Population Parameters aren’t known

\[ \mu - 1.96 \cdot SE \leq \bar{X} \leq \mu + 1.96 \cdot SE \]

\[ \bar{X} - 1.96 \cdot SE \leq \mu \leq \bar{X} + 1.96 \cdot SE \]

  • This is for a 95% confidence level, where \(\alpha=.05\). You can use any confidence level and only you will just need to change the 1.96 part.
  • The stuff that you add to either side is called the error bound.

\[ \bar{X} - EB \leq \mu \leq \bar{X} + EB \]

  1. Suppose a population unknown population mean and population standard deviation \(2.5\). A sample of \(450\) observations has a sample mean of \(\bar{x}=62\). Find a \(95 \%\) confidence interval.
# Collect all known values
sigma <- 2.5
n <- 450
barX <- 62

# Compute the standard error of the sampling distribution
SE <- sigma/sqrt(n)

# Add and subtract your z-score times the standard error (error bound)
EB <- qnorm(.975)*SE
barX + EB
## [1] 62.23098
barX - EB
## [1] 61.76902
# We are 95% confident that the population mean is between 61.77 and 62.23
# WE CANNOT say that there is a 95% chance that the population mean is in that range.
# 95% of all possible sample means will have a range that captures the population mean.
  1. What are the \(z\)-scores corresponding to a \(70 \%\) confidence level?
qnorm(.15)
## [1] -1.036433
qnorm(.85)
## [1] 1.036433
  1. A study is conducted to analyze internet privacy for teens. A group of \(50\) teens is randomly sampled and \(17\) report clicking on links from unknown email addresses. Construct a \(90 \%\) confidence interval for the true proportion of teens that click on links from unknown email addresses.
# collecting known values
n <- 50
phat <- 17/50

# Find standard error
SE <- sqrt(phat*(1-phat)/n)

# Add and subtract the error bound to phat
EB <- qnorm(.95)*SE
phat + EB
## [1] 0.4501929
phat - EB
## [1] 0.2298071
# We are 90% confident that the true proportion of teens clicking on bad links is between .23 and .45
  1. Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of \(3\) points. A random sample of \(36\) scores is taken with an mean score of \(72\). Find a \(98 \%\) confidence interval for the mean exam score of the population.
se <- 3/sqrt(36)

72 + qnorm(.99)*se
## [1] 73.16317
72 - qnorm(.99)*se
## [1] 70.83683
  1. Political experts claim that only \(35 \%\) of registered voters vote in local elections. A recent political poll (assume they are using methods that ensure independence) asks \(1000\) registered voters if they plan to vote in their next local election. The poll finds that \(840\) registered voters will vote in the next election.