Answer:
[tex](1024.69,\ 1127.31)[/tex]
Step-by-step explanation:
We know that the sample size was:
[tex]n = 300[/tex]
The average was:
[tex]{\displaystyle {\overline {x}}}=1,076[/tex]
The standard deviation was:
[tex]\s = 345[/tex]
The confidence level is
[tex]1-\alpha = 0.99[/tex]
[tex]\alpha=1-0.99\\\alpha=0.01[/tex]
The confidence interval for the mean is:
[tex]{\displaystyle{\overline {x}}} \± Z_{\frac{\alpha}{2}}*\frac{s}{\sqrt{n}}[/tex]
Looking at the normal table we have to
[tex]Z_{\frac{\alpha}{2}}=Z_{\frac{0.01}{2}}=Z_{0.005}=2.576[/tex]
Therefore the confidence interval for the mean is:
[tex]1,076\± 2.576*\frac{345}{\sqrt{300}}[/tex]
[tex]1,076\± 51.31[/tex]
[tex](1024.69,\ 1127.31)[/tex]
This means that the mean planned spending of all Americans who take long summer vacations in 2009 is between $ 1024.69 and $ 1127.31
