Recall from chapter 5 that statistical inference is the use of a subset of a population (the sample) to draw conclusions about the entire population. In chapter 5 you studied one kind of inference called estimation. In this chapter, we study a second kind of inference called hypothesis testing.
The validity of inference is related to the way the data are obtained, and to the stationarity of the process producing the data.
One stage of a manufacturing process involves a manually-controlled grinding operation. Management suspects that the grinding machine operators tend to grind parts slightly larger rather than slightly smaller than the target diameter, 0.75 inches while still staying within specification limits, which are 0.75 0.01 inches. To verify their suspicions, they sample 150 within-spec parts. We will use this example to illustrate the components of a statistical hypothesis testing problem.
H_{0}: | = | 0.75 | |
H_{a}: | > | 0.75 |
For the grinding problem, since H_{a} states that , large values of will provide evidence against H_{0} and in favor of H_{a}. Therefore any value of as large or larger than the observed value will provide as much or more evidence against H_{0} and in favor of H_{a} as does the observed test statistic value. Thus, the p-value is , where P_{0} is the probability computed under the assumption that H_{0} is true: that is, .
To calculate the p-value, we standardize the test statistic by subtracting its mean (remember we're assuming H_{0} is true, so we take ) and dividing by its estimated standard error:
If H_{0} is true, the result will have a t_{n-1}=t_{149} distribution. Putting this all together, the p-value is
In all examples we'll look at, H_{0} will be simple (i.e. will state that the parameter has a single value.) as opposed to compound. Alternate hypotheses will be one-sided (that the parameter be larger the null value, or smaller than the null value) or two-sided (that the parameter not equal the null value).
In the grinding example, we had
H_{0}: | = | 0.75 ( simple) | |
H_{a}: | > | 0.75 ( compound, one-sided) |
H_{0}: | = | 0.75 (simple) | |
H_{a}: | 0.75 (compound, two-sided) |
In this case, evidence against H_{0} and in favor of H_{a} is provided by both large and small values of .
Statistical hypothesis testing is modeled on scientific investigation. The two hypotheses represent competing scientific hypotheses.
For this reason the null hypothesis is given favored treatment.
Check out the appendix 6.1, p. 346, with me!
First, check out the appendix 6.1, p. 347, with me!
Example:
Back at the grinding operation, management has decided on another characterization of the scientific hypothesis that ``there is a tendency to grind the parts larger than the target diameter.'' They decide to make inference about p, the population proportion of in-spec parts with diameters larger than the target value. The hypotheses are
H_{0}: | p | = | 0.5 |
H_{a}: | p | > | 0.5 |
The datum is Y, the number of the 150 sampled parts with diameters larger than the target value.
Of the 150 parts, y^{*}=93 (a proportion 0.62) have diameters greater than the target value 0.75.
We will first perform an exact test of these hypotheses. Under H_{0}, , so the p-value is
Now, for illustration, we will use the large-sample test. This is valid since np_{0} and n(1-p_{0}) both equal 75>10.
The observed standardized test statistic is
The approximate p-value is thenWe assume that there are n_{1} measurements from population 1 generated by the C+E model
and n_{2} measurements from population 2 generated by the C+E modelWe want to compare and .
Sometimes each observation from population 1 is paired with another observation from population 2. For example, each student may take a pre- and post-test. In this case n_{1}=n_{2} and by looking at the pairwise differences, D_{i}=Y_{1,i}-Y_{2,i}, we transform the two population problem to a one population problem for C+E model , where and . Therefore, an hypothesis test for the difference is obtained by performing a one sample hypothesis test for based on the differences D_{i}.
In 1993 the National League expanded by adding the Florida and Colorado teams. Many experts predicted that this expansion would dilute the quality of pitching and inflate team batting statistics. Others pointed out that the batting level would also decline, and that the result would be little or no difference. To assess who was right, we have collected the team batting averages for 1992 and 1993 for all 12 teams that were in the league in 1992. We assume that each team's batting average each year follows a C+E model centered about an overall (and unknown) league average.
Since most personnel on a team stay the same from one year to the next, we feel that paired comparisons are appropriate. Thus, we compute the differences in 1993 and 1992 averages for each team.
Thus, for each team, we will compute D, the difference between the 1993 and 1992 team batting average. We will test the hypotheses
H_{0}: | = | 0 | |
H_{a}: | > | 0 |
The data (found in SASDATA.NLAVG923) are:
TEAM | AVG92 | AVG93 | DIFFAVG |
ATL | 0.254 | 0.262 | 0.008 |
CHI | 0.254 | 0.270 | 0.016 |
CIN | 0.260 | 0.264 | 0.004 |
HOU | 0.246 | 0.267 | 0.021 |
LA | 0.248 | 0.261 | 0.013 |
MON | 0.252 | 0.257 | 0.005 |
NY | 0.235 | 0.248 | 0.013 |
PHI | 0.253 | 0.274 | 0.021 |
PIT | 0.255 | 0.267 | 0.012 |
SD | 0.255 | 0.252 | -0.003 |
SF | 0.244 | 0.276 | 0.032 |
STL | 0.262 | 0.272 | 0.010 |
An inspection of the differences shows no evidence of nonnormality or outliers, so we proceed with the test. For these data, , and s_{d}=0.0092. Then , so the observed value of the standardized test statistic is
resulting in a p-valueLet and denote the sample means from populations 1 and 2, S_{1}^{2} and S_{2}^{2} the sample variances. The point estimator of , is . We will test
H_{0}: | = | ||
Versus one of | |||
H_{a-}: | < | , | |
H_{a+}: | < | , | |
. |
Equal Variances
If the population variances are equal (), then we estimate by the pooled variance estimator
The estimated standard error of is then given byThen, if H_{0} is true,
has a t_{n1+n2-2} distribution. Suppose t^{(p)*} is the observed value of t^{(p)}. Then the p-value of the test of H_{0} versus H_{a-} isp_{-}=P(t_{n1+n2-2}<t^{(p)*}),
versus H_{a+} isp^{+}=P(t_{n1+n2-2}>t^{(p)*}),
and versus isIf , then the standardized test statistic
approximately follows a distribution model, where is the largest integer less than or equal to andIf t^{(ap)*} denotes the observed value of t^{(ap)}, the p-values for H_{0} versus H_{a-}, H_{a+} and ,respectively, are , and .
A company buys grinding wheels used in its manufacturing process from two suppliers. In order to decide if there is a difference in wheel life, the lifetimes of 10 wheels from manufacturer 1 and 13 wheels from manufacturer 2 used in the same application are compared. A summary of the data shows the following (units are hours): (The data are in SASDATA.GRIND2)
Manufacturer | n | s | |
1 | 10 | 118.4 | 26.9 |
2 | 13 | 134.9 | 18.4 |
H_{0}: | = | 0 | |
H_{a}: | 0 |
The results for the two t-tests are not much different.
and are observations from two independent populations. The estimator of p_{1}-p_{2} is
We wish to test a null hypothesis that the two population proportions differ by a known amount ,
H_{0}: | p_{1}-p_{2} | = | , |
H_{a+}: | p_{1}-p_{2} | > | |
H_{a-}: | p_{1}-p_{2} | < | |
p_{1}-p_{2} |
Case 1:
Suppose H_{0} is p_{1}-p_{2}=0. Then, let p=p_{1}=p_{2} denote the common value of the two population proportions. If H_{0} is true, the variance of equals p(1-p)/n_{1} and that of equals p(1-p)/n_{2}. This implies the standard error of equals
Since we don't know p, we estimate it using the data from both populations:
The estimated standard error of is then
The standardized test statistic is then
which has a N(0,1) distribution if H_{0} is true.Case 2:
If ,the (by now) standard reasoning gives the standardized test statistic
where is the estimated standard error of .In a recent survey on academic dishonesty 24 of the 200 female college students surveyed and 26 of the 100 male college students surveyed agreed or strongly agreed with the statement ``Under some circumstances academic dishonesty is justified.'' Suppose p_{f} denotes the proportion of all female and p_{m} the proportion of all male college students who agree or strongly agree with this statement.
H_{0}: | p_{f} - p_{m} | = | 0 |
H_{a}: | p_{f} - p_{m} | 0 |
Since Y_{f}=24, 200-Y_{f}=176, Y_{m}=26, and 100-Y_{m}=74 all exceed 10, we may use the normal approximation.
The point estimate of p_{f} - p_{m} is
and the estimate of the common value of p_{f} and p_{m} under H_{0} is . Thus, and From this, we obtain , , and , this last being the p-value we want.H_{0}: | p_{f} - p_{m} | = | -0.10 |
H_{a}: | p_{f} - p_{m} | < | -0.10 |
The estimated standard error of p_{f} - p_{m} is
=0.05,
which gives and a p-value of .