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Examples

Example 1. Suppose that an ad for Western Michigan University claims that `a majority of WMU students come from outside the Southwest Michigan area'. A sample of 25 students revealed that only 9 out of 25, or 36%, come from outside Southwest Michigan. Can this just be chance error, or is this significant evidence that the advertisement is wrong?



Solution: The data consists of a sample percentage (or proportion); i.e. 36% of 25 students come from outside Southwest Michigan. Therefore, the ESTIMATE is a either a sample percentage or proportion (we will choose a proportion). At the end of the test, the conclusion will be one of two things: the data will provide evidence either for or against against the claim `a majority of WMU students come from outside the Southwest Michigan area'. Therefore the competing hypotheses are $p \geq .50$ versus p < .50. The hypothesis with the equal sign is null, so we have $H_0: p \geq .50$. The H0 Value is .50.

1.
Hypotheses: $H_0: p \geq .50$ versus H1: p < .50
2.
Test Statistic: $z = \frac{\hat{p}-.50}{\sqrt{.50(1-.50)/25}} =
\frac{ .36 -.50}{\sqrt{.50(1-.50)/25}} = -1.40$
3.
P-value: Presuming H0 is true, the likelihood of chance variation yielding a z-statistic as low as -1.40 is .08.

4.
Conclusion: Since P-value > .05, the observed sample value $\hat{p}=.36$ is not significantly different from .50. Hence, we do not reject $H_0: p \geq .50$. The sample does not provide enough evidence to prove the ad wrong.

Example 2 Is there "grade inflation" in WMU? A random sample of 100 student records from 10 years ago yields a sample average GPA of 2.90 with a standard deviation of .40. A random sample of 100 current students today yields a sample average of 2.98 with a standard deviation of .45. Is the difference between 2.90 and 2.98 just chance variation, or a statistically significant increase?

Solution: The evidence consists of two sample averages. The ESTIMATE of interest is the difference: 2.98-2.90=.08 (i.e average today minus average 10 years ago). At the end of the test, the conclusion will be one of two things: the data either support or provide evidence against the hypothesis of 'grade inflation'. Therefore the competing hypotheses are 'average grade today is higher than 10 years ago' or 'not'. In terms of the population averages, we write these as $\mu_2-\mu_1 > 0$, or $\mu_2-\mu_1 \leq 0$. The hypothesis with the equal sign is null, so we have $H_0: \mu_2-\mu_1 \leq 0$. The H0 Value is 0.

1.
Hypotheses: $H_0: \mu_2-\mu_1 \leq 0$ versus $H_1: \mu_2-\mu_1 > 0$
2.
Test Statistic: $z = \frac{(\overline{X}_2-\overline{X}_1) - 0 }
{\sqrt{\mbox{SE}_1^2 + \mbox{SE}_2^2}} =
\frac{ (2.98-2.90)-0}{\sqrt{(.045)^2 + (.040^2)}} = 1.33$
3.
P-value: Presuming H0 is true, a z-statistic as large as 1.33 results from chance variation 9% of the time (P-value=.09).

4.
Conclusion: Since P-value > .05, the observed increase in average GPA is not statistically significant, and may be explained by chance variation. Do not reject H0.

If the sample sizes were small and the two variances were equal, a pooled-SD t-test would have been the more appropriate test. We illustrate the procedure here. The SD's for the two samples S1=.40 and S2=.45 are pooled into one SD:

\begin{displaymath}S_p=\sqrt{ \frac{ (n_1-1)S_1^2+(n_2-1)S_2^2}{ n_1+n_2-2 }} =
...
...t{ \frac{ (100-1)(.45^2) + (100-1)(.40^2)}{100+100-2}} = .426.
\end{displaymath}

The SE's for the two averages are $\mbox{SE}_1=.426/\sqrt{100}=.0426$ and $\mbox{SE}_2=.426/\sqrt{100}=.0426$. Now we conduct the test.

1.
Hypotheses: $H_0: \mu_2-\mu_1 \leq 0$ versus $H_1: \mu_2-\mu_1 > 0$
2.
Test Statistic: $t = \frac{(\overline{X}_2-\overline{X}_1) - 0 }
{\sqrt{\mbox{SE}_1^2 + \mbox{SE}_2^2}} =
\frac{(2.98-2.90)-0}{\sqrt{(.0426)^2 + (.0426^2)}} = 1.33$
3.
P-value: Presuming H0 is true, a t-statistic with 198 degrees of freedom will by chance be as large as 1.33 around 9% of the time (P-value=.09).

4.
Conclusion: Since P-value > .05, the observed increase in average GPA is not statistically significant, and may be explained by chance variation. Do not reject H0.

Example 3. The paired t-test   Table  8.2 shows stock prices for six selected food service companies. Is there a significant difference between average Jan. 2002 and Jan. 2003 stock prices?


  
Table 8.2: Stock Prices of Selected Food Companies
\begin{table}
\begin{tex2html_preform}\begin{verbatim}Jan Jan
Company 2002 2003...
....16 28.07 3.09
SD 7.10 8.95 5.74\end{verbatim}\end{tex2html_preform}\end{table}

Solution:

At first glance, the evidence looks like it consists of two averages 31.12 and 28.07. However, remember that these two averages are not independent, because the data consists of two measurements on only ONE sample, rather than two separate samples. (Remember the Paired-t confidence interval?) The correct analysis here takes the differences between the two measurements, as shown on the table. If average 2002 and 2003 prices are significantly different, then the difference column will have an average that is significantly different from 0.

Ignoring the original two columns and focusing instead on the difference column, the evidence consists of one average. At the end of the test, the conclusion will be one of two things: the average difference is either significantly different from 0 or not ($\mu\neq 0$ versus $\mu=0$). Suppose we want a test that can detect either a positive or negative change in average stock prices, then we can choose to do a two-tailed test. (Of course, a one-tailed test that detects change in only one direction can also be done.)

1.
Hypotheses: $H_0: \mu=0$ versus $H_1: \mu \neq 0$
2.
Test Statistic: $t = \frac{\overline{X} - 0 }{S/\sqrt{n}} =
\frac{ 3.09-0}{5.74/\sqrt{6}} = 1.32$
3.
P-value: By chance alone, a t-statistic with 5 d.f. may get as large as 1.32 on either side of 0 around 24% of the time (P-value=.244).
4.
Conclusion: Since P-value > .05, the observed change in stock prices is not statistically significant, and may be explained by chance variation. Do not reject H0.

Example 4. Has retention rate at WMU been changing? Suppose that a random sample of 200 entering students in 1989 showed 74% were still enrolled 3 years later. Another random sample of 200 entering students in 1999 showed that 66% were still enrolled 3 years later. Is this drop of 8% statistical evidence of changing retention rate, or can it be just chance error (i.e luck of the draw in the students selected for the samples)?

Solution:

The evidence consists of two sample proportions , or more precisely, the difference between two sample proportions (.74-.66=.08). At the end of the test, the conclusion will be one of two things: the data either support or provide evidence against 'changing retention rates'. Since an increase or decrease will be interpreted as a change (this was discussed at the start of the study), we will conduct a two-tailed test. The competing hypotheses are 'change' versus 'no change'. In terms of the population, these may be written as $p_1-p_2\ neq 0$ versus p1-p2=0. The latter is assigned to H0 because it contains the equal sign. The H0 Value is 0.

1.
Hypotheses: H0: p1-p2=0 versus $H_1: p_1-p_2 \neq 0$
2.
Test Statistic: $z = \frac{(\hat{p_1}-\hat{p_2}) - 0 }
{\sqrt{\mbox{SE}_1^2 + \mbox{SE}_2^2}} =
\frac{ (.74-.66) -0}{\sqrt{(.031)^2 + (.033^2)}} = 1.78$
3.
P-value: Presuming H0 is true, a z-statistic as extreme as 1.78 on either side of 0 results from chance variation 7.5% of the time (P-value=.075).

4.
Conclusion: Since P-value > .05, the observed change in retention rate is not statistically significant, and may be explained by chance variation. Do not reject H0.


next up previous contents index
Next: Type I and Type Up: Test of Significance involving Previous: Generalizing Tests of Significance

2003-09-08