lecture: sampling distribution of a...
TRANSCRIPT
Lecture: Sampling Distribution of a Proportion
MA 217 - Stephen Sawin
Fairfield University
August 8, 2017
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
Sampling Distribution for a Proportion
Start with a population, adult Americans and a binary variable,whether they believe in God.The key parameter is the population proportion p. In this case letus suppose 82% of Americans believe in God.Take a sample of 400 Americans, ask if they believe in God. Thekey statistic is the sample proportion p̂, (“pee-hat”) the number ofyes answers divided by the total (400). The proportion of thesample who believe in God. Each sample has a different p̂. If weconsider all possible samples we can make a histogram of thosevalues, the sampling distribution of the random variable P̂.The sampling distribution of P̂ has
1. A mean of µP̂ = p
2. A standard deviation (standard error) of σP̂ =√
p(1−p)n
3. A normal distribution if n is big enough.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
The Fine Print
There were 3 assumptions underlying last slide. None exactly truein real situations, but Rules of Thumb say when close enough.
(a) SRS The sample is assumed to be a Simple Random Sample.The formula for standard deviation assumes sampling withreplacement so successive individuals sampled areindependent. This is close enough if population is much largerthan the sample:
(b) Independence/Large Population Assumption Thepopulation size is at least 20 times the sample size.As n gets larger the distribution gets more normal, but ithappens faster if p is close to .5. We can use the normal dist.to model sample proportion if
(c) Normality Assumption/Rule of 15 : the numbers np andn(1 − p) are both at least 15.
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
An Example82% of adult Americans believe in God. Take a SRS of 400 adultAmericans and ask if they believe. What are mean and standarddeviation of proportion in your sample who do? What is chanceless than 80% in your sample will believe in God? Between 80 and90%?It says simple random sample, so SRS assumption:Met. The meanis
µP̂ = p = .82.
Check the large population assumption: Need there to be morethan 400 · 20 = 8000 adult Americans: obviously true so Met.
σP̂ =
√p(1 − p)
n=
√.82 · .18
400= 0.0192.
Check normality assumption/ rule of 15 :np = 400 · .82 = 328 ≥ 15. n(1 − p) = 400 · .18 = 72 ≥ 15 so P̂ isnormal.Met
More Example
82% of adult Americans believe. Take a SRS of 400 adultAmericans and ask if they support him. What are mean andstandard deviation of proportion in your sample who do? What ischance less than 80% in your sample will believe? Between 80 and90%?To find the probability it is less than 80% since P̂ is normal
P(P̂ < .8) = normdist(.8, .82,√.82 ∗ .18/400, 1) = 14.9%
Between 80% and 90% :
P(.8 < P̂ < .9) = normdist(.9, .82,√.82 ∗ .18/400, 1)
− normdist(.8, .82,√.82 ∗ .18/400, 1) = 85.1%
More Example
82% of adult Americans believe. Take a SRS of 400 adultAmericans and ask if they support him. What are mean andstandard deviation of proportion in your sample who do? What ischance less than 80% in your sample will believe? Between 80 and90%?To find the probability it is less than 80% since P̂ is normal
P(P̂ < .8) = normdist(.8, .82,√.82 ∗ .18/400, 1) = 14.9%
Between 80% and 90% :
P(.8 < P̂ < .9) = normdist(.9, .82,√.82 ∗ .18/400, 1)
− normdist(.8, .82,√.82 ∗ .18/400, 1) = 85.1%
More Example
82% of adult Americans believe. Take a SRS of 400 adultAmericans and ask if they support him. What are mean andstandard deviation of proportion in your sample who do? What ischance less than 80% in your sample will believe? Between 80 and90%?To find the probability it is less than 80% since P̂ is normal
P(P̂ < .8) = normdist(.8, .82,√.82 ∗ .18/400, 1) = 14.9%
Between 80% and 90% :
P(.8 < P̂ < .9) = normdist(.9, .82,√.82 ∗ .18/400, 1)
− normdist(.8, .82,√.82 ∗ .18/400, 1) = 85.1%
More Example
82% of adult Americans believe. Take a SRS of 400 adultAmericans and ask if they support him. What are mean andstandard deviation of proportion in your sample who do? What ischance less than 80% in your sample will believe? Between 80 and90%?To find the probability it is less than 80% since P̂ is normal
P(P̂ < .8) = normdist(.8, .82,√.82 ∗ .18/400, 1) = 14.9%
Between 80% and 90% :
P(.8 < P̂ < .9) = normdist(.9, .82,√.82 ∗ .18/400, 1)
− normdist(.8, .82,√.82 ∗ .18/400, 1) = 85.1%
More Example
82% of adult Americans believe. Take a SRS of 400 adultAmericans and ask if they support him. What are mean andstandard deviation of proportion in your sample who do? What ischance less than 80% in your sample will believe? Between 80 and90%?To find the probability it is less than 80% since P̂ is normal
P(P̂ < .8) = normdist(.8, .82,√.82 ∗ .18/400, 1) = 14.9%
Between 80% and 90% :
P(.8 < P̂ < .9) = normdist(.9, .82,√.82 ∗ .18/400, 1)
− normdist(.8, .82,√.82 ∗ .18/400, 1) = 85.1%
The Example - The Big Picture
So we saw that if we take many samples of n = 400 from apopulation with proportion of successes p = .82 and compute thesample proportion p̂ for each one, these values of P̂ will have anormal distribution with mean µ = .82 and standard deviationσ = .019
The Example - The Big Picture
So we saw that if we take many samples of n = 400 from apopulation with proportion of successes p = .82 and compute thesample proportion p̂ for each one, these values of P̂ will have anormal distribution with mean µ = .82 and standard deviationσ = .019
The Example - The Big Picture
So we saw that if we take many samples of n = 400 from apopulation with proportion of successes p = .82 and compute thesample proportion p̂ for each one, these values of P̂ will have anormal distribution with mean µ = .82 and standard deviationσ = .019
The Example - The Big Picture
So we saw that if we take many samples of n = 400 from apopulation with proportion of successes p = .82 and compute thesample proportion p̂ for each one, these values of P̂ will have anormal distribution with mean µ = .82 and standard deviationσ = .019
The Example - The Big Picture
So we saw that if we take many samples of n = 400 from apopulation with proportion of successes p = .82 and compute thesample proportion p̂ for each one, these values of P̂ will have anormal distribution with mean µ = .82 and standard deviationσ = .019
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Sampling Distribution
In general consider a population and a variable. Take a simplerandom sample and compute some statistic like mean orproportion. Each time you do this you get a different answer, so itis a Random variable!If you consider the value of this statistic for every possible sample,you get a distribution, the sampling distribution. We want to knowits mean, standard deviation, and shape of its histogram.
I The population distribution is the values of the variable inthe population (the 82% of all adult Americans who believe)
I The data distribution is the values of the variable in oneparticular sample (maybe 320 yes answers in a sample of 400)
I The sampling distribution is the different values of thestatistic (like P̂) in different samples
Lecture on Sampling Distribution for a Proportion KeyPoints
After watching this lecture you should be able to
I Know we mean by the sampling distribution of P̂, and what itrepresents
I Calculate the mean and standard deviation of P̂
I Check the Independence/ Large Population assumption andwhat it tells you (that the s.d. formula is correct)
I Check the Normality / Rule of 15 assumption and what it tellsyou (can use normdist)
I Calculate probabilities of P̂ using normdist.