Question

A random sample of 300 Americans planning long summer vacations in 2009 revealed an average planned expenditure of $1,076. The expenditure for all Americans planning long summer vacations has a normal distribution with a standard deviation 345. Give a 99% confidence interval for the mean planned expenditure by all Americans taking long summer vacations in 2009. Explain your answer in relation to this context.

Answer 1

Answer:

$(1024.69,\ 1127.31)$

Step-by-step explanation:

We know that the sample size was:

$n = 300$

The average was:

${\displaystyle {\overline {x}}}=1,076$

The standard deviation was:

$\s = 345$

The confidence level is

$1-\alpha = 0.99$

$\alpha=1-0.99\\\alpha=0.01$

The confidence interval for the mean is:

${\displaystyle{\overline {x}}} \± Z_{\frac{\alpha}{2}}*\frac{s}{\sqrt{n}}$

Looking at the normal table we have to

$Z_{\frac{\alpha}{2}}=Z_{\frac{0.01}{2}}=Z_{0.005}=2.576$

Therefore the confidence interval for the mean is:

$1,076\± 2.576*\frac{345}{\sqrt{300}}$

$1,076\± 51.31$

$(1024.69,\ 1127.31)$

This means that the mean planned spending of all Americans who take long summer vacations in 2009 is between $ 1024.69 and $ 1127.31