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The Central Limit Theorem for Proportions
Section 7.4
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Objectives Construct the sampling distribution for a sample proportion
Use the Central Limit Theorem to compute probabilities for sample proportions
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Construct the sampling distribution for a sample proportion
Objective 1 Construct the sampling distribution for a sample proportion
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Sample and Population Proportion
In a population, the proportion who have a certain characteristic is called the population proportion. The population proportion is denoted by 𝑝. In a simple random sample of 𝑛 individuals, let 𝑥 be the number in the sample who have the characteristic. The sample proportion is 𝑝 = 𝑥 𝑛 . Example: A computer retailer wants to estimate the proportion of people in her city who own laptop computers. She cannot survey everyone in the city, so she draws a sample of 100 people and surveys them. It turns out that 35 out of the 100 people in the sample own laptops. The sample proportion is 𝑝 = The proportion of people in the entire city who own laptops is the population proportion, 𝑝.
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Sampling Distribution of 𝑝
If several samples are drawn from a population, they are likely to have different values for 𝑝 . Because the value of 𝑝 varies each time a sample is drawn, 𝒑 is a random variable, and it has a probability distribution. The probability distribution of 𝑝 is called the sampling distribution of 𝑝 .
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An Example of a Sampling Distribution
Consider tossing a fair coin five times. This produces a sample of size 𝑛 = 5, where each item in the sample is either a head or a tail. The proportion of times the coin lands on heads is the sample proportion 𝑝 . The probability that it lands heads each time is 0.5. Therefore, the population proportion of heads is 𝑝 = 0.5. All 32 possible samples are presented below. The table displays all possible samples of size 5 and their sample proportion 𝑝 . The mean of all of values of 𝑝 is 𝜇 𝑝 =0.5 and the standard deviation of all values of 𝑝 is 𝜎 𝑝 = Next, we compare these values to the population proportion (0.5) and the sample size (5). Sample 𝑝 TTTTT 0.0 THTHH 0.6 HTHHT TTTTH 0.2 THHTT 0.4 HTHHH 0.8 TTTHT THHTH HHTTT TTTHH THHHT HHTTH TTHTT THHHH HHTHT TTHTH HTTTT HHTHH TTHHT HTTTH HHHTT TTHHH HTTHT HHHTH THTTT HTTHH HHHHT THTTH HTHTT HHHHH 1.0 THTHT HTHTH
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Mean and Standard Deviation of a Sampling Distribution
In the previous table, the mean of the sampling distribution is 𝜇 𝑝 =0.5, which is the same as the population proportion, 𝑝=0.5. This relation always holds. The standard deviation of the sampling distribution is 𝜎 𝑝 = It is not immediately obvious how this is related to the population proportion 𝑝 and the sample size 𝑛. Note, however, that 𝜎 𝑝 =0.2236= 0.5 1−0.5 5 = 𝑝 1−𝑝 𝑛 The mean of the sampling distribution of 𝒑 is denoted by 𝝁 𝒑 and equals the population proportion: 𝝁 𝒑 =𝒑 The standard deviation of the sampling distribution of 𝒑 is denoted by 𝝈 𝒑 , is: 𝝈 𝒑 = 𝒑 𝟏−𝒑 𝒏
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Example – Sampling Distribution
The soft-drink cups at a certain fast-food restaurant have tickets attached to them. Customers peel off the tickets to see whether they win a prize. The proportion of tickets that are winners is 𝑝 = A total of 𝑛 = 70 people purchase soft drinks between noon and 1:00 PM on a certain day. Let 𝑝 be the proportion that win a prize. Find the mean and standard deviation of 𝑝 . Solution: The population proportion is 𝑝 = 0.25 and the sample size is 𝑛 = 70. The mean of 𝑝 is: 𝜇 𝑝 =𝑝=0.25 The standard deviation of 𝑝 is: 𝜎 𝑝 = 𝑝 1−𝑝 𝑛 = − =
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Probability Histogram for a Sampling Distribution of 𝑝
The probability histogram for the sampling distribution of 𝑝 for the proportion of heads in five tosses of a fair coin is presented. We can see that the distribution is reasonably well approximated by a normal curve. The experiment is conducted again but with 50 tosses of a fair coin. As we increased the number of tosses, the sampling distribution of 𝑝 is more closely approximated by a normal curve. When 𝑝 = 0.5, the sampling distribution of 𝑝 is somewhat close to normal even for a small sample size like 𝑛 = 5. When 𝑝 is close to 0 or close to 1, a larger sample size is needed before the distribution of 𝑝 is close to normal. A common rule of thumb is that the distribution may be approximated with a normal curve whenever 𝒏𝒑 and 𝒏(𝟏− 𝒑) are both at least 10.
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The Central Limit Theorem for Proportions
Let 𝒑 be the sample proportion for a sample of size 𝒏 from a population with population proportion 𝒑. If 𝒏𝒑≥𝟏𝟎 and 𝒏 𝟏−𝒑 ≥𝟏𝟎, then the distribution of 𝒑 is approximately normal, with mean 𝝁 𝒑 =𝒑 and standard deviation 𝝈 𝒑 = 𝒑 𝟏−𝒑 𝒏 . The Central Limit Theorem for Proportions
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Examples – The Central Limit Theorem
Example 1: A sample of size 20 is drawn from a population with population proportion 𝑝 = 0.7. Is it appropriate to use the normal distribution to find probabilities for 𝑝 ? Solution: No. 𝑛𝑝= =14 and 𝑛 1−𝑝 = =6. Since 𝑛(1−𝑝) is not at least 10, we cannot be certain that the distribution of 𝑝 is approximately normal. Example 2: A sample of size 55 is drawn from a population with population proportion 𝑝 = 0.8. Is it appropriate to use the normal distribution to find probabilities for 𝑝 ? Yes, since 𝑛𝑝= =44 and 𝑛 1−𝑝 = =11 are both at least 10, the distribution of 𝑝 is approximately normal.
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Objective 2 Use the Central Limit Theorem to compute probabilities for sample proportions
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Example 1 – Using the Central Limit Theorem
According to a Harris poll taken in September 2013, chocolate is the favorite ice cream flavor for 27% of Americans. If a sample of 100 Americans is taken, what is the probability that the sample proportion of those who prefer chocolate is greater than 0.30? Solution: We first check the assumptions. 𝑛𝑝=100(0.27) =27≥10 and 𝑛 1−𝑝 =100 1−0.27 =73≥10. We may use the normal curve. We compute 𝜇 𝑝 and 𝜎 𝑝 𝜇 𝑝 =𝑝=0.27 and 𝜎 𝑝 = 𝑝 1−𝑝 𝑛 = − = Find the area under the normal curve. The probability that the sample proportion of those who prefer chocolate is greater than 0.30 is [Answer using tables is ]
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Example 2 – Using the Central Limit Theorem
In the 2012 U.S. presidential election, 51% of voters voted for Barack Obama. If a sample of 75 voters were polled, would it be unusual if less than 40% of them had voted for Barack Obama? Solution: We first check the assumptions. 𝑛𝑝=75(0.51) =38.25≥10 and 𝑛 1−𝑝 =75 1−0.51 =36.75 ≥10. We may use the normal curve. We compute 𝜇 𝑝 and 𝜎 𝑝 𝜇 𝑝 =𝑝=0.51 and 𝜎 𝑝 = 𝑝 1−𝑝 𝑛 = − = Find the area under the normal curve. The probability that the sample proportion of those who voted for Barack Obama is less than 0.40 is It would be unusual for the sample proportion to be less than [Answer using tables is ]
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You Should Know… The notation for sample and population proportions
How to construct the sampling distribution for a sample proportion How to find the mean and standard deviation of the sampling distribution of 𝑝 How to use the Central Limit Theorem for Proportions to compute probabilities for sample proportions
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