Chapter 7: Confidence Intervals

Inference
We are in the fourth and final part of the course - statistical inference, where we draw conclusions about the population based on the data obtained from a sample chosen from it. Confidence Intervals Chapter 7
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2 Confidence Intervals (CI) Our Goal in Inference
The goal: to give a range of plausible values for the estimate of the unknown population parameter
the population mean, μ,,
the population proportion, p
the population standard deviation, σ
We start with our best guess: the sample statistic If ALL the populations, whatever we are interested in, would be manageable in size, we would just figure out the population parameter. Then there would be no need for inference.
the sample mean x, $
the sample proportion p
the sample standard deviation, s
Sample statistic = point estimate
4 Confidence Intervals (CI) Confidence Intervals (CI) to estimate … CI = point estimate ± margin of error Population MEAN Population PROPORTION Population STANDARD DEVIATION Margin of error Point estimate: x
5 Point estimate: p$ Margin of error Point estimate: s
6 1 Margin of error Confidence Intervals (CI) for a Mean
Suppose a random sample of size n is taken from a normal
Shows how accurate we believe our estimate is population of values for a quantitative variable whose mean µ is unknown, unknown, when the population’s standard deviation σ is known. known.
A confidence interval (CI) for µ is:
The smaller the margin of error, the more precise our estimate of the true parameter
Formula: CI = point estimate ± margin of error  critical   standard deviation   ⋅   E =   value   of the statistic  σ x ± z *⋅ Point estimate n Margin of error (m or E)
8 So what’s z*??? Statement: (memorize!!)
A confidence interval is associated with a confidence We are ________% confident that the true mean context lies within the interval ______ and ______. level. We will say: “the 95% confidence interval for the population mean is …”
The most common choices for a confidence level are
90% :z* = 1.645
95% : z* = 1.96,
99% : z* = 2.576.
9 The Trade-off Using the calculator
Calculator: STAT→TESTS →7:ZInterval…
Inpt: Data Stats Use this when you have data in one of your lists
There is a tradetrade-off between the level of confidence and Use this when you know x and σ precision in which the parameter is estimated. estimated.
higher level of confidence -- wider confidence interval
lower level of confidence – narrower confidence interval
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12 2 The Margin of Error 95% confident means: In 95% of all possible σ
The width (or length) of the CI is exactly twice n the margin of error (E): samples of this size n, E µ will indeed fall in our E confidence interval. In only 5% of samples E would miss µ.
The margin of error is therefore "in charge" of the width of the confidence interval.
14 How can you make the margin of error smaller? Comment
z* smaller
The margin of error (E ) is E = z *⋅ (lower confidence level) σ n
σ smaller (less variation in the population) and since n, the sample size, appears in the denominator, increasing n will reduce the margin of error for a fixed z*.
n larger
Really cannot (to cut the margin of error in half, n must change! be 4 times as big)
15 Calculating the Sample Size Margin of Error and the Sample Size
In situations where a researcher has some flexibility as to the sample size, the researcher can calculate in advance what the sample size is that he/she needs in order to be able to report a confidence interval with a certain level of confidence and a certain margin of error. E = z *⋅ σ n σ  n =  z *⋅   E 2 Clearly, the sample size n must be an integer. Calculation may give us a non-integer result. In these cases, we should always round up to the next highest integer. integer
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18 3 σ Example Assumptions for the validity of x ± z * ⋅
IQ scores are known to vary normally with standard deviation 15. How many students should be sampled if we want to estimate population mean IQ at 99% confidence with a margin of error equal to 2? 2
The standard deviation, σ, is known
and either
the sample size must be large (n ≥ 30) or
for smaller sample the variable of interest 2 15  σ  n =  z *  =  2.576  = 373.26  E  2 ⇒ n
The sample must be random n = 374 must be normally distributed in the population. They should take a sample of 374 students.
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20 Steps to follow 1. Check conditions: conditions SRS, σ is known, and either n ≥ 30 or the population distribution is normal 2. Calculate the CI for the given confidence level 3. Interpret the CI
21 Example 1
A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. Form past studies, the standard deviation is known to be 1.5 years and the population is normally distributed. Construct a 90% confidence interval of the population mean age.
22 Step 1: Check conditions Step 2: Calculate the 90% CI using the formula
A college admissions director wishes to estimate x = 22.9 the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. Form past studies, the standard deviation is known to be 1.5 years and the population is normally distributed. distributed σ = 15 . n = 20 z * = 1645 . x ± z*
SRS σ n = 22.9 ± 1645 . 15 . = 22.9 ± 0.6 = (22.3,235 .) 20
σ is known
The population is normally distributed
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24 4 Step 2: Calculate the 90% CI using the calculator Step 3: Interpretation
Calculator: STAT→TESTS →7:ZInterval…
Inpt: Data Stats
σ = 1.5
x = 22.9
n = 20
C-Level: .90
Calculate
We are 90% confident that the mean age of all students at that college is between 22.3 and 23.5 years. ZInterval : (22.3, 23.5)
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26 Example 1 Example 2
How many students should he ask if he wants the A scientist wants to know the density of bacteria in a certain solution. He makes measurements of 10 randomly selected sample: margin of error to be no more than 0.5 years with 99% confidence? 2 24, 31, 29, 25, 27, 27, 32, 25, 26, 29 *106 bacteria/ml. 2 σ 15 .    n =  z * ⋅  =  2.576  = 59.72   E 0.5 From past studies the scientist knows that the distribution of bacteria level is normally distributed and the population standard deviation is 2*106 bacteria/ml.
Thus, he needs to have at least 60 students in his a. What is the point estimate of μ? ? sample. x =27.5 *106 bacteria/ml.
27 Example 2 Example 2 Using the calculator:
Enter the number into on of the lists, say L1
STAT  TESTS 7: ZInterval
b. Find the 95% confidence interval for the mean level of bacteria in the solution.
Step 1: check conditions: SRS, normal
Inpt: Data distribution, σ is known. All satisfied.
σ: 2
Step 2: CI: x ± z* σ n = 27.5 ± 1.96
List: L1 2 = 27.5 ±1.24 = (26.26,28.74) 10
Freq: 1 (it’s always 1)
C-Level: .95
Step 3: Interpret: we are 95% confident that the
Calculate mean bacteria level in the whole solution is between 26.26 and 28.74 *106 bacteria/ml.
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(26.26, 28.74)
30 5 Example 2 Example 2
c. What is the margin of error?
d. How many measurements should he make to obtain a margin of error of at most 0.5*106 bacteria/ml with a confidence level of 95%? From part b: x ± z* σ = 27.5 ± 1.96 n 2 = 27.5 ±1.24 = (26.26,28.74) 10 2 2  σ 2 × 10 6   n =  z * ⋅  =  196 . .  = 614656  E 0.5 × 10 6   Thus, the margin of error is
Thus, he needs to take 62 measurements. E=1.24 *106 bacteria/ml.
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32 Assumptions for the validity of
The sample must be random σ x ± z *⋅ n
The standard deviation, σ, is known and either
The sample size must be large (n≥30) or
For smaller sample the variable of interest must be normally distributed in the population.
The only situation when we cannot use this confidence interval, then, is when the sample size is small and the variable of interest is not known to have a normal distribution. In that case, other methods called nonparameteric methods need to be used. Example 3 In a randomized comparative experiment on the effects of calcium on blood pressure, researchers divided 54 healthy, white males at random into two groups, takes calcium or placebo. The paper reports a mean seated systolic blood pressure of 114.9 with standard deviation of 9.3 for the placebo group. Assume systolic blood pressure is normally distributed. Can you find a z-interval for this problem? Why or why not?
33 So what if σ is unknown?
Well, there is some good news and some bad news! σ The good news is that we can easily replace the population standard deviation, σ, with the sample standard deviation s.
35 And the bad news is…
that once σ has been replaced by s, we lose the Central Limit Theorem together with the normality of X and therefore the confidence multipliers z* for the different levels of confidence are (generally) not accurate any more.
The new multipliers come from a different distribution called the "t distribution" and are therefore denoted by t* (instead of z*).
36 6 CI for the population mean when σ is unknown z* vs. t*
There is an important difference between the
The confidence interval for the population mean µ when σ is confidence multipliers we have used so far (z*) and those needed for the case when σ is unknown (t*).
z*, depends only on the level of confidence,
t* depend on both the level of confidence and on the sample size (for example: the t* used in a 95% confidence when n=10 is different from the t* used when n=40). unknown is therefore: x ± t *⋅ s n
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38 t-distribution t-distribution
There is a different t distribution for each
The t-distribution is bell shaped and symmetric about the mean. sample size. We specify a particular t distribution by giving its degrees of freedom. The degrees of freedom for the one-sample t statistic come from the sample standard error s in the denominator of t. Since s has n-1 degrees of freedom, the tdistribution has n-1 degrees of freedom.
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The total area under the t-curve is 1
The mean, median, and mode of the t-distribution are equal to zero.
The tails in the t-distribution are “thicker” than those in the standard normal distribution.
As the df (sample size) increases, the t-distribution approaches the normal distribution. After 29 df the tdistribution is very close to the standard normal zdistribution.
40 Historical Reference Density of the t-distribution (red and green) for 1, 2, 3, 5, 10, and 30 df compared to normal distribution (blue)
William Gosset (1876-1937) developed the t-distribution while employed by the Guinness Brewing Company in Dublin, Ireland. Gosset published his findings using the name “Student”. The tdistribution is, therefore, sometimes referred to as “Student’s t-distribution”.
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42 7 Calculator Example
To study the metabolism of insects,
Calculator: researchers fed cockroaches measured amounts of a sugar solution. After 2, 5, and 10 hours, they dissected some of the cockroaches and measured the amount of sugar in various tissues. Five roaches fed the sugar solution and dissected after 10 hours had the following amounts of sugar in their hindguts:
STAT→TESTS →8:TInterval…
Inpt: Data Stats Use this when you have data in one of your lists Use this when you know x and s
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44 Example Example
55.95, 68.24, 52.73, 21.50, 23.78
The large margin of error is due to the small sample size
Find the 95% CI for the mean amount of sugar in cockroach and the rather large variation among the cockroaches. hindguts: x = 44.44
Calculator: s = 20.741
Put the data in L1. The degrees of freedom, df=n-1=4, and from the table we find that for the 95% confidence, t*=2.776.
Then x ± t *⋅
STAT→TESTS →8:TInterval…
Inpt: Data Stats
List: L1
Freq:1
C-level: .95 s 20.741 = 44.44 ± 2.776 ⋅ = (18.69, 7019 . ) n 5
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46 Examples: You take:
24 samples, the data are normally distributed, σ is known σ x ± z*
normal distribution with σ n
14 samples, the data are normally distributed, σ is unknown x ± t *⋅
t-distribution with s s n
34 samples, the data are not normally distributed, σ is unknown
normal distribution with s x ±t* s n
12 samples; the data are not normally distributed, σ is unknown
cannot use the normal distribution or the t-distribution
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48 8 Some Cautions: Estimating a Population Proportion
The data MUST be a SRS from the When the variable of interest is categorical, the population parameter that we will infer about is a population proportion (p) associated with that variable. population
The formula is not correct for more complex sampling designs, i.e., stratified, etc.
No way to correct for bias in data
Outliers can have a large effect on confidence interval
Must know σ to do a z-interval – which is unrealistic in practice
For example, if we are interested in studying opinions about the death penalty among U.S. adults, and thus our variable of interest is "death penalty (in favor/against)," we'll choose a sample of U.S. adults and use the collected data to make inference about p the proportion of US adults who support the death penalty.
50 Example 2 Example 2
Suppose that we are interested in the opinions of U.S.
If we wanted to estimate p, the population proportion by a adults regarding legalizing the use of marijuana. In particular, we are interested in the parameter p, the proportion of U.S. adults who believe marijuana should be legalized.
Suppose a poll of 1000 U.S. adults finds that 560 of them believe marijuana should be legalized. single number based on the sample, it would make intuitive sense to use the corresponding quantity in the sample, the sample proportion p$ = 560/1000 = 0.56 . We say in this case that 0.56 is the point estimate for p, and that in general, we'll always use p$ as the point estimator for p.
Note, again, that when we talk about the specific value (.56), we use the term estimate, and when we talk in general about the statistic we use the term estimator. Here is a visual summary of this example:
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51 Example 2 Back to Example 2
Suppose a poll of 1000 U.S. adults finds that 560 of them believe marijuana should be legalized.
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54 9 The CI for p Calculator:
Thus, the confidence interval for p is
STAT→TESTS →A:1-PropZInt… p$ ± E = p$ ± z * ⋅ p$ (1 − p$ ) n x = np$
x is the number of successes:
For a 95% CI use z*=1.96
For a 90% CI use z*=1.645
For a 99% CI use z*=2.576
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56 Conditions Example
The CI is reasonably accurate when three conditions are met:
Suppose you have a random sample of 40 buses from a large city and find that 24 buses have a safety violation. Find the 90% CI for the proportion of all buses that have a safety violation.
Conditions:
SRS
The sample was a simple random sample (SRS) from a binomial population
Both and n(1 − p$) ≥ 10
The size of the population is at least 10 times the size of the sample np$ ≥ 10
both np$ = 40( 24 40 ) = 24 ≥ 10 and n(1 − p$) = 40(1 − 24 40 ) = 16 ≥ 10
The size of the population (all the buses) is at least 10 times the size of the sample (40)
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58 90% CI p$ = 24 = 0.6 40
For 90% CI z*=1.645 p$ ± E = p$ ± z * ⋅
59 p$(1 − p$) 0.6(1 − 0.6) = 0.6 ± 1645 . ⋅ = 0.6 ± 013 . = (0.47, 0.73) n 40 Interpretation
1. What is it that you are 90% sure is in the confidence interval? The proportion of all of the buses in this population that have safety violations if we could check them all.
2. What is the meaning (or interpretation) of the confidence interval of 0.47 to 0.73? We are 90% confident that if we could check all of the buses in this population, between 47% and 73% of them would have safety violations.
3. What is the meaning of 90% confidence? If we took 100 random samples of buses from this population and computed the 90% confidence interval from each sample, then we would expect that 90 of these intervals would contain the proportion of all buses in this population that have safety violations. In other words, we are using a method that captures the true population proportion 90% of the time.
60 10 Margin of Error and Sample Size Margin of Error and Sample Size
When we have some level of flexibility in determining the sample size, we can set a desired margin of error for estimating the population proportion and find the sample size that will achieve that.
For example, a final poll on the day before an election would want the margin of error to be quite small (with a high level of confidence) in order to be able to predict the election results with the most precision. This is particularly relevant when it is a close race between the candidates. The polling company needs to figure out how many eligible voters it needs to include in their sample in order to achieve that.
Let's see how we do that.
The confidence interval for p is p$ ± E = p$ ± z * ⋅ p$(1 − p$) n E = z *⋅ p$(1 − p$) n
Thus, the margin of error is Using some algebra we have 2  z * n =   p$(1 − p$)  E
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62 2  z * n =   p$(1 − p$)  E What sample size should we use for a survey if we want a margin of error to be at most 3%? Let’s use the 95% confidence here, so z*=1.96. Also, since we don’t have an estimate of p, we will use p$ = 0.5 .Then
If you have a good estimate p$ of p, use it in this formula, otherwise take the conservative approach by setting p$ = 1 2 .
You have to decide on a level of confidence so you know what 2 value of z* to use (most common one is the 95% level).
Also, obviously, you have to set the margin of error (the most common one is 3%). 2 .   z *  196 n =   p$ (1 − p$ ) =   ( 0.5)(1 − 0.5) = 1067.111  E  0.03 Because you must have a sample size of at least 1067.111, round up to 1068. So n should be at least 1068.
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64 Summary: CI for … a population proportion p$ ± z * ⋅ a population mean, σ is known and normally distributed population or n≥30 a population mean, σ is unknown and normally distributed population or n≥30 p$(1 − p$) n σ x ± z* x ± z* n x ± t *⋅ s n s n
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