Harold Nelson

October 27, 2015

Use the code snippets in the Module 6 notes to answer the following questions.

A sample of size 100 is taken from a population with a known standard deviation of 12.The sample estimate of the mean is 34. Construct a 95% connfidence interval for the population mean.

```
# Code snippet to construct a confidence interval
xbar = 34 # Sample Mean
sd = 12 # Population Standard Deviation
n = 100 # Sample size
CL = .95 # Required Confidence Level
zstar <- qnorm(CL+.5*(1-CL)) # Obtain Z-score for this confidence level
sd.xbar <- sd/sqrt(n) # Compute standard error of sample mean
ME <- zstar * sd.xbar # Compute margin of error
lb <- xbar - ME # Compute lower bound of CI
ub <- xbar + ME # Compute upper bound of CI
CI <- c(CL,lb,xbar,ub,ME) # Put our results in a vector
names(CI) <- c("Confidence Level","Lower Bound","Xbar","Upper Bound",
"Margin of Error") # Name the vector elements
CI # Display the vector
```

```
## Confidence Level Lower Bound Xbar Upper Bound
## 0.950000 31.648043 34.000000 36.351957
## Margin of Error
## 2.351957
```

Repeat problem 1, but provide an 80% confidence interval. How does the width of the 80% CI compare with that of the 95% CI?

```
# Code snippet to construct a confidence interval
xbar = 34 # Sample Mean
sd = 12 # Population Standard Deviation
n = 100 # Sample size
CL = .80 # Required Confidence Level
zstar <- qnorm(CL+.5*(1-CL)) # Obtain Z-score for this confidence level
sd.xbar <- sd/sqrt(n) # Compute standard error of sample mean
ME <- zstar * sd.xbar # Compute margin of error
lb <- xbar - ME # Compute lower bound of CI
ub <- xbar + ME # Compute upper bound of CI
CI <- c(CL,lb,xbar,ub,ME) # Put our results in a vector
names(CI) <- c("Confidence Level","Lower Bound","Xbar","Upper Bound",
"Margin of Error") # Name the vector elements
CI # Display the vector
```

```
## Confidence Level Lower Bound Xbar Upper Bound
## 0.800000 32.462138 34.000000 35.537862
## Margin of Error
## 1.537862
```

The 80% CI is narrower than the 95% CI.

Considering the common parameters of problems 1 and 2, how large a sample size would be required to obtain a margin of error of .5, with a confidence level of 95%?

```
CL = .95
zstar = qnorm(CL+.5*(1-CL))
ME = .5
sigma = 12
n = (zstar^2*sigma^2)/ME^2
ceiling(n)
```

`## [1] 2213`

A proportion is estimated based on a sample size of 150. The estimated proportion is .3. Construct a 90% confidence for the true value of the proportion.

```
# Code snippet to compute a confidence interval for a proportion
phat <- .3 # Estimated proportion
CL <- .90 # Required confidence level
n <- 150 # Sample size
zstar <- qnorm(CL+.5*(1-CL))
se.phat <-sqrt(phat*(1-phat)/n)
ME = zstar * se.phat
lb <- phat - ME
ub <- phat + ME
CI <- c(CL,ME,lb,phat,ub)
names(CI) <- c("Confidence Level", "Margin of Error", "lower Bound","phat","Upper Bound")
CI
```

```
## Confidence Level Margin of Error lower Bound phat
## 0.90000000 0.06154479 0.23845521 0.30000000
## Upper Bound
## 0.36154479
```

If a proportion is known to be approximately .3, how large a sample would be required to obtain a margin of error of .01 with a 95% confidence level?

```
# Code Snippet to compute a required sample size for a proportion
phat = .3
ME = .01
CL = .95
zstar = qnorm(CL+.5*(1-CL))
n = (phat*(1-phat)*zstar^2)/ME^2
ceiling(n)
```

`## [1] 8068`

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval.

```
# Code snippet to compute a confidence interval for a proportion
phat <- 408/865 # Estimated proportion
CL <- .95 # Required confidence level
n <- 865 # Sample size
zstar <- qnorm(CL+.5*(1-CL))
se.phat <-sqrt(phat*(1-phat)/n)
ME = zstar * se.phat
lb <- phat - ME
ub <- phat + ME
CI <- c(CL,ME,lb,phat,ub)
names(CI) <- c("Confidence Level", "Margin of Error", "lower Bound","phat","Upper Bound")
CI
```

```
## Confidence Level Margin of Error lower Bound phat
## 0.95000000 0.03326688 0.43840942 0.47167630
## Upper Bound
## 0.50494318
```

Of 90 adults selected randomly from one town, 69 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.

```
# Code snippet to compute a confidence interval for a proportion
phat <- 69/90 # Estimated proportion
CL <- .90 # Required confidence level
n <- 90 # Sample size
zstar <- qnorm(CL+.5*(1-CL))
se.phat <-sqrt(phat*(1-phat)/n)
ME = zstar * se.phat
lb <- phat - ME
ub <- phat + ME
CI <- c(CL,ME,lb,phat,ub)
names(CI) <- c("Confidence Level", "Margin of Error", "lower Bound","phat","Upper Bound")
CI
```

```
## Confidence Level Margin of Error lower Bound phat
## 0.9000000 0.0733327 0.6933340 0.7666667
## Upper Bound
## 0.8399994
```