Central Limit Theorem for Proportions - Comprehensive Notes

Central Limit Theorem for Proportions

Population Proportion

• The proportion of a population with a certain characteristic is the population proportion, denoted by $P$. Central to statistical inference and hypothesis testing.

Sample Proportion

• In a simple random sample of $N$ individuals, let $X$ be the number in the sample with the characteristic.
• The sample proportion, denoted by $\hat{P}$, is calculated as $\hat{P} = \frac{X}{N}$. It is an estimate of the population proportion.

Example

• A retailer surveys 100 people and finds 35 own laptops.
• The sample proportion is $\hat{P} = \frac{35}{100} = 0.35$.
• The population proportion, $P$, is the proportion of all people in the city who own laptops.

Sampling Distribution of $\hat{P}$

• If several samples are drawn, the values of $\hat{P}$ are likely to vary.
• $\hat{P}$ is a random variable and has a probability distribution.
• The probability distribution of $\hat{P}$ is called the sampling distribution of $\hat{P}$.

Example: Tossing a Fair Coin

• Toss a fair coin five times (sample size $n = 5$).
• The proportion of times the coin lands on heads is the sample proportion $\hat{P}$.
• The probability of heads is 0.5, so the population proportion is $P = 0.5$.
• There are $2^5 = 32$ possible samples.
• The table displays all possible samples of size five and their sample proportion $\hat{P}$.
• The mean of all values of $\hat{P}$ is $\mu_{\hat{P}} = 0.5$.
• The standard deviation of all values of $\hat{P}$ is $\sigma_{\hat{P}} = 0.2236$.

Mean and Standard Deviation of the Sampling Distribution

• The mean of the sampling distribution, $\mu<em>{\hat{P}}$, equals the population proportion $P$. $\mu</em>{\hat{P}} = P$
• The standard deviation of the sampling distribution, $\sigma<em>{\hat{P}}$, is given by: $\sigma</em>{\hat{P}} = \sqrt{\frac{P(1-P)}{n}}$

Example: Soft Drink Cups

• The proportion of winning tickets is $P = 0.25$.
• $n = 70$ people purchase soft drinks.
• The mean of $\hat{P}$ is $\mu_{\hat{P}} = P = 0.25$.
• The standard deviation of $\hat{P}$ is $\sigma_{\hat{P}} = \sqrt{\frac{0.25(1-0.25)}{70}} = 0.0518$.

Probability Histogram

• The probability histogram for the sampling distribution of $\hat{P}$ for the proportion of heads in five tosses of a fair coin is presented.
• The distribution is reasonably well approximated by a normal curve.
• As the number of tosses increases, the sampling distribution of $\hat{P}$ is more closely approximated by a normal curve.
• When $P = 0.5$, the sampling distribution of $\hat{P}$ is somewhat close to normal even for a small sample size like $n = 5$.
• When $P$ is close to 0 or 1, a larger sample size is needed before the distribution of $\hat{P}$ is close to normal.
• A common rule of thumb is that the distribution may be approximated with a normal curve whenever $n \times P \geq 10$ and $n \times (1-P) \geq 10$.

Central Limit Theorem for Proportions

• Let $\hat{P}$ be the sample proportion for a sample of size $n$ from a population with population proportion $P$.
• If $n \times P \geq 10$ and $n \times (1-P) \geq 10$, then the distribution of $\hat{P}$ is approximately normal with:
  • Mean: $\mu_{\hat{P}} = P$
  • Standard Deviation: $\sigma_{\hat{P}} = \sqrt{\frac{P(1-P)}{n}}$

Examples

• 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 $\hat{P}$?
  • $n \times P = 20 \times 0.7 = 14$
  • $n \times (1-P) = 20 \times 0.3 = 6$
  • Since $n \times (1-P)$ is not at least 10, we cannot be certain that the distribution of $\hat{P}$ is approximately normal.
• 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 $\hat{P}$?
  • $n \times P = 55 \times 0.8 = 44$
  • $n \times (1-P) = 55 \times 0.2 = 11$
  • These are both at least 10, so the distribution of $\hat{P}$ is approximately normal.

Calculating Probabilities Using Excel

Example: Ice Cream Preference

• According to a Harris Poll, 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 0.3?
• Here we are asked to find a probability involving a sample proportion. We check to make sure that the central limit theorem for proportions applies.

Verifying the applicability of CLT

• Central Limit Theorem states that if $\hat{P}$ is a sample proportion for a sample of size $n$ from a population with proportion $P$, then if $n \times P$ and $n \times (1-P)$ are both at least 10, then the distribution of $\hat{P}$ is approximately normal.
• The mean of this distribution is $\mu<em>{\hat{P}}$ which equals $P$, the population proportion, and the standard deviation is $\sigma</em>{\hat{P}}$, which equals $\sqrt{\frac{P(1-P)}{n}}$.
• In this example, $n = 100$, and the population proportion is $P = 0.27$. Since $n \times P = 27$ and $n \times (1-P) = 73$, and both of these are at least 10, we know that the normal curve can be used to find the probability that the sample proportion is greater than 0.3.

Calculations

• $\mu_{\hat{P}} = P = 0.27$
• $\sigma_{\hat{P}} = \sqrt{\frac{0.27 \times (1-0.27)}{100}} = 0.0444$
• In Excel we enter 1 - NORM.DIST(0.3, 0.27, 0.0444, TRUE).
• The result is 0.2496.

Additional examples

• A simple random sample of size 80 is drawn from a population with population proportion $P = 0.24$. We wish to find the probability that $\hat{P}$ is between 0.20 and 0.23.

  • $n = 80, P = 0.24$
  • $n \times P = 19.2$
  • $n \times (1-P) = 60.8$
  • Since both of these are at least 10, we may use the normal distribution to find the probability.

• A simple random sample of size 145 is drawn from a population with population proportion $P = 0.05$. We wish to find the probability that $\hat{P}$ is between 0.03 and 0.08.

  • $n = 145, P = 0.05$
  • $n \times P = 145 \times 0.05 = 7.25$
  • Note that this value is less than 10, which means that the assumptions are not satisfied. So we stop at this point.

Example: Smartphone ownership

• 73% of teenagers own samrtphones. A sample of 150 teenagers is drawn. Would it be unusual if less than 68% of the the sample teenagers own smartphones?
  • To determine wheter it would be unusual, we will find the area and compare to standard cutoff value of 0.05.
  • $n = 150, P = 0.73$
  • $n \times P = 109.5$
  • $n \times (1-P) = 40.5$
  • Since both quantitites are at least 10, the assumptions are satisfied and we may proceed.
  • $\mu_{\hat{P}} = P = 0.73$
  • $\sigma_{\hat{P}} = \sqrt{\frac{0.73 \times (1-0.73)}{150}} = 0.036249$
• To find the probability that less than 68% of the sample teenagers own smartphones, we find the area under the normal curve to the left of 0.68.
• Find the area to be approximately 0.0839. Since this is greater than 0.05, we conclude that it is not unusual that less than 68% of the sample teenagers own smartphones.