A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $23.71,$17.79,$29.87,$18.78,$28.76 Construct the 95% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Step 3 of 4 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

(16.904 ; 30.660)

Step-by-step explanation:

Given the data :

X = 23.71,17.79,29.87,18.78,28.76

Sample size = 5

Mean = (23.71+17.79+29.87+18.78,+28.76) / 5 = 23.782

Using a calculator, sample standard deviation :

Standard deviation, s = 5.54

Tcritical at α = 0.05, df = 5 - 1 = 4

Using tables ; Tcritical value = 2.7763

Confidence interval :

Mean ± Margin of Error

Margin of Error (MOE) = Tcritical * SE

Margin of Error = Tcritical * s/sqrt (n)

MOE = 2.7763 * 5.54/Sqrt(5) = 6.878

Confidence interval : 23.782 ± 6.878

Lower boundary : 23.782 - 6.878 = 16.904

Upper boundary = 23.782 + 6.878 = 30.66

(16.904 ; 30.660)

soo this can be solve by using the formula:

f = p( 1 + i)^ n

where f is the future profit

p is the initial profit

i is the percent increase

n is the number of years

a . f = 30000 ( 1.05)^n

b. f = 30000 ( 1.05)^15

f = $62,368