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Genrish500 [490]
3 years ago
5

A random sample of 300 Americans planning long summer vacations in 2009 revealed an average planned expenditure of $1,076. The e

xpenditure 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.
Mathematics
1 answer:
Rufina [12.5K]3 years ago
3 0

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 <em>the mean planned spending of all Americans who take long summer vacations in 2009 is between $ 1024.69 and $ 1127.31</em>

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