Testing confidence intervals in R
I currently have constructed a 95% confidence interval and have then used replicate() to randomly generate 1000 confidence intervals. I want to measure how many of the intervals contain my mean. I know in theory it should be in 950 of them but how do I get a definite answer? The function I used and the mean are listed below.
z <- function(a,b,c){
error <- rnorm(a, b, c) * c / sqrt(a)
left <- b - error
right <- j + error
paste("[",round(left,2),";",round(right,2),"]")
}
set.seed(123)
replicate(1000, z(10,1,1))
Where do I go from here?
r confidence-interval
edited Nov 23 '18 at 11:57
Roland
99k6111181
asked Nov 23 '18 at 11:11
John Fitz
63
|
show 3 more comments
I currently have constructed a 95% confidence interval and have then used replicate() to randomly generate 1000 confidence intervals. I want to measure how many of the intervals contain my mean. I know in theory it should be in 950 of them but how do I get a definite answer? The function I used and the mean are listed below.
z <- function(a,b,c){
error <- rnorm(a, b, c) * c / sqrt(a)
left <- b - error
right <- j + error
paste("[",round(left,2),";",round(right,2),"]")
}
set.seed(123)
replicate(1000, z(10,1,1))
Where do I go from here?
r confidence-interval
edited Nov 23 '18 at 11:57
Roland
99k6111181
asked Nov 23 '18 at 11:11
John Fitz
63
I can't make much sense of yourz()function. Could you format it better/tidy it up?
– AkselA
Nov 23 '18 at 11:16
@AkselA is that better?
– John Fitz
Nov 23 '18 at 11:22
It still looks like non-sense :). Have you written functions in R before?
– AkselA
Nov 23 '18 at 11:28
No I'm new to R. It generates the 1000 confidence intervals but how do I then test if the mean is contained in each confidence interval?
– John Fitz
Nov 23 '18 at 11:31
Confidence interval of what? Estimating the mean of a normal distribution?
– AkselA
Nov 23 '18 at 11:45
|
show 3 more comments
I currently have constructed a 95% confidence interval and have then used replicate() to randomly generate 1000 confidence intervals. I want to measure how many of the intervals contain my mean. I know in theory it should be in 950 of them but how do I get a definite answer? The function I used and the mean are listed below.
z <- function(a,b,c){
error <- rnorm(a, b, c) * c / sqrt(a)
left <- b - error
right <- j + error
paste("[",round(left,2),";",round(right,2),"]")
}
set.seed(123)
replicate(1000, z(10,1,1))
Where do I go from here?
r confidence-interval
edited Nov 23 '18 at 11:57
Roland
99k6111181
asked Nov 23 '18 at 11:11
John Fitz
63
I currently have constructed a 95% confidence interval and have then used replicate() to randomly generate 1000 confidence intervals. I want to measure how many of the intervals contain my mean. I know in theory it should be in 950 of them but how do I get a definite answer? The function I used and the mean are listed below.
z <- function(a,b,c){
error <- rnorm(a, b, c) * c / sqrt(a)
left <- b - error
right <- j + error
paste("[",round(left,2),";",round(right,2),"]")
}
set.seed(123)
replicate(1000, z(10,1,1))
Where do I go from here?
r confidence-interval
r confidence-interval
edited Nov 23 '18 at 11:57
Roland
99k6111181
asked Nov 23 '18 at 11:11
John Fitz
63
edited Nov 23 '18 at 11:57
Roland
99k6111181
asked Nov 23 '18 at 11:11
John Fitz
63
edited Nov 23 '18 at 11:57
Roland
99k6111181
edited Nov 23 '18 at 11:57
Roland
99k6111181
edited Nov 23 '18 at 11:57
Roland
99k6111181
99k6111181
asked Nov 23 '18 at 11:11
John Fitz
63
asked Nov 23 '18 at 11:11
John Fitz
63
asked Nov 23 '18 at 11:11
John Fitz
63
63
I can't make much sense of yourz()function. Could you format it better/tidy it up?
– AkselA
Nov 23 '18 at 11:16
@AkselA is that better?
– John Fitz
Nov 23 '18 at 11:22
It still looks like non-sense :). Have you written functions in R before?
– AkselA
Nov 23 '18 at 11:28
No I'm new to R. It generates the 1000 confidence intervals but how do I then test if the mean is contained in each confidence interval?
– John Fitz
Nov 23 '18 at 11:31
Confidence interval of what? Estimating the mean of a normal distribution?
– AkselA
Nov 23 '18 at 11:45
|
show 3 more comments
I can't make much sense of yourz()function. Could you format it better/tidy it up?
– AkselA
Nov 23 '18 at 11:16
@AkselA is that better?
– John Fitz
Nov 23 '18 at 11:22
It still looks like non-sense :). Have you written functions in R before?
– AkselA
Nov 23 '18 at 11:28
No I'm new to R. It generates the 1000 confidence intervals but how do I then test if the mean is contained in each confidence interval?
– John Fitz
Nov 23 '18 at 11:31
Confidence interval of what? Estimating the mean of a normal distribution?
– AkselA
Nov 23 '18 at 11:45
I can't make much sense of your
z() function. Could you format it better/tidy it up?– AkselA
Nov 23 '18 at 11:16
I can't make much sense of your
z() function. Could you format it better/tidy it up?– AkselA
Nov 23 '18 at 11:16
@AkselA is that better?
– John Fitz
Nov 23 '18 at 11:22
@AkselA is that better?
– John Fitz
Nov 23 '18 at 11:22
It still looks like non-sense :). Have you written functions in R before?
– AkselA
Nov 23 '18 at 11:28
It still looks like non-sense :). Have you written functions in R before?
– AkselA
Nov 23 '18 at 11:28
No I'm new to R. It generates the 1000 confidence intervals but how do I then test if the mean is contained in each confidence interval?
– John Fitz
Nov 23 '18 at 11:31
No I'm new to R. It generates the 1000 confidence intervals but how do I then test if the mean is contained in each confidence interval?
– John Fitz
Nov 23 '18 at 11:31
Confidence interval of what? Estimating the mean of a normal distribution?
– AkselA
Nov 23 '18 at 11:45
Confidence interval of what? Estimating the mean of a normal distribution?
– AkselA
Nov 23 '18 at 11:45
|
show 3 more comments
1 Answer
1
active
oldest
votes
Maybe this is what you're trying to do?
This z() will return the confidence interval for the population mean of a normal distribution.
z <- function(N, mu, std, cl=95) {
alpha <- (1-cl/100)/2
# CI for population mean
sep <- std/sqrt(N)
z_s <- qnorm(1 - alpha)
pop_lower <- mu - z_s*sep
pop_upper <- mu + z_s*sep
c(lower=pop_lower, upper=pop_upper)
}
Meaning that if I produce a random variate mean(rnorm(20, 0, 1)), then we expect the value of that to lie within z(20, 0, 1, 95) with probability 0.95.
To test this we can do
# specify parameters
N <- 20
mu <- 0
std <- 1
# produce a good number (10,000) of population means
set.seed(1)
r <- replicate(1e4, mean(rnorm(N, mu, std)))
# calculate confidence interval
ci <- z(N, mu, std)
# find which are below, within and above the interval
rc <- cut(r, c(min(r), ci, max(r)), c("below", "within", "above"))
# create a proportion table
round(prop.table(table(rc))*100, 2)
# below within above
# 2.59 95.08 2.33
answered Nov 23 '18 at 13:07
AkselA
4,25521125
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
Maybe this is what you're trying to do?
This z() will return the confidence interval for the population mean of a normal distribution.
z <- function(N, mu, std, cl=95) {
alpha <- (1-cl/100)/2
# CI for population mean
sep <- std/sqrt(N)
z_s <- qnorm(1 - alpha)
pop_lower <- mu - z_s*sep
pop_upper <- mu + z_s*sep
c(lower=pop_lower, upper=pop_upper)
}
Meaning that if I produce a random variate mean(rnorm(20, 0, 1)), then we expect the value of that to lie within z(20, 0, 1, 95) with probability 0.95.
To test this we can do
# specify parameters
N <- 20
mu <- 0
std <- 1
# produce a good number (10,000) of population means
set.seed(1)
r <- replicate(1e4, mean(rnorm(N, mu, std)))
# calculate confidence interval
ci <- z(N, mu, std)
# find which are below, within and above the interval
rc <- cut(r, c(min(r), ci, max(r)), c("below", "within", "above"))
# create a proportion table
round(prop.table(table(rc))*100, 2)
# below within above
# 2.59 95.08 2.33
answered Nov 23 '18 at 13:07
AkselA
4,25521125
add a comment |
Maybe this is what you're trying to do?
This z() will return the confidence interval for the population mean of a normal distribution.
z <- function(N, mu, std, cl=95) {
alpha <- (1-cl/100)/2
# CI for population mean
sep <- std/sqrt(N)
z_s <- qnorm(1 - alpha)
pop_lower <- mu - z_s*sep
pop_upper <- mu + z_s*sep
c(lower=pop_lower, upper=pop_upper)
}
Meaning that if I produce a random variate mean(rnorm(20, 0, 1)), then we expect the value of that to lie within z(20, 0, 1, 95) with probability 0.95.
To test this we can do
# specify parameters
N <- 20
mu <- 0
std <- 1
# produce a good number (10,000) of population means
set.seed(1)
r <- replicate(1e4, mean(rnorm(N, mu, std)))
# calculate confidence interval
ci <- z(N, mu, std)
# find which are below, within and above the interval
rc <- cut(r, c(min(r), ci, max(r)), c("below", "within", "above"))
# create a proportion table
round(prop.table(table(rc))*100, 2)
# below within above
# 2.59 95.08 2.33
answered Nov 23 '18 at 13:07
AkselA
4,25521125
add a comment |
Maybe this is what you're trying to do?
This z() will return the confidence interval for the population mean of a normal distribution.
z <- function(N, mu, std, cl=95) {
alpha <- (1-cl/100)/2
# CI for population mean
sep <- std/sqrt(N)
z_s <- qnorm(1 - alpha)
pop_lower <- mu - z_s*sep
pop_upper <- mu + z_s*sep
c(lower=pop_lower, upper=pop_upper)
}
Meaning that if I produce a random variate mean(rnorm(20, 0, 1)), then we expect the value of that to lie within z(20, 0, 1, 95) with probability 0.95.
To test this we can do
# specify parameters
N <- 20
mu <- 0
std <- 1
# produce a good number (10,000) of population means
set.seed(1)
r <- replicate(1e4, mean(rnorm(N, mu, std)))
# calculate confidence interval
ci <- z(N, mu, std)
# find which are below, within and above the interval
rc <- cut(r, c(min(r), ci, max(r)), c("below", "within", "above"))
# create a proportion table
round(prop.table(table(rc))*100, 2)
# below within above
# 2.59 95.08 2.33
answered Nov 23 '18 at 13:07
AkselA
4,25521125
Maybe this is what you're trying to do?
This z() will return the confidence interval for the population mean of a normal distribution.
z <- function(N, mu, std, cl=95) {
alpha <- (1-cl/100)/2
# CI for population mean
sep <- std/sqrt(N)
z_s <- qnorm(1 - alpha)
pop_lower <- mu - z_s*sep
pop_upper <- mu + z_s*sep
c(lower=pop_lower, upper=pop_upper)
}
Meaning that if I produce a random variate mean(rnorm(20, 0, 1)), then we expect the value of that to lie within z(20, 0, 1, 95) with probability 0.95.
To test this we can do
# specify parameters
N <- 20
mu <- 0
std <- 1
# produce a good number (10,000) of population means
set.seed(1)
r <- replicate(1e4, mean(rnorm(N, mu, std)))
# calculate confidence interval
ci <- z(N, mu, std)
# find which are below, within and above the interval
rc <- cut(r, c(min(r), ci, max(r)), c("below", "within", "above"))
# create a proportion table
round(prop.table(table(rc))*100, 2)
# below within above
# 2.59 95.08 2.33
answered Nov 23 '18 at 13:07
AkselA
4,25521125
answered Nov 23 '18 at 13:07
AkselA
4,25521125
answered Nov 23 '18 at 13:07
AkselA
4,25521125
answered Nov 23 '18 at 13:07
AkselA
4,25521125
4,25521125
add a comment |
add a comment |
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I can't make much sense of your
z()function. Could you format it better/tidy it up?– AkselA
Nov 23 '18 at 11:16
@AkselA is that better?
– John Fitz
Nov 23 '18 at 11:22
It still looks like non-sense :). Have you written functions in R before?
– AkselA
Nov 23 '18 at 11:28
No I'm new to R. It generates the 1000 confidence intervals but how do I then test if the mean is contained in each confidence interval?
– John Fitz
Nov 23 '18 at 11:31
Confidence interval of what? Estimating the mean of a normal distribution?
– AkselA
Nov 23 '18 at 11:45