswego. It’s a parameter.

It’d be difficult to get this value – you’d have to census virtually every student. b) 0.70 is a

statistic – it describes a sample. c) For a 90% confidence interval the error margin will be

smaller. (See #7.) d) To get an error margin that is 5 times smaller will require a sample size that

is 52 = 25 times larger. (See exercises 11 – 13.) That’s 2000 students.

20. The error margin for a larger sample size will be smaller. It will not be three times smaller. It

will be 3 = 1.732 times smaller: 0.07 / 1.732 = 0.0404 – about 4%. (See exercises 11 – 13.)

21. a) No. b) The error margin is smallest for Canada and largest for the United States. (Go back

and examine #14.)

22. a) The symbol is p. p = the proportion of all science majors at this university who are left

handed. b) 25.6005.05.004.0

96.1 2

n . Select 601 science majors.

c) 09.2169.01.004.0

96.1 2

n . Select 217 science majors. d) 217 is the better choice. The lefty

rate for scientists is going to be fairly close to that for the general population. (Not only that, but

the sample size is smaller!) e) 0.1106 0.0417. Pretty close to 0.04 for the error margin. It

missed a little because the actual lefty rate was slightly closer to 0.5 than the guess of 0.10 that

was used to determine the sample size.



23. a) The weekend constitutes 2 days out of 7. We wouldn’t expect half of all births to occur in

2/7th

s of all days. b) A better guess would be 2/7 = 0.286 (anything from 0.25 to 0.30 is

reasonable; anything outside of this is not).

c) 19.41475.025.0035.0

645.1 2

n , so sample 415. d) 0.035. The interval is (0.215, 0.285).

The error margin is 0.035 – right on the target. The interval does not include 0.286 = 2/7. So we

have some evidence that the 2/7 hypothesis is false. e) It will be smaller. When the actual result

falls further from 0.5, the error margin is smaller. (In fact, if 0.20 is the result, the error margin is

0.032.)

24. a) It’s a media poll, so the confidence is 95%. The required sample size is then 1068. (This

number is well known to pollsters. Many polls have 3% error rates because they use a sample

size of around 1000.) b) Absolutely not. Exercises 18 and 21 covered this.

25. You shouldn’t guess anything except 0.25. The required sample size is then

3.1014475.025.001.0

326.2

2

n , so sample 10,145 obituaries.

CI for a Proportion

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