8.4 #7 - Confidence Interval Review

A random sample of 100 students is selected from a certain school. They are given an IQ test which has a known standard deviation of 11. The sample mean is found to be 112. Determine a 98% confidence interval for estimating

the mean school intelligence. Round to the nearest tenth (one decimal place).

Answer explanation

This is means with known standard deviation, so a z statistic is used (z*).
Mean = 112
sigma = 11
n = 100
z* for a 98% CI = 2.32
$112\pm2.326\ \left(\frac{11}{\sqrt{100}}\right)$

In a random sample of 250 high school students, it was found that 85% of them later graduated. Find the 98% confidence interval for the true proportion of all high school students who graduate. Round to the nearest tenth (one decimal place).

Answer explanation

This is categorical (proportions) so a z-test is used.
p-hat = 0.85
sigma = $\sqrt{\frac{.85\left(.15\right)}{250}}$

z* for 98% confidence interval = 2.326
$.85\ \pm\ 2.326\sqrt{\frac{.85\left(.15\right)}{250}}$

A survey of hospital records of 35 randomly selected patients suffering from a particular disease indicated that the average hospital stay was 10 days with a standard deviation of 2.1 days. Find a 99% confidence interval. Round to the nearest hundredth (two decimal places).

Answer explanation

This data is quantitative (means), but the standard deviation comes from the sample, so a t-interval is used.
mean = 10
s = 2.1
n = 35
df = 34
t* at 99% Conf. level = 2.750

$10\pm2.75\left(\frac{2.1}{\sqrt{35}}\right)$

A forester wishes to estimate the mean growth of seedlings in a large timber plot since last year. A random sample of 100 seedlings is taken and it is found that the one-year growth yields an average of 12.8 cm and s = 2.5 cm. Give a 95% confidence interval estimate for Mu, the average growth of all seedlings in the plot. Round to the nearest tenth (one decimal place).

Answer explanation

This is quantitative data (means) with the sample standard deviation (population SD is unknown); therefore it is a t-interval.
mean = 12.8
s = 2.5
n = 100
df = 99
t* at 95% conf. level = 1.99

 $12.8\pm1.99\left(\frac{2.5}{\sqrt{100}}\right)$  

A simple random sample of 730 registered voters shows that 65% of the registered voters favor raising taxes to pay for state parks. Construct a 90% confidence interval for the percent of the population of all registered voters who favor raising taxes to pay for state parks. Round to the nearest hundredth (two decimal places).

Answer explanation

This is categorical data (proportions) so this will be a z interval.
p-hat = 0.65
sigma = $\sqrt{\frac{.65\left(1-.65\right)}{730}}$

z* at a 90% conf. level = 1.645
$0.65\pm1.645\left(\sqrt{\frac{.65\left(.35\right)}{730}}\right)$