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 we 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 of 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.
where the random error, , follows a distribution model.
For the grinding problem, since Ha states that , large values of will provide evidence against H0 and in favor of Ha. Therefore any value of as large or larger than the observed value will provide as much or more evidence against H0 and in favor of Ha as does the observed test statistic value. Thus, the p-value is , where P0 is the probability computed under the assumption that H0 is true: that is, .
To calculate the p-value, we standardize the test statistic by subtracting its mean (remember we're assuming H0 is true, so we take ) and dividing by its estimated standard error:
If H0 is true, the result will have a tn-1=t149 distribution.
Putting this all together, the p-value is
In all examples we'll look at, H0 will be simple (i.e. will state that the parameter has a single value.) as opposed to compound. Alternative 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
|H0:||=||0.75 ( simple)|
|Ha:||>||0.75 ( compound, one-sided)|
Suppose in the grinding problem that management wanted to see if the mean diameter was off target. Then appropriate hypotheses would be:
|Ha:||0.75 (compound, two-sided)|
In this case, evidence against H0 and in favor of Ha is provided by both large and small values of .
To compute the p-value of the two-sided test, we first compute the standardized test statistic t, and its observed value, t*:
Recall that under H0, . By the symmetry of the t distribution about 0, we compute the p-value as
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 Appendix 6.1, p. 346, with me!
First, check out Appendix 6.1, p. 347, with me!
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
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 H0, , so the p-value is
Now, for illustration, we will use the large-sample test. This is valid since np0 and n(1-p0) both equal 75>10.
The observed standardized test statistic is
The approximate p-value is then
We assume that there are n1 measurements from population 1 generated by the C+E model
and n2 measurements from population 2 generated by the C+E model
We 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 n1=n2 and by looking at the pairwise differences, Di=Y1,i-Y2,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 Di.
The manufacturer of a new warmup bat wants to test its efficacy. To do so, it selects a random sample of 12 baseball players from among a larger number who volunteer to try the bat. For each player, company researchers compute D, the difference between the player's test year average and his previous year's average. Assuming that these differences follow a C+E model, they want to test
The data (found in SASDATA.BATTING) are:
An inspection of the differences shows no evidence of nonnormality or outliers, so we proceed with the test. For these data, , and sd=0.0092. Then , so the observed value of the standardized test statistic is
resulting in a p-value
Let and denote the sample means from populations 1 and 2, S12 and S22 the sample variances. The point estimator of , is . We will test
|Versus one of|
If the population variances are equal (), then we estimate by the pooled variance estimator
The estimated standard error of is then given by
Then, if H0 is true,
has a tn1+n2-2 distribution.
Suppose t(p)* is the observed value of t(p). Then the p-value of the test of H0 versus Ha- is
versus Ha+ is
and versus is
If , then the standardized test statistic
approximately follows a distribution model, where is the largest integer less than or equal to
If t(ap)* denotes the observed value of t(ap), the p-values for H0 versus Ha-, Ha+ and , respectively, are , and .
A company buys cutting blades used in its manufacturing process from two suppliers. In order to decide if there is a difference in blade life, the lifetimes of 10 blades from manufacturer 1 and 13 blades 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.BLADE2)
The experimenters want to test
The experimenters generated histograms and normal quantile plots of the two data sets and found no evidence of nonnormality or outliers. The estimate of is .
So the standard error estimate of is
Therefore, t(p)*=-16.52/9.44=-1.75, with 21 degrees of freedom. So , , and the p-value for this problem is .
The observed value of the standardized test statistic is t(ap)*=-16.52/9.92=-1.67. The degrees of freedom is computed as the greatest integer less than or equal to
Therefore, , ,
and the p-value for this problem is
The results for the two t-tests are not much different.
and are observations from two independent populations. The estimator of p1-p2 is
We wish to test a null hypothesis that the two population proportions differ by a known amount ,
Case 1: 0
Suppose H0 is p1-p2=0. Then, let p=p1=p2 denote the common value of the two population proportions. If H0 is true, the variance of equals p(1-p)/n1 and that of equals p(1-p)/n2. 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 H0 is true.
Case 2: 0
If , the (by now) standard reasoning gives the standardized test statistic
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 pf denotes the proportion of all female and pm the proportion of all male college students who agree or strongly agree with this statement.
Since Yf=24, 200-Yf=176, Ym=26, and 100-Ym=74 all exceed 10, we may use the normal approximation.
The point estimate of pf-pm is
and the estimate of the common value of pf and pm under H0 is .
From this, we obtain , , and , this last being the p-value we want.
The estimated standard error of pf-pm is
and a p-value of .
Steps, illustrated using grinding example:
and under H0 it has a N(0,1) distribution.
meaning H0 will be rejected if and only if the observed value of Z is greater than or equal to 1.645.
which falls in the critical region, so H0 is rejected in favor of Ha.
In the grinding example, the power is
This document was generated using the LaTeX2HTML translator Version 97.1 (release) (July 13th, 1997)
Copyright © 1993, 1994, 1995, 1996, 1997, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
The command line arguments were:
latex2html -split 0 lect6.
The translation was initiated by Joseph D Petruccelli on 10/27/1999