Estimating 𝑝 In the Binomial Distribution The binomial distribution is completely determined by the number of trials 𝑛 and the probability 𝑝 of success on a single trial. For most experiments, the number of trials is chosen in advance and the distribution is completely determined by 𝑝.

Estimating 𝑝 In the Binomial Distribution The binomial distribution is completely determined by the number of trials 𝑛 and the probability 𝑝 of success on a single trial. For most experiments, the number of trials is chosen in advance and the distribution is completely determined by 𝑝. In this section, we will consider the problem of estimating 𝑝 under the assumption that 𝑛 has already been selected. We are using what are called “large – sample methods”. We will assume that the normal curve is a good approximation to the binomial distribution. We will also use sample estimates for the standard deviation.

Estimating 𝑝 In the Binomial Distribution The binomial distribution is completely determined by the number of trials 𝑛 and the probability 𝑝 of success on a single trial. For most experiments, the number of trials is chosen in advance and the distribution is completely determined by 𝑝. In this section, we will consider the problem of estimating 𝑝 under the assumption that 𝑛 has already been selected. We are using what are called “large – sample methods”. We will assume that the normal curve is a good approximation to the binomial distribution. We will also use sample estimates for the standard deviation. Again, these methods are proven to be quite good when : 𝑛𝑝>5 , 𝑛𝑞>5 , where 𝑞=1−𝑝

Estimating 𝑝 In the Binomial Distribution The binomial distribution is completely determined by the number of trials 𝑛 and the probability 𝑝 of success on a single trial. For most experiments, the number of trials is chosen in advance and the distribution is completely determined by 𝑝. In this section, we will consider the problem of estimating 𝑝 under the assumption that 𝑛 has already been selected. We are using what are called “large – sample methods”. We will assume that the normal curve is a good approximation to the binomial distribution. We will also use sample estimates for the standard deviation. Again, these methods are proven to be quite good when : 𝑛𝑝>5 , 𝑛𝑞>5 , where 𝑞=1−𝑝 Point estimates for 𝑝 and 𝑞 are : 𝑝 = 𝑟 𝑛 𝑞 =1− 𝑝 - where 𝑛= number of trials and 𝑟= number of successes

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ?

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? 1. First estimate 𝑝 for the entire student body 𝑝 = 𝑟 𝑛 = 600 800 =0.75 - this is the point estimate for 𝑝

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? 1. First estimate 𝑝 for the entire student body 𝑝 = 𝑟 𝑛 = 600 800 =0.75 - this is the point estimate for 𝑝 2. Like our other sections, we need to factor in some margin of error

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? 1. First estimate 𝑝 for the entire student body 𝑝 = 𝑟 𝑛 = 600 800 =0.75 - this is the point estimate for 𝑝 2. Like our other sections, we need to factor in some margin of error -- in this case, the margin of error = 𝑝 −𝑝

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? 1. First estimate 𝑝 for the entire student body 𝑝 = 𝑟 𝑛 = 600 800 =0.75 - this is the point estimate for 𝑝 2. Like our other sections, we need to factor in some margin of error -- in this case, the margin of error = 𝑝 −𝑝 3. In large samples, the distribution of 𝑝 is well approximated by a normal curve with mean 𝜇=𝑝 AND standard error 𝜎= 𝑝𝑞 𝑛

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? 1. First estimate 𝑝 for the entire student body 𝑝 = 𝑟 𝑛 = 600 800 =0.75 - this is the point estimate for 𝑝 2. Like our other sections, we need to factor in some margin of error -- in this case, the margin of error = 𝑝 −𝑝 3. In large samples, the distribution of 𝑝 is well approximated by a normal curve with mean 𝜇=𝑝 AND standard error 𝜎= 𝑝𝑞 𝑛 4. Our bounds then become 𝑃 −𝑧 𝑐 𝑝𝑞 𝑛 < 𝑝 −𝑝< 𝑧 𝑐 𝑝𝑞 𝑛 =𝑐

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? Confidence interval for 𝑝 : 𝑝 −𝐸<𝑝< 𝑝 +𝐸 where 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 OR 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? Confidence interval for 𝑝 : 𝑝 −𝐸<𝑝< 𝑝 +𝐸 where 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 OR 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 Find a 99% confidence interval for 𝑝 in our example

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? Confidence interval for 𝑝 : 𝑝 −𝐸<𝑝< 𝑝 +𝐸 where 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 OR 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 Find a 99% confidence interval for 𝑝 in our example 𝑧 0.99 =2.58 ( from the critical value table – level of confidence = 0.99 )

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? Confidence interval for 𝑝 : 𝑝 −𝐸<𝑝< 𝑝 +𝐸 where 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 OR 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 Find a 99% confidence interval for 𝑝 in our example 𝑧 0.99 =2.58 ( from the critical value table – level of confidence = 0.99 ) 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 = 2.58 0.75 1−0.75 800 = 2.58 0.015 =0.0395

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? Confidence interval for 𝑝 : 𝑝 −𝐸<𝑝< 𝑝 +𝐸 where 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 OR 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 Find a 99% confidence interval for 𝑝 in our example 𝑧 0.99 =2.58 ( from the critical value table – level of confidence = 0.99 ) 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 = 2.58 0.75 1−0.75 800 = 2.58 0.015 =0.0395 𝑝 −𝐸<𝑝< 𝑝 +𝐸 0.75−0.0395<𝑝<0.75+0.0395 0.71<𝑝<0.79

Estimating 𝑝 In the Binomial Distribution EXAMPLE : Suppose 800 students are selected at random from a student body of 20,000 and that they are each given a shot to prevent a certain type of flu. These 800 students are then exposed to the flu, and 600 of them do not get the flu. What is the probability 𝑝 that the shot will be successful for any single student selected at random from the entire population of 20,000 students ? Confidence interval for 𝑝 : 𝑝 −𝐸<𝑝< 𝑝 +𝐸 where 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 OR 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 Find a 99% confidence interval for 𝑝 in our example 𝑧 0.99 =2.58 ( from the critical value table – level of confidence = 0.99 ) 𝐸= 𝑧 𝑐 𝑝 (1− 𝑝 ) 𝑛 = 2.58 0.75 1−0.75 800 = 2.58 0.015 =0.0395 𝑝 −𝐸<𝑝< 𝑝 +𝐸 0.75−0.0395<𝑝<0.75+0.0395 0.71<𝑝<0.79 We are 99% confident that the probability a flu shot will be effective for a student selected at random is between 0.71 and 0.79

Estimating 𝑝 In the Binomial Distribution EXAMPLE #2 : A random sample of 188 books purchased at a local bookstore showed that 66 of the books were murder mysteries. Let 𝑝 represent the proportion of books sold by this store that are murder mysteries. Find a 90% confidence interval for 𝑝.

Estimating 𝑝 In the Binomial Distribution EXAMPLE #2 : A random sample of 188 books purchased at a local bookstore showed that 66 of the books were murder mysteries. Let 𝑝 represent the proportion of books sold by this store that are murder mysteries. Find a 90% confidence interval for 𝑝. SOLUTION : 1. Find the point estimate 𝑝 = 𝑟 𝑛 = 66 188 =0.35 𝑞 =1− 𝑝 =1−0.35=0.65

Estimating 𝑝 In the Binomial Distribution EXAMPLE #2 : A random sample of 188 books purchased at a local bookstore showed that 66 of the books were murder mysteries. Let 𝑝 represent the proportion of books sold by this store that are murder mysteries. Find a 90% confidence interval for 𝑝. SOLUTION : 1. Find the point estimate 𝑝 = 𝑟 𝑛 = 66 188 =0.35 𝑞 =1− 𝑝 =1−0.35=0.65 2. Find 𝑧 𝑐 𝑧 𝑐 =1.645 ( from table when 𝑐=0.90 )

Estimating 𝑝 In the Binomial Distribution EXAMPLE #2 : A random sample of 188 books purchased at a local bookstore showed that 66 of the books were murder mysteries. Let 𝑝 represent the proportion of books sold by this store that are murder mysteries. Find a 90% confidence interval for 𝑝. SOLUTION : 1. Find the point estimate 𝑝 = 𝑟 𝑛 = 66 188 =0.35 𝑞 =1− 𝑝 =1−0.35=0.65 2. Find 𝑧 𝑐 𝑧 𝑐 =1.645 ( from table when 𝑐=0.90 ) 3. Find 𝐸 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 =1.645 (0.35)(0.65) 188 = 1.645 0.035 =0.058

Estimating 𝑝 In the Binomial Distribution EXAMPLE #2 : A random sample of 188 books purchased at a local bookstore showed that 66 of the books were murder mysteries. Let 𝑝 represent the proportion of books sold by this store that are murder mysteries. Find a 90% confidence interval for 𝑝. SOLUTION : 1. Find the point estimate 𝑝 = 𝑟 𝑛 = 66 188 =0.35 𝑞 =1− 𝑝 =1−0.35=0.65 2. Find 𝑧 𝑐 𝑧 𝑐 =1.645 ( from table when 𝑐=0.90 ) 3. Find 𝐸 𝐸= 𝑧 𝑐 𝑝 𝑞 𝑛 =1.645 (0.35)(0.65) 188 = 1.645 0.035 =0.058 4. Find the confidence interval 𝑝 −𝐸<𝑝< 𝑝 +𝐸 0.35−0.058<𝑝<0.35+0.058 0.29<𝑝<0.41

Estimating 𝑝 In the Binomial Distribution How to Find Sample Size 𝑛 for Estimating a Proportion 𝑝 𝑛=𝑝(1−𝑝) 𝑧 𝑐 𝐸 2 - if you have a preliminary estimate for 𝑝 𝑛= 1 4 𝑧 𝑐 𝐸 2 - if you do not have a preliminary estimate for 𝑝 where 𝐸= specified maximal error of estimate 𝑧 𝑐 = critical value from the normal distribution for the desired confidence level c. Also : increase sample size 𝑛 to ensure that 𝑛𝑝>5 and 𝑛𝑞>5

Estimating 𝑝 In the Binomial Distribution How to Find Sample Size 𝑛 for Estimating a Proportion 𝑝 𝑛=𝑝(1−𝑝) 𝑧 𝑐 𝐸 2 - if you have a preliminary estimate for 𝑝 𝑛= 1 4 𝑧 𝑐 𝐸 2 - if you do not have a preliminary estimate for 𝑝 where 𝐸= specified maximal error of estimate 𝑧 𝑐 = critical value from the normal distribution for the desired confidence level c. Also : increase sample size 𝑛 to ensure that 𝑛𝑝>5 and 𝑛𝑞>5 Find the sample size needed if there is no preliminary estimate for 𝑝 for 95% confidence level that is not in error by 0.01 either way.

Estimating 𝑝 In the Binomial Distribution How to Find Sample Size 𝑛 for Estimating a Proportion 𝑝 𝑛=𝑝(1−𝑝) 𝑧 𝑐 𝐸 2 - if you have a preliminary estimate for 𝑝 𝑛= 1 4 𝑧 𝑐 𝐸 2 - if you do not have a preliminary estimate for 𝑝 where 𝐸= specified maximal error of estimate 𝑧 𝑐 = critical value from the normal distribution for the desired confidence level c. Also : increase sample size 𝑛 to ensure that 𝑛𝑝>5 and 𝑛𝑞>5 Find the sample size needed if there is no preliminary estimate for 𝑝 for 95% confidence level that is not in error by 0.01 either way. 𝑛=0.25 𝑧 𝑐 𝐸 2 =0.25 1.96 0.01 2 =0.25 38,416 =9604