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 pˆ X n. The mean and standard deviation for pˆ are
Mean of pˆ : pˆ p= 0.74 Standard deviation of pˆ :
06203 50 26 . 0 74 . 0 1 ˆ 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 pˆ is approximately Normal. Using the Normal approximation we know that the probability is approximately 95% that pˆ falls within 1.96 standard deviations of the mean.
95% of all samples have a sample proportion pˆ 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 “ pˆ is within 0.122 of 0.74.” For this situation approximately 95% of all possible samples yield a proportion pˆ 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 pˆ within 1.96 p
1p
n 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 pˆ is within 1.96 p
1p
n of p. Flip-flopping p and pˆ yields the following statement (which is true):For approximately 95% of all samples p is within 1.96 pˆ
1 pˆ
n of pˆ .In this second statement the interval is random. Different samples yield different values for pˆ , which result in different intervals.
The Result – a 95% Confidence Interval for p
For one sample yielding a result pˆ , the interval E
pˆ where E1.96 pˆ
1 pˆ
nforms an approximate 95% confidence interval for p. pˆ 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 the confidence 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
pˆ ˆ
pˆ E, pˆE
pˆE topˆ E pˆEThe 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,
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 pˆ = 0.3430. Then
1 ˆ
1.96 0.343
0.657
1000 1.96 0.01501 0.0294 ˆ 96 . 1 p p n 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 ± 0.0294 The 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 pˆ 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! 1.96 0.343
0.657
1000is 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%.”
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 Confidence Interval for a Proportion
An approximate C% confidence interval for p is E pˆ where E is the error margin
n p p z E 2 ˆ 1 ˆ ,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 ˆ 1 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” described 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 in
application 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, z2 z0.005 = 2.576. The estimated/observed proportion of falsifiers is pˆ = 15/84 = 0.1786. Then the error margin is
1076 . 0 0418 . 0 576 . 2 84 8214 . 0 1786 . 0 576 . 2 84 1786 . 0 1 1786 . 0 576 . 2 ˆ 1 ˆ 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
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 is not 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.
pˆ 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.
z2 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. 2E = W.
n p pˆ 1 ˆ
is often called the “(estimated) standard error of pˆ .” A standard error is essentially a standard deviation.5 Recall: Standard deviation measures “typical deviation from mean.” The deviation of pˆ 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 p
SE ˆ ˆ 1 ˆ .
The error margin E for this interval can be expressed E z2SE
pˆ .
5 This quantity really is an estimated standard deviation, as the standard deviation of
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 qˆ
1 pˆ
nx
n. When this is done
n q p n p pˆ1 ˆ ˆˆ .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. Before it'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 (pˆ )
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 to take 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 72.0 160.0
The 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 amounts of 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 polls choose 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
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 pˆ 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, then ultimately 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 pˆ , 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 p SE z E 2 ˆ 2 ˆ 1 ˆ . This is equivalent to p
p
E z n ˆ 1 ˆ 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 (pˆ ), then an
appropriate minimum sample size for the study is p
p
E z n ˆ 1 ˆ 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 pˆ 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 pˆ = 0.45. The required sample size is 0.45
0.55
2376.99 02 . 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 pˆ 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 Ed. (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
0.40
637.03 60 . 0 05 . 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 0.9
0.1 238.89 05 . 0 576 . 2 2 n ). Notice that 239 1 . 0 9 . 0 576 .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 sam p le s iz 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 3
Whenever 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.
0 200 400 600 800 1000 0 0.2 0.4 0.6 0.8 1 proportion p sam p le s iz 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 same shape 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 confidence intervals
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(pˆ ) =
np pˆ 1 ˆ
, yielding
the error margin
n p p z p SE z E * ˆ * ˆ 1 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 pˆ .
b) Which of p and pˆ 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 pˆ ? 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 pˆ .
b) Which of p and pˆ 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 the
following:
a) The point estimate. b) The critical value (Z/2).
c) The error margin.
d) Write the interval bounds in this format: pˆ 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) pˆ: Parameter or Statistic? p: Parameter or Statistic?
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 pˆ - you don’t have to compute it: pˆ = 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 pˆ = 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 pˆ 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 pˆ = 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
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 and sample 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:
np p z
E 2 ˆ 1 ˆ . 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 pˆ = 0.10 and pˆ = 0.90? How about pˆ = 0.20 and pˆ = 0.80? pˆ = 0.30 and pˆ = 0.70? Why does this make sense?
c) Write a single sentence describing the relationship between the proportion pˆ and the error margin of the confidence interval.
Parents’ Status # of grads pˆ 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 point estimate for the proportion of green peas is pˆ = 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 pˆ = 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 pˆ= 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. The same 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%.
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 0.25 to obtain a sample size for this study.
d) If the actual prevalence is pˆ = 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 people who 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) pˆ = 60/92 = 0.6522. p = 0.5956 is a parameter; pˆ = 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
0.5956
0.4044
142 = 0.0412.2. a) p = 0.15. This is a parameter. X = 2. pˆ = 2/30 = 0.06667. p = 0.15 is a parameter; pˆ is are statistic. 3. a) pˆ = 85/212 = 0.4009. b) 1.96. c) E = 0.0660 3. 0660 . 0 212 5991 . 0 4009 . 0 96 . 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) pˆ: 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.
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 0.054 c) As the confidence increases, the error margin increases.
9. a) 4276(0.940) = 4019.44 – so 4019 have telephones. b) pˆ = 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 k2 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 pˆ 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 pˆ = 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 margin depends only upon the prevalence pˆ 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) 0.5
0.5 600.25 04 . 0 96 . 1 2 n . Select 601 science majors.
c) 0.1
0.9 216.09 04 . 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 2/7ths 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) 0.25
0.75
414.19 035 . 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
