Arandom 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.

Step-by-step explanation:

We know that the sample size was:

The average was:

The standard deviation was:

The confidence level is

The confidence interval for the mean is:

Looking at the normal table we have to

Therefore the confidence interval for the mean is:

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

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6