This chapter presents two common tests of the hypothesis that a population mean equals a particular value and of the hypothesis that two population means are equal: the z test and the t test. These tests are approximate: They are based on approximations to the probability distribution of the test statistic when the null hypothesis, so their significance levels are not exactly what they claim to be. If the sample size is reasonably large and the population from which the sample is drawn has a nearly normal distribution—a notion defined in this chapter—the nominal significance levels of the tests are close to their actual significance levels. If these conditions are not met, the significance levels of the approximate tests can differ substantially from their nominal values. The z test is based on the normal approximation; the t test is based on Student's t curve, which approximates some probability histograms better than the normal curve does. The chapter also presents the deep connection between hypothesis tests and confidence intervals, and shows how to compute approximate confidence intervals for the population mean of nearly normal populations using Student's t-curve.
In we constructed the z test for equality of two percentages using independent random samples from the two populations. The original test statistic was the difference φ^{t−c} between the two independent sample percentages. If the null hypothesis that the two population means are equal is true, the expected value of the test statistic, E(φ^{t−c}), is zero. If, in addition, the sample sizes are large, we can estimate SE(φ^{t−c}) accurately using the pooled bootstrap estimate of the SD of the "null box:"
SD(box)~ s^{*}=(φ×(1− φ))^{½},
where φ is the pooled sample percentage of the two samples. The estimate of SE(φ^{t−c}) under the null hypothesis is
se = s^{*}×(1/n_{t} + 1/n_{c})^{½},
where n_{t} and n_{c} are the sizes of the two samples. If the null hypothesis is true, the Z statistic,
Z=φ^{t−c}/se,
is the original test statistic φ^{t−c} in approximately standard units, and Z has a probability histogram that is approximated well by the normal curve, which allowed us to select the rejection region for the approximate test.
This strategy—transforming a test statistic approximately to standard units under the assumption that the null hypothesisis true, and then using the normal approximation to determine the rejection region for the test—works to construct approximate hypothesis tests in many other situations, too. The resulting hypothesis test is called a z test. Suppose that we are testing a null hypothesis using a test statistic X, and the following conditions hold:
Then, under the null hypothesis, the probability histogram of the Z statistic
Z = (X − E(X))/se
is approximated well by the normal curve, and we can use the normal approximation to select the rejection region for the test using Z as the test statistic. If the null hypothesis is true,
P(Z<z_{a}) ~ a
P(Z>z_{1−a}) ~ a,
and
P(|Z|>z_{1−a/2}) ~ a.
These three approximations yield three different z tests of the hypothesis that μ = μ_{0} at approximate significance level a:
The word "tail" refers to the tails of the normal curve: In a left-tail test, the probability of a Type I error is approximately the area of the left tail of the normal curve, from minus infinity to z_{a}. In a right-tail test, the probability of a Type I error is approximately the area of the right tail of the normal curve, from z_{1−a} to infinity. In a two-tail test, the probability of a Type I error is approximately the sum of the areas of both tails of the normal curve, the left tail from minus infinity to z_{a/2} and the right tail from z_{1−a/2} to infinity. All three of these tests are called z tests. The observed value of Z is called the z score.
Which of these three tests, if any, should one use? The answer depends on the probability distribution of Z when the alternative hypothesis is true. As a rule of thumb, if, under the alternative hypothesis, E(Z) < 0, use the left-tail test. If, under the alternative hypothesis, E(Z) > 0, use the right-tail test. If, under the alternative hypothesis, it is possible that E(Z) < 0 and it is possible that E(Z) > 0, use the two-tail test. If, under the alternative hypothesis, E(Z) = 0, use a different approach: Consult a statistician. Generally (but not always), this rule of thumb selects the test with the most power for a given significance level.
Each of the three z tests gives us a family of procedures for testing the null hypothesis at any (approximate) significance level a between 0 and 100%—we just use the appropriate quantile of the normal curve. This makes it particularly easy to find the P value for a z test. Recall that the P value is the smallest significance level for which we would reject the null hypothesis, among a family of tests of the null hypothesis at different significance levels.
Suppose the z score (the observed value of Z) is x. In a left-tail test, the P value is the area under the normal curve to the left of x: Had we chosen the significance level a so that z_{a}=x, we would have rejected the null hypothesis, but we would not have rejected it for any smaller value of a, because for all smaller values of a, z_{a}<x. Similarly, for a right-tail z test, the P value is the area under the normal curve to the right of x: If x=z_{1−a} we would reject the null hypothesis at approximate significance level a, but not at smaller significance levels. For a two-tail z test, the P value is the sum of the area under the normal curve to the left of −|x| and the area under the normal curve to the right of |x|.
Finding P values and specifying the rejection region for the z test involves the probability distribution of Z under the assumption that the null hypothesis is true. Rarely is the alternative hypothesis sufficiently detailed to specify the probability distribution of Z completely, but often the alternative does help us choose intelligently among left-tail, right-tail, and two-tail z tests. This is perhaps the most important issue in deciding which hypothesis to take as the null hypothesis and which as the alternative: We calculate the significance level under the null hypothesis, and that calculation must be tractable.
How close the normal approximations to the significance and power are to the true significance level and power depends on how well the normal curve approximates the probability histogram of the test statistic in standard units. If the original test statistic is a sample sum or a sample mean of draws with replacement (or a sum or difference of independent sample sums or sample means ), its probability histogram can be approximated accurately by a normal curve if the sample size is large; this is a consequence of the central limit theorem.
However, to construct a z test, we need to know the expected value and SE of the test statistic under the null hypothesis. Usually it is easy to determine the expected value, but often the SE must be estimated from the data. Later in this chapter we shall see what to do if the SE cannot be estimated accurately, but the shape of the distribution of the numbers in the population is known. The next section develops z tests for the population percentage and mean, and for the difference between two population means.
The central limit theorem assures us that the probability histogram of the sample mean of random draws with replacement from a box of tickets—transformed to standard units—can be approximated increasingly well by a normal curve as the number of draws increases. In the previous section, we learned that the probability histogram of a sum or difference of independent sample means of draws with replacement also can be approximated increasingly well by a normal curve as the two sample sizes increase. We shall use these facts to derive z tests for population means and percentages and differences of population means and percentages.
Suppose we have a population of N units of which G are labeled "1" and the rest are labeled "0." Let p = G/N be the population percentage. Consider testing the null hypothesis that p = p_{0} against the alternative hypothesis that p ≠ p_{0}, using a random sample of n units drawn with replacement. (We could assume instead that N >> n and allow the draws to be without replacement.)
Under the null hypothesis, the sample percentage
φ = (# tickets labeled "1" in the sample)/n
has expected value E(φ) = p_{0} and standard error
SE(φ) = ( p_{0} × (1 − p_{0})/n )^{½}.
Let Z be φ transformed to standard units:
Z = (φ − p_{0})/SE(φ).
Provided n is large and p_{0} is not too close to zero or 100% (say n×p > 30 and n×(1−p) > 30), the probability histogram of Z will be approximated well by the normal curve, and we can use it as the Z statistic in a z test. For example, if we reject the null hypothesis when |Z| > 1.96, the significance level of the test will be about 95%.
The approach in the previous subsection applies, mutatis mutandis, to testing the hypothesis that the population mean equals a given value, even when the population contains numbers other than just 0 and 1. However, in contrast to the hypothesis that the population percentage equals a given value, the null hypothesis that a more general population mean equals a given value does not specify the SD of the population, which poses difficulties that are surmountable (by approximation and estimation) if the sample size is large.
Consider testing the null hypothesis that the population mean μ is equal to a specific null value μ_{0}, against the alternative hypothesis that μ<μ_{0}, on the basis of a random sample with replacement of size n. Recall that the sample mean M of n random draws with or without replacement from a box of numbered tickets is an unbiased estimator of the population mean μ: If
M = (sum of sample values)/n,
then
E(M) = μ = (sum of population values)/N,
where N is the size of the population. The population mean determines the expected value of the sample mean. The SE of the sample mean of a random sample with replacement is
SD(box)/n^{½},
where SD(box) is the SD of the list of all the numbers in the box, and n is the sample size. As a special case, the sample percentage φ of n independent random draws from a 0-1 box is an unbiased estimator of the population percentage p, with SE equal to
(p×(1−p))^{½}/n^{½}.
In testing the null hypothesis that a population percentage p equals p_{0}, the null hypothesis specifies not only the expected value of the sample percentage φ, it automatically specifies the SE of the sample percentage as well, because the SD of the values in a 0-1 box is determined by the population percentage p:
SD(box) = (p×(1−p))^{½}.
The null hypothesis thus gives us all the information we need to standardize the sample percentage under the null hypothesis. In contrast, the SD of the values in a box of tickets labeled with arbitrary numbers bears no particular relation to the mean of the values, so the null hypothesis that the population mean μ of a box of tickets labeled with arbitrary numbers equals a specific value μ_{0} determines the expected value of the sample mean, but not the standard error of the sample mean. To standardize the sample mean to construct a z test for the value of a population mean, we need to estimate the SE of the sample mean under the null hypothesis. When the sample size is large, the sample standard deviation s is likely to be close to the SD of the population, and
se=s/n^{½}
is likely to be an accurate estimate of SE(M). The central limit theorem tells us that when the sample size n is large, the probability histogram of the sample mean, converted to standard units, is approximated well by the normal curve. Under the null hypothesis,
E(M) = μ_{0},
and thus when n is large
Z = (M−μ_{0})/(s/n^{½})
has expected value zero, and its probability histogram is approximated well by the normal curve, so we can use Z as the Z statistic in a z test. If the alternative hypothesis is true, the expected value of Z could be either greater than zero or less than zero, so it is appropriate to use a two-tail z test. If the alternative hypothesis is μ > μ_{0}, then under the alternative hypothesis, the expected value of Z is greater than zero, and it is appropriate to use a right-tail z test. If the alternative hypothesis is μ < μ_{0}, then under the alternative hypothesis, the expected value of Z is less than zero, and it is appropriate to use a left-tail z test.
Consider the problem of testing the hypothesis that two population means are equal, using random samples from the two populations. Different sampling designs lead to different hypothesis testing procedures. In this section, we consider two kinds of random samples from the two populations: paired samples and independent samples, and construct z tests appropriate for each.
Consider a population of N individuals, each of whom is labeled with two numbers. For example, the N individuals might be a group of doctors, and the two numbers that label each doctor might be the annual payments to the doctor by an HMO under the terms of the current contract and under the terms of a proposed revision of the contract. Let the two numbers associated with individual i be c_{i} and t_{i}. (Think of c as control and t as treatment. In this example, control is the current contract, and treatment is the proposed contract.) Let μ_{c} be the population mean of the N values
{c_{1}, c_{2}, …, c_{N}},
and let μ_{t} be the population mean of the N values
{t_{1}, t_{2}, …, t_{N}}.
Suppose we want to test the null hypothesis that
μ = μ_{t} − μ_{c} = μ_{0}
against the alternative hypothesis that μ<μ_{0}. With μ_{0}=$0, this null hypothesis is that the average annual payment to doctors under the proposed revision would be the same as the average payment under the current contract, and the alternative is that on average doctors would be paid less under the new contract than under the current contract. With μ_{0}=−$5,000, this null hypothesis is that the proposed contract would save the HMO an average of $5,000 per doctor, compared with the current contract; the alternative is that under the proposed contract, the HMO would save even more than that. With μ_{0}=$1,000, this null hypothesis is that doctors would be paid an average of $1,000 more per year under the new contract than under the old one; the alternative hypothesis is that on average doctors would be paid less than an additional $1,000 per year under the new contract—perhaps even less than they are paid under the current contract. For the remainder of this example, we shall take μ_{0}=$1,000.
The data on which we shall base the test are observations of both c_{i} and t_{i} for a sample of n individuals chosen at random with replacement from the population of N individuals (or a simple random sample of size n<<N): We select n doctors at random from the N doctors under contract to the HMO, record the current annual payments to them, and calculate what the payments to them would be under the terms of the new contract. This is called a paired sample, because the samples from the population of control values and from the population of treatment values come in pairs: one value for control and one for treatment for each individual in the sample. Testing the hypothesis that the difference between two population means is equal to μ_{0} using a paired sample is just the problem of testing the hypothesis that the population mean μ of the set of differences
d_{i} = t_{i} − c_{i}, i= 1, 2, …, N,
is equal to μ_{0}. Denote the n (random) observed values of c_{i} and t_{i} by {C_{1}, C_{2}, …, C_{n}} and {T_{1}, T_{2}, …, T_{n}}, respectively. The sample mean M of the differences between the observed values of t_{i} and c_{i} is the difference of the two sample means:
M = (sample mean of observed differences t_{i}−c_{i})
= ((T_{1}−C_{1}) + (T_{2}−C_{2})+ … + (T_{n}–C_{n}))/n
= (T_{1}+T_{2}+ … + T_{n})/n − (C_{1}+C_{2}+ … + C_{n})/n
= (sample mean of observed values of t_{i}) − (sample mean of observed values of c_{i}).
M is an unbiased estimator of μ, and if n is large, the normal approximation to its probability histogram will be accurate. The SE of M is the population standard deviation of the N values {d_{1}, d_{2}, …, d_{N}}, which we shall denote SD_{d}, divided by the square root of the sample size, n^{½}. Let sd denote the sample standard deviation of the n observed differences
(T_{i}−C_{i}), i=1, 2, …, n:
sd = ( ((T_{1}−C_{1}−M)^{2} + (T_{2}−C_{2}−M)^{2} + … + (T_{1}−C_{n}−M)^{2})/(n−1))^{½}
(recall that M is the sample mean of the observed differences). If the sample size n is large, sd is very likely to be close to SD(d), and so, under the null hypothesis,
Z = (M−μ_{0})/(sd/n^{½})
has expected value zero, and when n is large the probability histogram of Z can be approximated well by the normal curve. Thus we can use Z as the Z statistic in a z test of the null hypothesis that μ=μ_{0}. Under the alternative hypothesis that μ<μ_{0} (doctors on the average are paid less than an additional $1,000 per year under the new contract), the expected value of Z is less than zero, so we should use a left-tail z test. Under the alternative hypothesis μ≠μ_{0} (on average, the difference in average annual payments to doctors is not an increase of $1,000, but some other number instead), the expected value of Z could be positive or negative, so we would use a two-tail z test. Under the alternative hypothesis that μ<μ_{0} (on average, under the new contract, doctors are paid less than an additional $1,000 per year), the expected value of Z would be less than zero, so we should use a left-tail z test.
Consider two separate populations of numbers, with population means μ_{t} and μ_{c}, respectively. Let μ=μ_{t}−μ_{c} be the difference between the two population means. We would like to test the null hypothesis that μ=μ_{0} against the alternative hypothesis that μ>0. For example, let μ_{t} be the average annual payment by an HMO to doctors in the Los Angeles area, and let μ_{c} be the average annual payment by the same HMO to doctors in the San Francisco area. Then the null hypothesis with μ_{0}=0 is that the HMO pays doctors in the two regions the same amount annually, on the average; the alternative hypothesis is that the average annual payment by the HMO to doctors differs between the two areas. Suppose we draw a random sample of size n_{t} with replacement from the first population, and independently draw a random sample of size n_{c} with replacement from the second population. Let M_{t} and M_{c} be the sample means of the two samples, respectively, and let
M = M_{t} − M_{c}
be the difference between the two sample means. Because the expected value of M_{t} is μ_{t} and the expected value of M_{c} is μ_{c}, the expected value of M is
E(M) = E(M_{t} − M_{c}) = E(M_{t}) − E(M_{c}) = μ_{t} − μ_{c} = μ.
Because the two random samples are independent, M_{t} and −M_{c} are independent random variables, and the SE of their sum is
SE(M) = (SE^{2}(M_{t}) + SE^{2}(M_{c}))^{½}.
Let s_{t} and s_{c} be the sample standard deviations of the two samples, respectively. If n_{t} and n_{c} are both very large, the two sample standard deviations are quite likely to be close to the standard deviations of the two populations, and so
s_{t}/n_{t}^{½}
is likely to be close to SE(M_{t}), and
s_{c}/n_{c}^{½}
is likely to be close to SE(M_{c}). Therefore, the pooled estimate of the standard error
se_{p} = ( (s_{t}/n_{t}^{½})^{2} + (s_{c}/n_{c}^{½})^{2})^{½} = ( st_{2}/n_{t} + sc_{2}/n_{c} )^{½}
is likely to be close to SE(M). Under the null hypothesis, the statistic
has zero expected value, and its probability histogram is approximated well by the normal curve, so we can use it as the Z statistic in a z test.
Under the alternative hypothesis
μ = μ_{t} − μ_{c} > μ_{0},
the expected value of Z is greater than zero, so it is appropriate to use a right-tail z test.
If the alternative hypothesis were μ≠μ_{0}, under the alternative the expected value of Z could be greater than zero or less than zero, so it would be appropriate to use a two-tail z test. If the alternative hypothesis were μ<μ_{0}, under the alternative the expected value of Z would be less than zero, so it would be appropriate to use a right-tail z test.
The following exercises check that you can compute the z test for a population mean or a difference of population means. The exercises are dynamic: the data will tend to change when you reload the page.
For the nominal significance level of the z test for a population mean to be approximately correct, the sample size typically must be large. When the sample size is small, two factors limit the accuracy of the z test: the normal approximation to the probability distribution of the sample mean can be poor, and the sample standard deviation can be an inaccurate estimate of the population standard deviation, so se is not an accurate estimate of the SE of the test statistic Z. For nearly normal populations, defined in the next subsection, the probability distribution of the sample mean is nearly normal even when the sample size is small, and the uncertainty of the sample standard deviation as an estimate of the population standard deviation can be accounted for by using a curve that is broader than the normal curve to approximate the probability distribution of the (approximately) standardized test statistic. The broader curve is Student's t curve. Student's t curve depends on the sample size: The smaller the sample size, the more spread out the curve.
A list of numbers is nearly normally distributed if the fraction of values in any range is close to the area under the normal curve for the corresponding range of standard units—that is, if the list has mean μ and standard deviation SD, and for every pair of values a < b,
(the fraction of numbers in the list between a and b)
is approximately equal to
(the area under the normal curve between (a − μ)/SD and (b − μ)/SD ).
A list is nearly normally distributed if the normal curve is a good approximation to the histogram of the list transformed to standard units. The histogram of a list that is approximately normally distributed is (nearly) symmetric about some point, and is (nearly) bell-shaped.
No finite population can be exactly normally distributed, because the normal curve has positive area between every two distinct values—no matter how large or small the values. No population that contains only a finite number of distinct values can be exactly normally distributed, for the same reason. In particular, populations that contain only zeros and ones are not approximately normally distributed, so results for the sample mean of samples drawn from nearly normally distributed populations need not apply to the sample percentage of samples drawn from 0-1 boxes. Such results will be more accurate for the sample percentage when the population percentage is close to 50% than when the population percentage is close to 0% or 100%, because then the histogram of population values is more nearly symmetric.
Suppose a population is nearly normally distributed. Then a histogram of the population is approximately symmetric about the mean of the population. The fraction of numbers in the population within ±1 SD of the mean of the population is about 68%, the fraction of numbers within ±2 SD of the mean of the population is about 95%, and the fraction of numbers in the population within ±3 SD of the mean of the population is about 99.7%.
The following exercises check that you understand what it means for a list to be nearly normally distributed. The exercises are dynamic: the data tend to change when you reload the page.
Student's t curve is similar to the normal curve, but broader. It is positive, has a single maximum, and is symmetric about zero. The total area under Student's t curve is 100%. Student's t curve approximates some probability histograms more accurately than the normal curve does. There are actually infinitely many Student t curves, one for each positive integer value of the degrees of freedom. As the degrees of freedom increases, the difference between Student's t curve and the normal curve decreases.
Consider a population of N units labeled with numbers. Let μ denote the population mean of the N numbers, and let SD denote the population standard deviation of the N numbers. Let M denote the sample mean of a random sample of size n drawn with replacement from a population, and let s denote the sample standard deviation of the sample. The expected value of M is μ, and the SE of M is SD/n^{½}. Let
Z = (M − μ)/(SD/n^{½}).
Then the expected value of Z is zero, the SE of Z is 1, and if n is large enough, the normal curve is a good approximation to the probability histogram of Z. The more nearly normal the distribution of values in the population, the smaller n needs to be for the normal curve to be a good approximation to the distribution of Z. Consider the statistic
T = (M − μ)/(s/n^{½}),
which replaces SD by its estimated value (the sample standard deviation s). If n is large enough, s is very likely to be close to SD, so T will be close to Z; the normal curve will be a good approximation to the probability histogram of T; and we can use T as the Z statistic in a z test of hypotheses about μ.
For many populations, when the sample size is small—say less than 25, but the accuracy depends on the population—the normal curve is not a good approximation to the probability histogram of T. For nearly normally distributed populations, when the sample size is intermediate—say 25–100, but again this depends on the population—the normal curve is a good approximation to the probability histogram of Z, but not to the probability histogram of T, because of the variability of the sample standard deviation s from sample to sample, which tends to broaden the probability distribution of T (to make SE(T)>1).
For nearly normally distributed populations, Student's t curve is a better approximation to the probability histogram of T than the normal curve is. Student's t curve is broader and flatter than the normal curve, which accounts for the extra variability in the distribution of T. Actually, Student's t curve is not one curve: It is a family of curves, one for each value of the degrees of freedom d.f., 1, 2, …. In approximating the probability histogram of T, the appropriate value of d.f. to use is n −1, one less than the sample size. When d.f. is small, Student's t curve is much broader and flatter than the normal curve. As d.f. grows, Student's t curve gets closer and closer to the normal curve; for d.f. over 200, the two curves are essentially indistinguishable. For every value of the degrees of freedom, the total area under Student's t curve is 100%, the curve has a single peak at zero, and the curve is symmetric about zero. plots Student's t curve for various values of the degrees of freedom, and gives the area under Student's t curve over any interval.
When you first load this page, the degrees of freedom will be set to 25, and the region from −1.96 to 1.96 will be hilighted. The area under the normal curve between ±1.96 is 95%, but for Student's t curve with 25 degrees of freedom, the area is about 93.9%: Student's t curve with d.f.=25 is broader than the normal curve. Increase the degrees of freedom to 200; you will see that the Student t curve gets slightly narrower, and the area under the curve between ±1.96 is about 94.9%.
We define quantiles of Student t curves in the same way we defined quantiles of the normal curve: For any number a between 0 and 100%, the a quantile of Student's t curve with d.f.=d, t_{d,a}, is the unique value such that the area under the Student t curve with d degrees of freedom from minus infinity to t_{d,a} is equal to a. For example, t_{d,0.5} = 0 for all values of d. Generally, the value of t_{d,a} depends on the degrees of freedom d. The probability calculator allows you to find quantiles of Student's t curve.
We can use Student's t curve to construct approximate tests of hypotheses about the population mean μ when the population standard deviation is unknown, for intermediate values of the sample size n. The approach is directly analogous to the z test, but instead of using a quantile of the normal curve, we use the corresponding quantile of Student's t curve for the appropriate value of degrees of freedom. However, for the test to be accurate when n is small or intermediate, the distribution of values in the population must be nearly normal, which is a somewhat bizarre restriction: It can require a very large sample to detect that the population is not nearly normal, but if the sample is very large, we can use the z test instead of the t test. It is my opinion that the t test is over-taught and overused—because its assumptions are not verifiable in the situations where it is potentially useful.
Consider testing the null hypothesis that μ=μ_{0} using the sample mean M and sample standard deviation s of a random sample of size n drawn with replacement from a population that is known to have a nearly normal distribution. Define
T = (M − μ_{0})/(s/n^{½}).
Under the null hypothesis, if n is not too small, Student's t curve with n−1 degrees of freedom will be an accurate approximation to the probability histogram of T, so
P(T < t_{n−1,a}),
P(T > t_{n−1,1−a}),
P(|T| > t_{n−1,1−a}/2)
all are approximately equal to a. As we saw earlier in this chapter for the Z statistic, these three approximations give three tests of the null hypothesis μ=μ_{0} at approximate significance level a—a left-tail t test, a right-tail t test, and a two-tail t test:
To decide which t test to use, we can apply the same rule of thumb we used for the z test:
P-values for t tests are computed in much the same way as P-values for z tests. Let t be the observed value of T (the t score). In a left-tail t test, the P-value is the area under Student's t curve with n−1 degrees of freedom, from minus infinity to t. In a right-tail t test, the P-value is the area under Student's t curve with n−1 degrees of freedom, from t to infinity. In a two-tail t test, the P-value is the total area under Student's t curve with n−1 degrees of freedom between minus infinity and −|t| and between |t| and infinity.
There are versions of the t test for comparing two means, as well. Just like for the z test, the method depends on how the samples from the two populations are drawn. For example, if the two samples are paired (if we are sampling individuals labeled with two numbers and for each individual in the sample, we observe both numbers), we just base the t test on the sample mean of the paired differences and the sample standard deviation of the paired differences. Let μ_{1} and μ_{2} be the means of the two populations, and let
μ = μ_{1} − μ_{2}.
The T statistic to test the null hypothesis that μ=μ_{0} is
T = ( (sample mean of differences) − μ_{0} )/((sample standard deviation of differences)/n^{½}),
and the appropriate curve to use to find the rejection region for the test is Student's t curve with n−1 degrees of freedom, where n is the number of individuals (differences) in the sample.
Two-sample t tests for a difference of means using independent samples depend on additional assumptions, such as equality of the two population standard deviations; we shall not present such tests here. The following exercises check your ability to compute t tests. The exercises are dynamic: the data tend to change when you reload the page.
There is a deep connection between hypothesis tests about parameters, and confidence intervals for parameters. If we have a procedure for constructing a level 100%×(1−a) confidence interval for a parameter μ, then the following rule is a two-sided significance level a test of the null hypothesis that μ = μ_{0}:
reject the null hypothesis if the confidence interval does not contain μ_{0}.
Similarly, suppose we have an hypothesis-testing procedure that lets us test the null hypothesis that μ=μ_{0} for any value of μ_{0}, at significance level a. Define
A = (all values of μ_{0} for which we would not reject the null hypothesis that μ = μ_{0}).
Then A is a 100%×(1−a) confidence set for μ:
P(A contains the true value of μ) = 100%×(1−a).
(A confidence set is a generalization of the idea of a confidence interval: a 1−a confidence set for the parameter μ is a random set that has probability 1−a of containing μ. As is the case with confidence intervals, the probability makes sense only before collecting the data.) The set A might or might not be an interval, depending on the nature of the test. If one starts with a two-tail z test or two-tail t test, one ends up with a confidence interval rather than a more general confidence set.
The t test lets us test the hypothesis that the population mean μ is equal to μ_{0} at approximate significance level a using a random sample with replacement of size n from a population with a nearly normal distribution. If the sample size n is small, the actual significance level is likely to differ considerably from the nominal significance level. Consider a two-sided t test of the hypothesis μ=μ_{0} at significance level a. If the sample mean is M and the sample standard deviation is s, we would not reject the null hypothesis at significance level a if
|(M−μ_{0})/(s/n^{½})| = t_{n−1,1−a}/2.
We rearrange this inequality:
−t_{n−1,1−a}/2 ≤ (M−μ_{0})/(s/n^{½}) ≤ t_{n−1,1−a}/2
−t_{n−1,1−a}/2 × s/n^{½} ≤ M − μ_{0} ≤ t_{n−1,1−a}/2 × s/n^{½}
−M − t_{n−1,1−a}/2 × s/n^{½} ≤ − μ_{0} ≤ −M + t_{n−1,1−a}/2 × s/n^{½}
M + t_{n−1,1−a}/2 × s/n^{½} ≤ μ_{0} ≤ M − t_{n−1,1−a}/2 × s/n^{½}
M − t_{n−1,1−a}/2 × s/n^{½} ≤ μ_{0} ≤ M + t_{n−1,1−a}/2 × s/n^{½}.
That is, we would not reject the hypothesis μ = μ_{0} provided μ_{0} is in the interval
[M − t_{n−1,1−a}/2 × s/n^{½}, M + t_{n−1,1−a}/2 × s/n^{½}].
Therefore, that interval is a 100%−a confidence interval for μ:
P([M − t_{n−1,1−a}/2 × s/n^{½}, M + t_{n−1,1−a}/2 × s/n^{½}] contains μ) ~ 1−a.
The following exercise checks that you can use Student's t curve to construct a confidence interval for a population mean. The exercise is dynamic: the data tend to change when you reload the page.
In hypothesis testing, a Z statistic is a random variable whose probability histogram is approximated well by the normal curve if the null hypothesis is correct: If the null hypothesis is true, the expected value of a Z statistic is zero, the SE of a Z statistic is approximately 1, and the probability that a Z statistic is between a and b is approximately the area under the normal curve between a and b. Suppose that the random variable Z is a Z statistic. If, under the alternative hypothesis, E(Z)<0, the appropriate z test to test the null hypothesis at approximate significance level a is the left-tailed z test: Reject the null hypothesis if Z<z_{a}, where z_{a} is the a quantile of the normal curve. If, under the alternative hypothesis, E(Z)>0, the appropriate z test to test the null hypothesis at approximate significance level a is the right-tailed z test: Reject the null hypothesis if Z>z_{1−a}. If, under the alternative hypothesis, E(Z)≠0 but could be greater than 0 or less than 0, the appropriate z test to test the null hypothesis at approximate significance level a is the two-tailed z test: reject the null hypothesis if |Z|>z_{1−a/2}. If, under the alternative hypothesis, E(Z)=0, a z test probably is not appropriate—consult a statistician. The exact significance levels of these tests differ from a by an amount that depends on how closely the normal curve approximates the probability histogram of Z.
Z statistics often are constructed from other statistics by transforming approximately to standard units, which requires knowing the expected value and SE of the original statistic on the assumption that the null hypothesis is true. Let X be a test statistic; let E(X) be the expected value of X if the null hypothesis is true, and let se be approximately equal to the SE of X if the null hypothesis is true. If X is a sample sum of a large random sample with replacement, a sample mean of a large random sample with replacement, or a sum or difference of independent sample means of large samples with replacement,
Z = (X−E(X))/se
is a Z statistic.
Consider testing the null hypothesis that a population percentage p is equal to the value p_{0} on the basis of the sample percentage φ of a random sample of size n with replacement. Under the null hypothesis, E(φ)=p_{0} and
SE(φ) = (p_{0}×(1−p_{0}))^{½}/n^{½},
and if n is sufficiently large (say n×p>30 and n×(1−p)>30, but this depends on the desired accuracy), the normal approximation to
Z = (φ−p_{0})/((p_{0} × (1−p_{0}))^{½}/n^{½})
will be reasonably accurate, so Z can be used as the Z statistic in a z test of the null hypothesis p=p_{0}.
Consider testing the null hypothesis that a population mean μ is equal to the value μ_{0}, on the basis of the sample mean M of a random sample of size n with replacement. Let s denote the sample standard deviation. Under the null hypothesis, E(M)=μ_{0}, and if n is large,
SE(M)=SD/n^{½}~s/n^{½},
and the normal approximation to
will be reasonably accurate, so Z can be used as the Z statistic in a z test of the null hypothesis μ=μ_{0}.
Consider a population of N individuals, each labeled with two numbers. The ith individual is labeled with the numbers c_{i} and t_{i}, i=1, 2, …, N. Let μ_{c} be the population mean of the N values {c_{1}, …, c_{N}} and let μ_{t} be the population mean of the N values {t_{1}, …, t_{N}}. Let μ=μ_{t}−μ_{c} be the difference between the two means. Consider testing the null hypothesis that μ=μ_{0} on the basis of a paired random sample of size n with replacement from the population: that is, a random sample of size n is drawn with replacement from the population, and for each individual i in the sample, c_{i} and t_{i} are observed. This is equivalent to testing the hypothesis that the population mean of the N values {(t_{1}−c_{1}), …, (t_{N}−c_{N})} is equal to μ_{0}, on the basis of the sample random sample of size n drawn with replacement from those N values. Let M_{t} be the sample mean of the n observed values of t_{i} and let M_{c} be the sample mean of the n observed values of c_{i}. Let sd denote the sample standard deviation of the n observed differences {(t_{i}−c_{i})}. Under the null hypothesis, the expected value of
(M_{t}−M_{c}) is μ_{0}, and if n is large,
SE(M_{t}−M_{c})~sd/n^{½},
and the normal approximation to the probability histogram of
Z = ((M_{t}−M_{c}−μ_{0})/(sd/n^{½})
will be reasonably accurate, so Z can be used as the Z statistic in a z test of the null hypothesis that μ_{t}−μ_{c}=μ_{0}.
Consider testing the hypothesis that the difference (μ_{t}−μ_{c}) between two population means, μ_{c} and μ_{t}, is equal to μ_{0}, on the basis of the difference (M_{t}−M_{c}) between the sample mean M_{c} of a random sample of size n_{c} with replacement from the first population and the sample mean M_{t} of an independent random sample of size n_{t} with replacement from the second population. Let s_{c} denote the sample standard deviation of the sample of size n_{c} from the first population and let s_{t} denote the sample standard deviation of the sample of size n_{t} from the second population. If the null hypothesis is true,
E(M_{t}−M_{c})=μ_{0},
and if n_{c} and n_{t} are both large,
SE(M_{t}−M_{c}) ~ (st_{2}/n_{t} + sc_{2}/n_{c})^{½}
Z = (M_{t}−M_{c}−μ_{0})/(st_{2}/n_{t} + sc_{2}/n_{c})^{½}
A list of numbers is nearly normally distributed if the fraction of numbers between any pair of values, a<b, is approximately equal to the area under the normal curve between (a−μ)/SD and (b−μ)/SD, where μ is the mean of the list and SD is the standard deviation of the list. Student's t curve with d degrees of freedom is symmetric about 0, has a single bump centered at 0, and is broader and flatter than the normal curve. The total area under Student's t curve is 1, no matter what d is; as d increases, Student's t curve grows closer and closer to the normal curve. Let M be the sample mean of a random sample of size n with replacement from a population with mean μ and a nearly normal distribution, and let s be the sample standard deviation of the random sample. For moderate values of n (n<100 or so), Student's t curve approximates the probability histogram of (M−μ)/(s/n^{½}) better than the normal curve does, which can lead to an approximate hypothesis test about μ that is more accurate than the z test: Consider testing the null hypothesis that the mean μ of a population with a nearly normal distribution is equal to μ_{0} from a random sample of size n with replacement. Let
T=(M−μ_{0})/(s/n^{½}),
where M is the sample mean and s is the sample standard deviation. The tests reject the null hypothesis if T<t_{n−1,a} (left-tail t test), reject the null hypothesis if T>t_{n−1,1−a} (right-tail t test), and reject the null hypothesis if |T|>t_{n−1,1−a}/2 (two-tail t test) all have approximate significance level a. How close the nominal significance level a is to the actual significance level depends on the distribution of the numbers in the population, the sample size n, and a. The same rule of thumb for selecting whether to use a left, right, or two-tailed z test (or not to use a z test at all) works to select whether to use a left, right, or two-tailed t test: If, under the alternative hypothesis, E(T)<0, use a left-tail test. If, under the alternative hypothesis, E(T)>0, use a right-tail test. If, under the alternative hypothesis, E(T) could be less than zero or greater than zero, use a two-tail test. If, under the alternative hypothesis, E(T)=0, consult an expert. Because the t test differs from the z test only when the sample size is small, and from a small sample it is not possible to tell whether the population has a nearly normal distribution, the t test should be used sparingly, if ever.
A 1−a confidence set for a parameter μ is like a 1−a confidence interval for a parameter μ: It is a random set of values that has probability 1−a of containing the true value of μ. The difference is that the set need not be an interval. There is a deep duality between hypothesis tests about a parameter μ and confidence sets for μ. Given a procedure for constructing a 1−a confidence set for μ, the rule reject the null hypothesis that μ=μ_{0} if the confidence set does not contain μ is a significance level a test of the null hypothesis that μ=μ_{0}. Conversely, given a family of significance level a hypothesis tests that allow one to test the hypothesis that μ=μ_{0} for any value of μ_{0}, the set of all values μ_{0} for which one would not reject the null hypothesis that μ=μ_{0} is a 1−a confidence set for μ.