The daytona beach tourism commission is interested in the average amount of money a typical college student spends per day during spring break. they survey 35 students and find that the mean spending is $63.57 with a standard deviation of $17.32. develop a 95% confidence interval for the population mean daily spending.

Answer & Step-by-step explanation:

The confidence interval formula is:  

I (1-alpha) (μ)= mean+- [(t(n-1))* S/sqrt(n)]  

alpha= is the proportion of the distribution tails that are outside the confidence interval. In this case, 5%.  

t(n-1)= t(34)= is the critical value of the t distribution with n-1 degrees of freedom for an area of alpha/2 (2.5%). In this case is 2.032  

S= sample standard deviation because the problem does not specify that the standard deviation is from the population. In this case $17.32

mean= $63.57

n= number of observations =35

Then, the confidence interval (90%):  

I 95%(μ)= 63.57+- [2.032*(17.32/sqrt(35))  

I 95%(μ)= 63.57+- [5.95)  

I 95%(μ)= [63.57-5.95; 63.57+5.95]  

I 95%(μ)= [57.62; 69.52]  

where r the answers

step-by-step explanation:

if she was to only spend a dollar she can get 1lb

step-by-step explanation: