Central Limit Theorem

Central Limit Theorem: If random samples of n observations are drawn from a nonnormal population with finite μ and standard deviation σ, then, when n is large, the sampling distribution of the sample mean (symbol : \(\bar{x}\)) is approximately normally distributed, with mean μ and standard deviation \(\frac{\sigma}{n}\). The approximation becomes more accurate as n becomes large

If the population is normal, then the sampling distribution of will also be normal, no matter what the sample size. When the population is approximately symmetric, the distribution becomes approximately normal for relatively small values of n. When the population is skewed, the sample size must be at least 30 before the sampling distribution of becomes approximately normal.

Illustration:

Finding Probabilities for the Sample Mean

Example

1. A random sample of size n = 16 from a normal distribution with μ = 10 and σ = 8. Find the probability of \(\bar{x}>12\).

Answer:

xbar=12
sd1=8/sqrt(16)
miu=10
pnorm(q = xbar, mean = miu, sd = sd1, lower.tail = FALSE)

## [1] 0.1586553

2. A soda filling machine is supposed to fill cans of soda with 12 fluid ounces. Suppose that the fills are actually normally distributed with a mean of 12.1 oz and a standard deviation of 0.2 oz. What is the probability that the average fill for a 6-pack of soda is less than 12 oz (the probability of \(\bar{x}<12\))?

Answer:

1. Confidence Interval for the Mean of One Population

100(1−α)% confidence interval for the population mean μ:

1. For Large Sample or Known 𝜎

Formula:

\(\bar{x}\pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\)

2. For Small Sample or Unknown 𝜎

Formula:

\(\bar{x}\pm t_{\alpha/2,df} \frac{s}{\sqrt{n}}\)

Example:

A random sample of 12 students of a certain school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all students of this school. [A 95% confidence interval for μ (small sample)]

\(\bar{x}\pm t_{\alpha/2,df} \frac{s}{\sqrt{n}}\)

mean1<-79.3
t_table<-qt(0.975,11,lower.tail = TRUE)
t_table

## [1] 2.200985

s_value<-7.8
n<-12
moe<-t_table*(s_value/sqrt(n))
cat(" Lower", mean1-moe,"\n","Upper:" ,mean1+moe)

##  Lower 74.34412 
##  Upper: 84.25588

2. Confidence Interval for the Mean Difference between Two Populations

LARGE-SAMPLE CASE (n1 > 30 AND n2 > 30)

1. Interval Estimate with \(\sigma_{1}\) and \(\sigma_{2}\) Known

\(\bar{x_{1}}-\bar{x_{2}}\pm z_{\alpha/2} \sigma_{\bar{x_{1}}-\bar{x_{2}}}\)

2. Interval Estimate with \(\sigma_{1}\) and \(\sigma_{2}\) Unknown

\(\bar{x_{1}}-\bar{x_{2}}\pm z_{\alpha/2} s_{\bar{x_{1}}-\bar{x_{2}}}\)

Example:

Compare the average daily intake of dairy products of men and women using a 95% confidence interval.Could you conclude, based on this confidence interval, that there is a difference in the average daily intake of dairy products for men and women?

mean_dif1<-756-762
n1<-50
n2<-50
s1<-35^2
s2<-30^2

z_table<-qnorm(0.975,lower.tail = TRUE)
z_table

## [1] 1.959964

s1_2<-sqrt(s1/n1+s2/n2)
s1_2

## [1] 6.519202

moe<-z_table*s1_2

cat(" Lower", mean_dif1-moe,"\n","Upper:" ,mean_dif1+moe)

##  Lower -18.7774 
##  Upper: 6.777402

The Confidence Interval contains 0. Therefore, it is possible that \(\mu_{1}=\mu_{2}\). We can conclude that there is NO DIFFERENCE in average daily intake of dairy products for men and women.