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Statistical inference is inferring information about the distribution of a population from information about a sample. We’re generally talking about one of two things:
1. ESTIMATING PARAMETERS (CONFIDENCE INTERVALS)
2. ANSWERING A YES/NO QUESTION ABOUT A PARAMETER (HYPOTHESIS TESTING)
1 Confidence intervals
A confidence interval is the interval within which a population parameter is believed to lie with a mea- surable level of confidence.
We want to be able to fill in the blanks in a statement like the following:
We estimate the parameter to be between and and this interval will contain the true value of the parameter approximately % of the times we use this method.
What does confidence mean? A CONFIDENCE LEVEL OF C INDICATES THAT IF REPEATED SAMPLES
(OF THE SAME SIZE) WERE TAKEN AND USED TO PRODUCE LEVEL C CONFIDENCE INTERVALS,
THE POPULATION PARAMETER WOULD LIE INSIDE THESE CONFIDENCE INTERVALS C PERCENT
OF THE TIME IN THE LONG RUN.
Where do confidence intervals come from? Let’s think about a generic example. Suppose that we take an SRS of size n from a population with mean μ and standard deviation σ. We know that the sampling
distribution for sample means ( ¯x) is (approximately) N(μ, √σn ).
By the 68-95-99.7 rule, about 95% of samples will produce a value for ¯x that is within about 2 standard
deviations of the true population mean μ. That is, about 95% of samples will produce a value for ¯x that
is within 2 √σn of μ. (Actually, it is more accurate to replace 2 by 1.96 , as we can see from a Normal
Distribution Table.)
But if x¯ is within (1.96) √σn of μ , then μ is within (1.96) √σn of x¯. This won’t be true of every sam-
ple, but it will be true of 95% of samples.
So we are fairly confident that μ is between x¯ − 1.96 √σn and x¯ + 1.96 √σn. We will express this by saying
that the 95% CI for μ is ¯x ± 1.96 √σn
Example. A sample of 25 Valencia oranges weighed an average (mean) of 10 oz per orange. The standard deviation of the population of weights of Valencia oranges is 2 oz. Find a 95% confidence interval (CI) for the population mean.
5
, OR (9.216, 10.784)
Of course, we are free to choose any level of confidence we like. The only thing that will change is the critical value 1.96. We denote the critical value by z∗. If we want a confidence level of C, we choose the critical value z∗^ so that
THE PROBABILITY P(Z > z∗) = C/ 2
and then
the level C confidence interval is ¯x ± z∗^
n
z∗^
σ √ n
is called THE MARGIN OF ERROR
Example. Now compute a 99% CI for the mean weight of the Valencia oranges based on our previous sample of size 25. (n = 25, ¯x = 10, σ = 2)
z∗^ = 2.576 CI: (8.970, 11.030)
Example. Now compute a 90% CI for the mean weight of the Valencia oranges based on our previous sample of size 25. (n = 25, ¯x = 10, σ = 2)
z∗^ = 1.645 CI: (9.342, 10.658)
Example. A random sample of 100 batteries has a mean lifetime of 15 hours. Suppose the standard deviation for battery life in the population of all batteries is 2 hours. Give a 95% CI for the mean battery life.
z∗^ = 1.96 n = 100 σ = 2 x¯ = 15 CI: (14.608, 15.392)
1.1 What can we learn from the formula?
- As n increases, the width of the CI DECREASES
- As σ increases, the width of the CI INCREASES
- As the confidence level increases, z∗^ INCREASES , so the width INCREASES
1.4 Some Examples
Here are some additional examples. In these examples we will give the sample standard deviation (s) rather than the population standard deviation σ which would probably be unavailable in most cases. For now we will just proceed using s in place of σ and a normal distribution. In several of the cases below, this simplification is unwarranted because the sample sizes are too small , but we will learn how to deal appropriately with these small sample sizes when we learn more about t-distributions. In these cases, our confidence intervals will be smaller using a normal distribution than they should be.
For each of these examples compute a 95% confidence interval and a confidence interval at some other confidence level.
- In a sample of 45 circuits, the mean breakdown voltage (in kV) under controlled circumstances was 54.7 and the sample standard deviation was 5.23.
95% CI: (53.172, 56.228), 99% CI: (52.692, 56.708)
- Ten new Discraft 175 gram discs were tested to see how much water they could hold. The mean volume of water was 1.936 liters and the standard deviation was 0.0259.
90% CI: (1.933, 1.939), 95% CI: (1.932, 1.940)
- Ten old Discraft 175 gram discs were also tested to see how much water they could hold. The mean volume of water was 1.778 liters and the standard deviation was 0.0582.
95% CI: (1.769, 1.787), 99% CI: (1.767, 1.789)
- An experiment was conducted to see whether it is easier to learn a list of words by looking at textual or graphical representations of the words. 10 people were given a list of words, and 10 were given a list of pictures. The subjects were given 30 seconds to study the list and then asked to recall as many of the items as they could. For those with picture lists, the mean was 8.9 with a standard deviation of 1.595. For those with word lists, the mean was 7.8 with a standard deviation of 1.874.
95% CI FOR PICTURE: (7.911, 9.889), 98% CI FOR WORD: (6.422, 9.178)
We will return to these examples and do them right once we know how to deal with such small samples.
1.5 One-Sided Confidence Intervals
Sometimes we are only interested in bounding our estimate for a parameter in one direction. That is, we want to be able to say with some level of confidence that we believe the parameter θ is greater than some value (or less than some value).
Example. Let’s return to our battery example. A producer of batteries would like to make a claim that on average the batteries last at least some length of time. (No one will complain, after all, if they happen to last longer than expected.) A random sample of 100 batteries has a mean lifetime of 15 hours. Suppose the standard deviation for battery life in the population of all batteries is known to be 2 hours. Give a 95% confidence lower bound for the mean battery life.
z∗^ = 1.645 n = 100 σ = 2 x¯ = 15 lower bound: 14.671 HOURS
For more practice, go back to the previous confidence interval examples and compute one-sided intervals instead.
1.6 Sampling distribution of sample means
For random samples from a normal population with mean μ and standard deviation σ,
- x¯ is normally distributed with mean μ and standard deviation √σn.
- (^) σx¯/−√μn has a normal distribution with mean 0 and standard deviation 1. (N(0, 1))
- (^) sx¯/−√μn has a t-distribution with n − 1 degrees of freedom. (d f = n − 1)
While the family of normal distributions is defined by the mean and standard deviation, the family of t-distributions is defined by degrees of freedom , usually abbreviated df.
1.7 The t -distributions
In many ways the t-distributions are a lot like the standard normal distribution. For example, all t- distributions are UNIMODAL and SYMMETRIC with a mean and median both equal to ZERO But the density curve for the t distributions is FLATTER and MORE SPREAD OUT than the density curve for the normal distribution. Statisticians like to say that the t distributions have HEAVIER TAILS , which means more of the values are farther from the center (0).
As df increases, the t-distribution becomes more and more like the standard normal distribution:
We can compute probabilities based on t-distributions just as easily as we can for normal distributions, but we need to use a computer or a different table.
Examples. Suppose the random variable t has a t-distribution with d f = 11. Use the t-distribution table to determine the following:
- P(t > 1.363) =0.
- P(t < 1.363) =0.
- P(t > −1.363) =0.
- P(t < −1.363) =0.
- P(t > 2.201) =0.
- P(t < −1.796) =0.
- P(−1.796 < t < 1.796) =0.
- P(t > 2.00) =BETWEEN 0.025 AND 0.
1.10 Some cautions about CIs for the mean of a population
- Data must be from an SRS.
- The formula is not correct for more complex designs (e.g., stratified random sampling).
- Because the mean is not resistant, outliers can have a large effect on the CI.
- If the sample size is small and the population not normal, we need different procedures.
- The margin of error in a CI covers only random sampling errors, not errors in design or data collec- tion.
1.11 Prediction Intervals
Prediction intervals are similar to confidence intervals, but they are estimating something different. The confidence intervals we have been looking at give an estimate for the mean of the population. But there is another type of estimate that is commonly done. For a prediction interval, we want to give a range in which we expect one randomly selected individual’s value to be.
Once again, the key to figuring out prediction intervals is understanding the sampling distribution in- volved. We consider our sample x 1 , x 2 ,... , xn to be n randomly selected values from a population dis- tribution, and we consider x to be one additional random seletion from this distribution. The idea is to make a prediction for x (in the form of a confidence interval) based on the results of the sample.
The obvious point estimate to use is ¯x, the sample mean. But what can we say about the quality of this estimate? Let’s look at the sampling distribution of ¯x − x (the differences between the sample mean and the value of an additional random selection). If the population has mean μ and standard deviation σ, then
E( x¯ − x) = E( x¯) − E(x) = μ − μ = 0
and
V( x¯ − x) = V( x¯) + V(x) =
σ^2 n
n
so the standard deviation of the sampling distribution is σ
1 + (^) n^1.
From this we see that a prediction interval would be
x¯ ± z∗σ
n
if we knew σ. As you might guess, when we do not know sigma, we will use instead
x¯ ± t∗s
n
using the sample standard deviation s in place of the population standard deviation σ and using a t- critical value (with n − 1 degrees of freedom) instead of a critical value from the normal distribution.
2 Hypothesis Testing
Our goal with tests of significance (hypothesis testing) is to ASSESS THE EVIDENCE PROVIDED BY DATA
WITH RESPECT TO SOME CLAIM ABOUT A POPULATION.
A hypothesis test is a formal procedure for comparing observed (sample) data with a hypothesis whose truth we want to ascertain.
The hypothesis is a STATEMENT about the PARAMETERS in a POPULATION or MODEL. The results of a test are expressed in terms of a probability that measures how well the data and the hypothesis agree. In other words, it helps us decide if a hypothesis is reasonable or unreasonable based on the likelihood of getting sample data similar to our data.
Just like confidence intervals, hypothesis testing is based on our knowledge of SAMPLING DISTRIBU-
TIONS.
2.1 4 step procedure for testing a hypothesis
Step 1: Identify parameters and state the null and alternate hypotheses.
State the hypothesis to be tested. It is called the null hypothesis , designated H 0.
The alternate hypothesis describes what you will believe if you reject the null hypothesis. It is designated Ha. This is the statement that we hope or suspect is true instead of H 0.
Example. The average LDL of healthy middle-aged women is 108.4. We suspect that smokers have a
higher LDL (LDL is supposed to be low). H 0 : μ = 108.4, Ha : μ > 108.
Example. The average Beck Depression Index (BDI) of women at their first post-menopausal visit is 5.1. We wonder if women who have quit smoking between their baseline and post visits have a BDI different
from the other healthy women. H 0 : μ = 5.1, Ha : μ 6 = 5.
Step 2: Compute the test statistic
A test statistic measures how well the sample data AGREES WITH THE NULL HYPOTHESIS.
When we are testing a hypothesis about the mean of a population, the test statistic has the form
(DATA VALUE) - (HYPOTHESIS VALUE)
SD OR SE
For our HWS examples (LDL: 159 smokers, ¯x = 115.7, sd=29.8; BDI: 27 quitters, ¯x = 5.6, sd=5.1):
LDL: t = x s¯/−√μn^0 = 115.729.8/−√108.4 159 = 3.
BDI: t = x s¯/−√μn^0 = (^) 5.15.6/−√5.1 27 = 0.
Step 4: State a conclusion
A decision rule is simply a statement of the conditions under which the null hypothesis is or is not rejected. This condition generally means choosing a significance level α. If p is less than the significance level α , we reject the null hypothesis , and decide that the sample data do not support our null hypothesis and the results of our study are then called statistically significant at significance level α.
If p > α , then we do not reject the null hypothesis. This doesn’t necessarily mean that the null hypoth- esis is true, only that our data do not give strong enough evidence to reject it. It is rather like a criminal trial. If there is strong enough evidence, we convict (guilty), but if there is reasonable doubt, we find the defendant “not guilty”. This doesn’t mean the defendant is innocent, only that we don’t have enough evidence to claim with confidence that he or she is guilty.
Looking back at our previous examples, if we select a significance level of α = 0.05, what do we conclude?
- LDL (p-value < 0.0025) Since 0.0025 < 0.05, we REJECT the null hypothesis and conclude that the smokers have a signifi- cantly higher LDL. “Smokers have higher LDL (t = 3.09, d f = 158, p < .0025)”.
- BDI (p-value > .5) This p-value is higher than .05, so we DO NOT REJECT the null hypothesis. It is quite possible that the null hypothesis is true and our data differed from the hypothesized value just based on random variation. In this case we say that the difference between the mean of 5.6 (those who quit smoking) and 5.1 (average healthy women) is not statistically significant. “Those who quit smoking do not have a significantly different average BDI than middle-aged healthy women (t = .51, d f = 26, p > .50).”
It’s good to report the actual p-value, rather than just the level of significance at which it was or was not significant. This is because:
IT IS USEFUL TO KNOW BY HOW MUCH THE NULL HYPOTHESIS HAS BEEN REJECTED OR NOT RE-
JECTED. IT ALSO LETS SOMEONE ELSE WHO HAS SET THE SIGNIFICANCE LEVEL α DIFFERENTLY
TO DRAW HER OWN CONCLUSION.
Example. Suppose that it costs $60 for an insurance company to investigate accident claims. This cost was deemed exorbitant compared to other insurance companies, and cost-cutting measures were instituted. In order to evaluate the impact of these new measures, a sample of 26 recent claims was selected at random. The sample mean and sd were $57 and $10, respectively. At the 0.01 level, is there a reduction in the average cost? Or, can the difference of three dollars ($60-$57) be attributed to the sample we happened to pick?
Hypotheses: H 0 : μ = 60, Ha : μ < 60
Test statistic: t =
= −1.530 p-value: 0.1 < P(|t| ≥ 1.530) < 0.
Conclusion: THE DATA IS NOT SIGNIFICANT AT THE 1% LEVEL TO REJECT THE NULL HYPOTHE-
SIS. RATHER, IT IS CONSISTENT WITH THE BELIEF THAT, DESPITE THE COST-CUTTING MEASURES
ALREADY TAKEN, THE AVERAGE COST PER ACCIDENT CLAIM IS STILL $60. (t = 1.53, d f =
25, P > 0.1)
Example. The amount of lead in a certain type of soil, when released by a standard extraction method, averages 86 parts per million (ppm). Developers of a new extraction method wondered if their method would extract a significantly different amount of lead. 41 specimens were obtained, with a mean of 83 ppm lead and a sd of 10 ppm.
Hypotheses: H 0 : μ = 86, Ha : μ 6 = 86
Test statistic: t =
= −1.921 p-value: 0.05 < P(|t| ≥ 1.921) < 0.
Conclusion: SAMPLES AS EXTREME AS THE ONE THESE DEVELOPERS TOOK OCCUR LESS THAN
10%, BUT MORE THAN 5% OF THE TIME WHEN THE MEAN FOR THE METHOD IS STILL 86 PARTS
PER MILLION. THUS, WE COULD REJECT H 0 AT THE 5% LEVEL, BUT NOT THE 10% LEVEL. (t =
1.921, d f = 40, 0.05 < P < 0.1)
Example. The final stage of a chemical process is sampled and the level of impurities determined. The final stage is recycled if there are too many impurities, and the controls are readjusted if there are too few impurities (which is an indication that too much catalyst is being added). If it is concluded that the mean impurity level=.01 gram/liter, the process is continued without interruption. A sample of n = 100 specimens is measured, with a mean of 0.0112 and a sd of 0.005 g/l. Should the process be interrupted? Use a level of significance of .05.
Hypotheses: H 0 : μ = 0.01, Ha : μ 6 = 0.
Test statistic: t =
= 2.4 p-value: 0.01 < P(|t| ≥ 2.4) < 0.
Conclusion: THE RESULTS ARE SIGNIFICANT AT THE 5% LEVEL TO CONCLUDE THE PROCESS
MEAN IS NOT AT THE TARGET LEVEL (0.01 G/L) AND SHOULD BE INTERRUPTED FOR ADJUST-
MENTS. (t = 2.4, d f = 99, P < 0.02)
