A. Closest proportion to .1000 is .1003; look in Column C→ z -score = −1.28 X = 90 + (−1.28)(12)
= 90 − 15.36 = 74.64
B. Closest proportion to .7500 is .7486; look in Column B → z -score = +.67 X = 74 + (+.67)(8)
= 74 + 5.36 = 79.36
VI. Locating Extreme Raw Scores in a Distribution X = 100 + (−1.96)(15) X = 100 + (+1.96)(15)
= 100 − 29.40 = 100 + 29.40 = 70.60 = 129.40
Answers to “Your Turn!” Problems (continued)
Additional Practice Problems
Answers to odd numbered problems are in Appendix C at the end of the book.
1. The Table of the Normal Distribution Curve is made up of four columns. Briefly describe what each column contains.
2. Locate the probabilities associated with the following:
a. p(z > +1.25) b. p(z > −2.10) c. p(z < −1.87) d. p(z < +2.68)
3. Locate the z-scores associated with the following:
a. lowest 40% of the normal distribution b. highest 56% of the normal distribution c. highest 12.5% of the normal distribution
4. What z-score separates the highest 35% of the normal distribution from the lowest 65%?
5. What z-score separates the lowest 25% of the normal distribution from the highest 75%?
6. Given a normal distribution with a μ = 82 and σ = 8, determine the following proportions:
a. p(X > 93) b. p(X > 72) c. p(X < 63) d. p(X < 104)
7. Given a normal distribution with a μ = 37 and σ = 4, what proportion of the population can be expected to score
a. greater than 40?
b. greater than 25?
c. less than 32?
d. less than 39?
8. Given a normal distribution with a μ = 112 and σ = 10, what percentage of the popula- tion can be expected to score
a. lower than 100?
b. higher than 90?
9. A normal distribution has a μ = 50 and σ = 5.
a. What proportion of the population can be expected to obtain scores between 42 and 53?
b. What is the proper notation for the above question and answer?
10. A normal distribution has a μ = 84 and σ = 6.
a. What proportion of the population can be expected to obtain scores between 72 and 89?
b. What is the proper notation for the above question and answer?
11. A normal distribution has a μ = 57 and σ = 8. What is the probability that a randomly selected score would fall between 59 and 71?
12. A normal distribution has a μ = 36 and σ = 3. What is the probability that a randomly selected score would fall between 28 and 32?
13. Given a normal distribution with a μ = 40 and σ = 4, what raw score separates the upper 68% of the distribution from the lower 32%?
14. Given a normal distribution with a μ = 96 and σ = 12, what raw score separates the lower 56% of the distribution from the upper 44%?
15. A normally shaped distribution has a μ = 71 and σ = 6. What raw scores are so extreme that they will be achieved by only 9% of the population?
16. A normally shaped distribution has a μ = 34 and σ = 4. What raw scores are so extreme that they will be achieved by only 3% of the population?
107
WHAT IS A SAMPLING DISTRIBUTION?
As we have discussed previously, inferential statistics are used to predict characteristics of a population based on sample data. We use sample statistics to estimate population param- eters. However, statistics vary from one sample to the next. If we drew ten random samples from the same population, measured them on some dependent variable, and computed their means, we could easily end up with ten different means. How do we know if our sample statistics are accurate estimates of population parameters? We determine this by using a sam- pling distribution, a theoretical probability distribution that represents a statistic (such as a mean) for all possible samples of a given size from a population of interest. Let us examine the sampling distribution of means to illustrate.
Imagine a population of interest and, from that population, we draw a random sample of a certain size. We will use n = 60. We measure our sample on some characteristic, calcu- late a mean, and record it. We then repeat the process of drawing additional samples of the same size from the same population and recording their means. We do this until all possible samples of size n = 60 have been drawn from the population and their means recorded. We then plot all of these means in a frequency distribution, called the sampling distribution of the means.
Of course, this is not done in actuality. The sampling distribution of means is a theoreti- cal distribution, and we only imagine theoretical distributions. We do not actually construct them. Neither does it make sense to undertake such an endeavor. If we could study every possible sample in a population, we could more easily just calculate the mean for the popu- lation to begin with. However, as you know, we usually do not have access to populations.
Fortunately, statisticians have been able to identify the important characteristics of sampling distributions so that we do not have to actually construct them. These characteristics are described in the central limit theorem.
THE CENTRAL LIMIT THEOREM
The central limit theorem states that (1) as the size of the sample (n) increases, the shape of the sampling distribution of means approximates the shape of the normal distribution; (2) the mean of the sampling distribution of means will equal the population mean (μ); and (3) the standard deviation will equal σ / n. These ideas are discussed more fully below.
7
Sampling Distribution of Means
• Shape. Even if the shape of the population distribution from which a sample is drawn is not normal, if the sample size is large (i.e., n = 60 or more), the sampling distribution itself will be normal in shape.
• Mean. If we literally added up all of the means from our sampling distribution of means and divided by whatever number constituted all possible samples, we would obtain the mean of the sampling distribution of means, called the expected value, which would be equal to μ. Remember that sample means are unbiased estimates of the population mean. They do not systematically either overestimate or under- estimate the population mean. Errors in one direction are balanced by errors in the other direction. Thus, when all sample means are averaged, the resulting value should be equal to the population mean (μ). Furthermore, any particular sample mean, in actual application, is expected to be close in value to the population mean, hence the expected value.
• Standard Deviation. The standard deviation of the sampling distribution of means is called the standard error of the mean (σM), or simply standard error. It is called error because, while any particular sample mean is expected to be close in value to the popu- lation mean, it is not expected to be exact. Most of the sample means will vary some- what in value from μ just because of chance, or random sampling error. The standard error (σM) represents the amount that a sample mean (M) is expected to vary from the population mean (μ).
The formula for the standard error of the mean is a function of the population standard deviation and sample size, as shown below:
M n
Note the fact that as the size of the sample increases, the amount of standard error decreases. For instance, the standard deviations shown below are identical, but the sample sizes are different. Look what happens to the size of the standard error when n goes from 25 to 100.
Give: σ = 16 σ = 16
n = 25 n = 100
M 16 25
3 2. M 16
100 1 6.
Thus, larger samples generally produce less sampling error.
PROBABILITIES, PROPORTIONS, AND PERCENTAGES OF SAMPLE MEANS The sampling distribution of means can be used to determine the probabilities, proportions, and percentages associated with particular sample means. This is similar to our previous use of the normal distribution curve to determine these figures for particular raw scores. But because we are now dealing with sample means rather than raw scores, the formulas involved are modified as follows:
z M
M
= −à
σ and M z M
the baseline and z-scores are in standard error units rather than standard deviation units. In the sampling distribution below, μ = 80, σ = 14, and n = 49. Using the appropriate formula, we find the standard error (σM) to be 2. Thus,
M 16 49 2
We can now use the sampling distribution of means to answer questions about probabili- ties, proportions, and percentages.
For Example
1. Given the previous normally shaped distribution with μ = 80 and σm = 2, A. What is the probability that an obtained sample mean will be below 81?
• Convert the given sample mean to a z-score:
z M
M
81 80 2 .50
• Then, locate in the z-table the probability associated with sample means below a z-score of +.50 (column B):
p M( 81) .6915
Notice that the correct notation now specifies a sample mean (M) rather than a raw score (X).
B. What proportion of the sample means can be expected to have a value greater than 83?
z M
M
83 80
2 1 50. p M( <83)=.0668
2. Given a normally shaped population distribution with μ = 95 and σ = 5, a sample size of n = 25 is drawn at random.
A. The probability is .05 that the mean of the sample will be above what value?
• First, find the standard error: M 5 25 = 1
• Next, find the z-score associated with the closest proportion above .0500, which is 1.65.
• Finally, use the formula to convert the z-score to a sample mean:
M
z M 95 1 65 1 96 65
. .
B. What range of sample means would be expected to occur in the middle of the dis- tribution 70% of the time?
Tip!
We need to identify the z-scores associated with the extreme 30% in the tails (i.e., .1500 at each end).
M z
M M
M
95 1 04 1
93 96 . .
95 1 04 1 96 04
. .
Thus, 70% of the time, we can expect to draw sample means that fall within the range of 93.96 and 96.04.
Tip!
When working out problems that deal with the sampling dis- tribution of means, be sure to pay attention to what kind of values you are being asked to find: proportions, probabilities, or sample means. this will determine which formula to use.
Your Turn!