Sampling Distribution Homework Answers
Question Set #1
1. The observations are random numbers between 50-100 and have an equal likelihood of
occurrence. There is no reason to expect that one number would occur more or less
frequently than another.
2. Chart 1 graphs the frequency of the random observations. It is relatively flat.
Chart 2 graphs the frequency of the average of two random observations. It is steeper
in the middle. This result is intuitive because averaging two observations will tend to
reduce the number of extreme occurrences. For example, a value of 50 is just as likely as
any other value if we select one random variable. However, if we average two variables,
you would have to draw the number 50 twice in a given sample. Any other observation
would result in an average greater than 50. Thus, the extreme values will occur with less
frequency than the values surrounding the mean.
3. Probably not. The max and min values are near the extremes of our sample
possibilities. The standard error is also large. This suggests that we will be very likely to
retrieve sample means that are not representative of the population mean.
Question Set #2
1. In this problem, the grand mean is the equivalent of a sample mean with 2000
observations. The grand mean closely approximates the population mean. The grand
means are all approx. the same. The probability of the grand mean is approximately zero,
because we are using continuous data. The grand mean will not equal the population
mean precisely.
2. They are never precisely zero, but are close in many instances. This uses the same
logic as question 1. The average sampling error is also close to zero, as it is simply the
population mean minus the grand mean.
3. It measures the dispersion of sample means around the population mean. It will tell
you how much the error deviates from 0 on average.
4. The sampling error is simply the sample mean minus the population mean. This means
that the average sampling error is equal to the average sample mean-the population mean.
Therefore, the deviations of the sample means and the sample errors from their respective
averages are equal. Hence, their standard deviations are equal.
5. Because the observations in chart 3 are normally distributed, whereas the observations
in chart 1 are randomly distributed with an equal likelihood of occurrence.
7. Charts 2 and 4 differ again because the data are normally distributed. The values
around the mean have a greater chance of occurring. Therefore, the data are more closely
concentrated around the mean.
Set #3
1. As n increases, the sample mean will be more likely converge to the population mean
and reduce the frequency of values further from the mean. This results in a lower
standard error.
2. They are all close because in all 3 instances the grand mean is an average of 2000
samples. While not exactly equal to the population mean, they should be close.
3. Chart 5 has a larger sample size. As n increases, the sample mean will be more likely
converge to the population mean and reduce the frequency of values further from the
mean.
4. As n increases, the sample mean will be more likely converge to the population mean
and reduce the frequency of values further from the mean. Thus, the max and min sample
means will also converge to the population mean as n becomes larger.
5. Hypothetically, if you have a value for x of 74.5 you calculate the Z-score using
x
. Your standard error would be 10/sqrt(5) or 4.47. So z equals -.5/4.47 or
Z
X
-.112.
To get the probability that z is less than -.112 enter normsdist(-.112) which equals .455.
To get the probability z is greater enter 1-normsdist(-.112) which equals .645.