AP Statistics / Mr. Hansen
4/16/2002

Name: ________________________

1.

 (a)

The choice of whether to run a 1-sided or 2-sided test depends on the research question of interest. Are you expecting to gather evidence to show that juniors are less likely to favor lowering the drinking age, or are you simply interested in ascertaining whether there is a difference of any type? You could make a case both ways, but a 2-sided approach is probably what would be more standard here. After performing the VHA(S)TPC steps, you will find that there is no evidence (z = 1.215, phat₁ = 0.6, phat₂ = 0.58, n₁ = 2200, n₂ = 1500, p = 0.224) that there is any difference between the true proportions of juniors and seniors that favor lowering the drinking age. Be sure to check your assumptions!

(b)

Show your work. Conclusion: We are 98% confident that the true difference between proportions is between –1.83 percentage points and +5.83 percentage points, where negative means juniors are more likely than seniors to favor lowering the drinking age, and positive means seniors are more likely than juniors.

2.

This is a one-sample t situation (matched pairs).

(a)

After showing work, we can say that we are 95% confident that the true mean change between week 1 and week 2 is between 4.20 and 5.80 inches. Equivalently, we could say that we are 95% confident that the true mean change in height is 5 ± 0.8 inches.

(b)

After performing VHA(S)TPC, we might conclude that there is good evidence (xbar = 5, t = 2.877, df = 8, p = 0.0103) that the true mean growth between week 1 and week 2 exceeds 4 inches. However, the evident skewness of the growth column should give us pause. The t test is not robust in the face of such skewness when the sample is this small (see p.606), and we might well prefer to conduct a nonparametric test instead. In class we discussed the sign test, in which we analyze how likely it is that (in this case) all 9 out of 9 plants show growth of more than 4 inches under the assumption H₀: m = 4, where m denotes the true mean growth. Using the B(9, 0.5) distribution, we see that the p-value is a most impressive 0.00195, which gives strong evidence indeed. The advantage of nonparametric tests is that they make no assumptions about the population distributions, and such tests are very useful in cases like this where the normality of the population of growth figures is in doubt.

(c)

For a question regarding 5 inches (which also happens to be the sample mean), common sense tells you that a sampling distribution centered on 5, which is what H₀ would claim, will produce values close to 5 a high percentage of the time. No, there is no evidence that the true mean growth exceeds 5 inches.

Under the sign test, which is not required for AP or even for our test, but which is again very useful because of the possible nonnormality of the growth population, we would be looking at the likelihood that chance alone would produce 3 or more (out of 9) growth figures that were on the high side of 5 under the assumption H₀: m = 5. By calc., p = 0.910, which is far from being statistically significant. Clearly, there is no evidence that the true mean growth exceeds 5 inches.

3.

The key words are "assuming that everything else remains unchanged."

(a)

s.e.: no change
p-value: doubles

(b)

s.e.: decreases by a factor of 3
p-value: decreases, though the exact amount of change varies with the problem

(c)

s.e.: increases by a factor of 9 (by definition of s.e.!)
p-value: increases, though the exact amount of change varies with the problem

4.

With the change, this is now a two-sample t situation. Unfortunately, the TI-83 does not allow us to perform this as a "button-pusher" because the TI-83 compares two means using a null hypothesis where the means are assumed to differ by 0, and here we have a null hypothesis in which the means differ by 4. The trick is to create a new list (perhaps called H2MOD) that equals the old week 2 height data minus 4. In other words, knock all the week 2 values down by 4. After performing VHA(S)TPC by manual procedures, and checking against the TI-83, we find good evidence (t = –2.415, df = 15.985, p = 0.014) that the true difference of means exceeds 4 inches. Be sure to check assumptions! We are in good shape this time, because there is ample biological reason to expect that the populations of 1-week-old and 2-week-old plant heights are normal, and the two-sample t procedures are quite robust when the sample sizes are approximately equal. NQPs of the height data considered individually show no startling departures from normality.

Why would we ever perform the experiment as in #2? In other words, why would we use matched pairs (one-sample) when the two-sample t procedures in #4 apparently work better? The reason is that we would see, especially for larger samples, that the matched pairs design is superior in removing the lurking variable of individual plant-to-plant variation. The matched pairs design allows us to focus on the growth figures alone. When we look at the plants from two independent populations, we will undoubtedly have other lurking variables as well. For example, the 2-week-old plants may have had different temperature or humidity conditions during their first week of growth that we could not possibly replicate in the other sample (because those plants weren’t even alive yet). The matched pairs design removes those lurking variables.