Inference and Sampling Distributions

Solutions to practice problems

1. The sample range would always be smaller than the population range, unless you just happened to get the shortest and tallest person in your sample, in which case the sample range would equal the population range. But the sample range could never exceed the population range, so the average sample range over many samples would be less than the true population range (parameter). Thus, the sample range is a biased estimator of the population range: it tends to underestimate the parameter.

2. 100 is a fairly large sample size, so the distribution of the sample proportion should be approximately normal. (The numbers provided pass the np>10 and n(1-p)>10 tests.) The mean of the sample proportion is equal to the population proportion, 0.29, and the standard deviation of the sample proportion is equal to root(p(1-p)/n), which is approximately 0.045. So the sample proportion of smokers is approximately normally distributed with mean 0.29 and standard deviation 0.045. (Incidentally, a consequence of this is that an observed sample proportion between about 0.20 and 0.38 would occur about 95% of the time.)

3. The mean of x-bar is always equal to the population mean: 2.1 inches, in this case. The standard error of x-bar is always equal to the population standard deviation divided by root n. In this case, that's 1.9/root(45)=0.28 inches. The distribution of x-bar is approximately normal because 45 is a fairly large sample size. From the TI-83, we can compute normalcdf(0, 1.7, 2.1, 0.28)=0.077. Or we can compute the z-score associated with 1.7 inches and look in the standard normal table. z=(1.7-2.1)/0.28=-1.43. P(z<-1.43)=0.0764.

Three remarks follow. First, despite the way the answer was written here, it is a good idea not to round off numbers until the final answer or until they must be rounded off for a table reference. So a better calculator entry would have been normalcdf(0,1.7, 2.1, 1.9/root(45)).

Second, note that although the calculator entry has been for the range from 0 to 1.7 inches (no sample mean should be less than 0 inches tall), the normal distribution technically has a tail extending out to minus infinity. But if we're using the normal approximation appropriately, then there should be almost 0 probability out beyond zero, and in this case that's true.

Third, the AP exam grading rubrics tend not to like "calculatorspeak", so students should get in the habit of writing their equations in more standard notation. However, it is not at all discouraged to have students using their calculators to do the normal calculations; no z-table is required at all. So a perfectly acceptable written answer to this question could include the statements: E(x-bar)=mu=2.1. SE(x-bar)=sd/root(n)=1.9/root(45)=0.28. 45 is a large sample size, so the CLT applies. x-bar has a normal distribution with mean 2.1 and standard error (or deviation) 0.28. P(x-bar<1.7)=0.077.

4. Although the mean and standard deviation are given in the problem, the shape of the distribution is not. You can’t determine the probability of being less than 1.7 inches tall without knowing the shape of the distribution. The reason this was possible in question 3—even without knowing the population distribution shape—was that we were averaging together many observations, and the Central Limit Theorem assures us that the sampling distribution of the sample mean will be approximately normal regardless of the distribution of the population.