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Authors: Deborah Rumsey
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As a result of the CLT, the distribution of X can be non-normal or even unknown and as long as n is large enough, you can still find approximate probabilities for using the standard normal ( Z ) distribution and the process described earlier. (That is, convert to a Z -value and find probabilities using the Z -table (Table A-1, appendix).)
When you do have to use the CLT to find a probability for you need to say that your answer is an approximation and that you've got a large enough n to proceed because of the CLT. (If n is not large enough for the CLT, you use the t -distribution in many cases — see Chapter 9.)
The Sampling Distribution of the Sample Proportion
The Central Limit Theorem (CLT) doesn't apply only to sample means. You can also use it with other statistics, including sample proportions. The population proportion , p , is the proportion of individuals in the population that have a certain characteristic of interest based on a binomial random variable (see Chapter 4). The sample proportion , denoted , is the proportion of individuals in the sample that have that same characteristic of interest. The sample proportion is the number of individuals in the sample who have that characteristic of interest divided by the total sample size ( n ). If you take a sample of 100 students and find 60 freshman, the sample proportion for freshman is 60/100 = 0.60. This section examines the sampling distribution of all possible sample proportions, , from samples of size n from a population.
The sampling distribution of has these properties:
Its mean is the population proportion, denoted by p .
Its standard error is . (Note that because n is
in the denominator, standard error decreases as n increases.)
Its shape is approximately normal, provided that the sample size is large enough. This is due to the CLT. That means you can use the normal distribution to find probabilities for . (See Chapter 5 for more.)
The larger the sample size ( n ), the closer the distribution of sample proportions is to a normal distribution.
How large is large enough for the CLT to work for categorical data? Most statisticians agree that both np and n (1 - p ) should be greater than or equal to 10. You want the average number of successes ( np ) and the average number of failures n( 1 - p) to be at least 10. (Note the second condition involves n (1 - p ), not np (1 - p ), the variance of the binomial distribution.)
What proportion of students need math help?
Suppose you want to know what proportion of incoming college students would like help in math. A student survey accompanies the ACT test each year, and one of the questions is whether the student would like some help with math skills. Assume (through past research) that 38% of the students taking the ACT respond yes. That means p = 0.38 in this case.
The original data has a binomial distribution where success = would like help. The yes responses ( p ) and no responses (1 - p ) for the population are shown in Figure 6-4 as a bar graph. (See Chapter 3 for more on bar graphs.)
Figure 6-4: Population percentages for responses to ACT math-help question.
Now take all possible samples of size 1,000 from this population and find the proportion in each who said they needed math help. The distribution of these sample proportions is in Figure 6-5. It has an approximate normal distribution with mean p = 0.38 and standard error equal to
(or about 1.5%). This approximation is valid because the two conditions for the CLT are met: 1) np = 1,000(0.38) = 380 (which is at least 10); and 2) n (1 - p ) = 1,000(0.62) = 620 (also at least 10).
Figure 6-5: Proportion of students responding yes to ACT math-help question for samples of size 1,000.
Finding Probabilities for
For the ACT test example, suppose it's reported that 0.38 or 38% of all the students taking the ACT test would like math help. Suppose you took a random sample of 1,000 students. What is the chance that more than 40 percent
When you do have to use the CLT to find a probability for you need to say that your answer is an approximation and that you've got a large enough n to proceed because of the CLT. (If n is not large enough for the CLT, you use the t -distribution in many cases — see Chapter 9.)
The Sampling Distribution of the Sample Proportion
The Central Limit Theorem (CLT) doesn't apply only to sample means. You can also use it with other statistics, including sample proportions. The population proportion , p , is the proportion of individuals in the population that have a certain characteristic of interest based on a binomial random variable (see Chapter 4). The sample proportion , denoted , is the proportion of individuals in the sample that have that same characteristic of interest. The sample proportion is the number of individuals in the sample who have that characteristic of interest divided by the total sample size ( n ). If you take a sample of 100 students and find 60 freshman, the sample proportion for freshman is 60/100 = 0.60. This section examines the sampling distribution of all possible sample proportions, , from samples of size n from a population.
The sampling distribution of has these properties:
Its mean is the population proportion, denoted by p .
Its standard error is . (Note that because n is
in the denominator, standard error decreases as n increases.)
Its shape is approximately normal, provided that the sample size is large enough. This is due to the CLT. That means you can use the normal distribution to find probabilities for . (See Chapter 5 for more.)
The larger the sample size ( n ), the closer the distribution of sample proportions is to a normal distribution.
How large is large enough for the CLT to work for categorical data? Most statisticians agree that both np and n (1 - p ) should be greater than or equal to 10. You want the average number of successes ( np ) and the average number of failures n( 1 - p) to be at least 10. (Note the second condition involves n (1 - p ), not np (1 - p ), the variance of the binomial distribution.)
What proportion of students need math help?
Suppose you want to know what proportion of incoming college students would like help in math. A student survey accompanies the ACT test each year, and one of the questions is whether the student would like some help with math skills. Assume (through past research) that 38% of the students taking the ACT respond yes. That means p = 0.38 in this case.
The original data has a binomial distribution where success = would like help. The yes responses ( p ) and no responses (1 - p ) for the population are shown in Figure 6-4 as a bar graph. (See Chapter 3 for more on bar graphs.)
Figure 6-4: Population percentages for responses to ACT math-help question.
Now take all possible samples of size 1,000 from this population and find the proportion in each who said they needed math help. The distribution of these sample proportions is in Figure 6-5. It has an approximate normal distribution with mean p = 0.38 and standard error equal to
(or about 1.5%). This approximation is valid because the two conditions for the CLT are met: 1) np = 1,000(0.38) = 380 (which is at least 10); and 2) n (1 - p ) = 1,000(0.62) = 620 (also at least 10).
Figure 6-5: Proportion of students responding yes to ACT math-help question for samples of size 1,000.
Finding Probabilities for
For the ACT test example, suppose it's reported that 0.38 or 38% of all the students taking the ACT test would like math help. Suppose you took a random sample of 1,000 students. What is the chance that more than 40 percent
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