Example 16-1 Section
As is often the case, we'll motivate the methods of this lesson by way of example. Specifically, we'll return to the question posed in the introduction to this lesson.
Suppose the Penn State student population is 60% female and 40%, male. Then, if a sample of 100 students yields 53 females and 47 males, can we conclude that the sample is (random and) representative of the population? That is, how "good" do the data "fit" the assumed probability model of 60% female and 40% male?
Answer
Testing whether there is a "good fit" between the observed data and the assumed probability model amounts to testing:
\(H_0 : p_F =0.60\)
\(H_A : p_F \ne 0.60\)
Now, letting \(Y_1\) denote the number of females selected, we know that \(Y_1\) follows a binomial distribution with n trials and probability of success \(p_1\). That is:
\(Y_1 \sim b(n,p_1)\)
Therefore, the expected value and variance of \(Y_1\) are, respectively:
\(E(Y_1)=np_1\) and \(Var (Y_1) =np_1(1-p_1)\)
And, letting \(Y_2\) denote the number of males selected, we know that \(Y_2 = n − Y_1\) follows a binomial distribution with n trials and probability of success \(p_2\). That is:
\(Y_2 = n-Y_1 \sim (b(n,p_2)=b(n,1-p_1)\)
Therefore, the expected value and variance of \(Y_2\) are, respectively:
\(E(Y_2)=n(1-p_1)=np_2\) and \(Var(Y_2)=n(1-p_1)(1-(1-p_1))=np_1(1-p_1)=np_2(1-p_2)\)
Now, for large samples (\(np_1 ≥ 5\) and \(n(1 − p_1) ≥ 5)\), the Central Limit Theorem yields the normal approximation to the binomial distribution. That is:
\(Z=\dfrac{(Y_1-np_1)}{\sqrt{np_1(1-p_1)}}\)
follows, at least approximately, the standard normal N(0,1) distribution. Therefore, upon squaring Z, we get that:
\(Z^2=Q_1=\dfrac{(Y_1-np_1)^2}{np_1(1-p_1)}\)
follows an approximate chi-square distribution with one degree of freedom. Now, we could stop there. But, that's not typically what is done. Instead, we can rewrite \(Q_1\) a bit. Let's start by multiplying \(Q_1\) by 1 in a special way, that is, by multiplying it by \(\left( \left( 1 − p_1 \right) + p_1 \right)\):
\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1(1-p_1)} \times ((1-p_1)+p_1)\)
Then, distributing the "1" across the numerator, we get:
\(Q_1=\dfrac{(Y_1-np_1)^2(1-p_1)}{np_1(1-p_1)}+\dfrac{(Y_1-np_1)^2p_1}{np_1(1-p_1)}\)
which simplies to:
\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(Y_1-np_1)^2}{n(1-p_1)}\)
Now, taking advantage of the fact that \(Y_1 = n − Y_2\) and \(p_1 = 1 − p_2\), we get:
\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(n-Y_2-n(1-p_2))^2}{np_2}\)
which simplifies to:
\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(-(Y_2-np_2))^2}{np_2}\)
Just one more thing to simplify before we're done:
\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(Y_2-np_2)^2}{np_2}\)
In summary, we have rewritten \(Q_1\) as:
\( Q_1=\sum_{i=1}^{2}\dfrac{(Y_i-np_i)^2}{np_i}=\sum_{i=1}^{2}\dfrac{(\text{OBSERVED }-\text{ EXPECTED})^2}{EXPECTED} \)
We'll use this form of \(Q_1\), and the fact that \(Q_1\) follows an approximate chi-square distribution with one degree of freedom, to conduct the desired hypothesis test.
Before we return to solving the problem posed by our example, a couple of points are worthy of emphasis.
- First, \(Q_1\) has only one degree of freedom, since there is only one independent count, namely \(Y_1\). Once \(Y_1\) is known, the value of \(Y_2 = n − Y_1\) immediately follows.
- Note that the derived approach requires the Central Limit Theorem to kick in. The general rule of thumb is that the expected number of successes must be at least 5 (that is, \(np_1 ≥ 5\)) and the expected number of failures must be at least 5 (that is, \(n(1−p_1) ≥ 5\)).
- The statistic \(Q_1\) will be large if the observed counts are very different from the expected counts. Therefore, we must reject the null hypothesis \(H_0\) if \(Q_1\) is large. How large is large? Large is determined by the values of a chi-square random variable with one degree of freedom, which can be obtained either from a statistical software package, such as Minitab or SAS or from a standard chi-square table, such as the one in the back of our textbook.
- The statistic \(Q_1\) is called the chi-square goodness-of-fit statistic.
Example 16-2 Section
Suppose the Penn State student population is 60% female and 40%, male. Then, if a sample of 100 students yields 53 females and 47 males, can we conclude that the sample is (random and) representative of the population? Use the chi-square goodness-of-fit statistic to test the hypotheses:
\(H_0 : p_F =0.60\)
\(H_A : p_F \ne 0.60\)
using a significance level of \(\alpha = 0.05\).
Answer
The value of the test statistic \(Q_1\) is:
\(Q_1=\dfrac{(53-60)^2}{60}+\dfrac{(47-40)^2}{40}=2.04\)
We should reject the null hypothesis if the observed number of counts is very different from the expected number of counts, that is if \(Q_1\) is large. Because \(Q_1\) follows a chi-square distribution with one degree of freedom, we should reject the null hypothesis, at the 0.05 level, if:
\(Q_1 \ge \chi_{0.05, 1}^{2}=3.84\)
Because:
\(Q_1=2.04 < 3.84\)
we do not reject the null hypothesis. There is not enough evidence at the 0.05 level to conclude that the data don't fit the assumed probability model.
As an aside, it is interesting to note the relationship between using the chi-square goodness of fit statistic \(Q_1\) and the Z-statistic we've previously used for testing a null hypothesis about a proportion. In this case:
\( Z=\dfrac{0.53-0.60}{\sqrt{\frac{(0.60)(0.40)}{100}}}=-1.428 \)
which, you might want to note that, if we square it, we get the same value as we did for \(Q_1\):
\(Q_1=Z^2 =(-1.428)^2=2.04\)
as we should expect. The Z-test for a proportion tells us that we should reject if:
\(|Z| \ge 1.96 \)
Well, again, if we square it, we should see that that's equivalent to rejecting if:
\(Q_1 \ge (1.96)^2 =3.84\)
And not surprisingly, the P-values obtained from the two approaches are identical. The P-value for the chi-square goodness-of-fit test is:
\(P=P(\chi_{(1)}^{2} > 2.04)=0.1532\)
while the P-value for the Z-test is:
\(P=2 \times P(Z>1.428)=2(0.0766)=0.1532\)
Identical, as we should expect!