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:[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