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1 Chapter 5
Contents
1 Outline 1
2 Introduction 1 2.1 Overview of Chapter 5...................... 1
3 Chapter 5 1 3.1 Sampling Proportion....................... 1 3.2 Law of Large Numbers for Sample Percentages......... 2 3.3 Sampling Distribution of ˆp is Approximately Normal..... 2 3.4 Estimating the population proportion p............. 3 3.5 Estimate the Standard Error of proportion p.......... 5 3.6 class exercise........................... 5
2 Introduction
2.1 Overview of Chapter 5
Goals and Objectives
- To learn about the sampling distribution of the proportion
Topics
- Sample Proportion
- Law of Large Numbers
- Sampling Distribution of ˆp
- Estimate the population proportion p
- Estimate the Standard Error of proportion p
3 Chapter 5
3.1 Sampling Proportion
Sampling Proportion Let us start with an example.
- Suppose TV World sells 60 extended warranties with 300 TV sets sold. The warranty sales rate is 30060 = 0.20.
- Therefore, let X denote the number of successes out of a sample of n observations.
- Where X is a binomial random variable with parameters n and p. Also, the proportion of successes in a sample is a random variable.
Sampling Proportion
- pˆ = Xn =(number of successes) ÷ (total number of observations in the sample)
- For the binomial, X is expected to be around np give or take
np(1 − p)
- For the proportion, ˆp is expected to be npn give or
take
np(1−p) n =
p(1−p) n
Sampling Proportion example
- TV World example: The number of warranties sold is expected to be around 60 ± 7
- The proportion of warranties sold is expected to be around
60 300 ±^
7 300 or^.^2 ±^0 .02.
3.2 Law of Large Numbers for Sample Percentages
Law of Large Numbers for Sample Percentages
- The sample percentage tends to get closer to the true percentage as sample size increases
- The population proportion p are generally unknown and are estimated from the data.
- Suppose we want to estimate the number of students planning to attend graduate school.
- Will the sample proportion equal the population proportion? Yes or No.
- If not, by how much will we miss it?
Estimating the population proportion p Consider this example: n = 40 graduating seniors, X = 6 is the number of graduating seniors planning to attend graduate school.
- What is the proportion of graduating seniors planning to attend grad- uate school?
- pˆ = Xn = 406 = 0. 15
- By how much will we miss the true population proportion?
- SEˆp =
pˆ(1−ˆp n =
.15(1−.15) 40 = 0.^05646
iClicker Question Recently, a random sample of 40 small retail businesses found that 32 had experienced cash flow problems in their first year of operation. What proportion of small retail businesses had cash flow problems in their first year of operation?
a. 0.
b. 1.
c. 25%
d. 1.
e. 0.
Estimating the population proportion p Consider this example: n = 40 graduating seniors, X = 6 is the number of graduating seniors planning to attend graduate school.
- If we take another sample of size 40 graduating seniors, what is the probability that 17 percent or less of the graduating seniors will attend graduate school?
- P [ˆp ≤ .17] = normCDF (− 99 ,. 17 ,. 15 , .05646) = 0. 6384
- If we take another sample of size 40 graduating seniors, what is the probability that 20 percent or more of the graduating seniors will attend graduate school?
- P [ˆp ≥ .2] = normCDF (. 2 , 99 ,. 15 , .05646) = 0. 1879
3.5 Estimate the Standard Error of proportion p
Summarizing the Estimate of the population proportion p
- pˆ is an estimate of the population proportion, i.e., E[ˆp] = p
- We will miss it by the standard error of the proportion = SEˆp =
pˆ(1−ˆp) n
Summarizing the Estimate of the population proportion p
- The population proportion p is estimated using the sample proportion pˆ, i.e., E[ˆp] = p. This estimate tends to miss by an amount called the SEˆp.
- The SEˆp is calculated as
pˆ(1−pˆ) n.
- As sample size increases, the SEˆp decreases like the square root of sample size.
0.15 0.20 0.25 0.30 0.35 0.40 0.
0
2
4
6
8
MS millionaires
Probability
class exercise If a random samples of 100 MS employees are selected at random, what proportion of the samples will be between 25 and 35% millionaires? Given: p = .3, n = 100, P [. 25 ≤ ˆp ≤ .35].
- P [. 25 ≤ pˆ ≤ .35] =
- normCDF (. 25 ,. 35 ,. 3 ,
.3(1−.3) 100 ) = 0.^7248
0.15 0.20 0.25 0.30 0.35 0.40 0.
0
2
4
6
8
MS millionaires
Probability
problem 1 page 68
- Given:
- p = 35 = .6,
- n = 49
- P [ˆp > 23 ] =
problem 1 page 68 answers Given: p = 35 = .6, n = 49, P [ˆp > 23 ]
- P [ˆp > 23 ] = normCDF (^23 , 99 ,. 6 ,
.6(1−.6) 49 ) = 0.^1704
0.4 0.5 0.6 0.7 0.
0
1
2
3
4
5
Lose Money
Probability
iClicker Question Recently, a random sample of 40 small retail businesses found that 32 had experienced cash flow problems in their first year of operation. what is the probability that between 70% and 85% of small retail businesses had cash flow problems in their first year of operation?
a. 0.
b. 0.
c. 0.
d. 0.
e. 0.
Questions Questions?
