Question

[ 14 pts) A fast food company wants to estimate the mean amount of time that it takes for customers to receive their food at its take out window. A random sample of 35 customers has a sample mean of 178.7 seconds and a sample standard deviation of 14.8 seconds. a) (2 pts) Find a point estimate for the mean amount of time at a take out window b) (12 pts) Without using the T-interval feature on the calculator construct a 95% confidence interval for the mean amount of time it takes to receive food

Answer 1

solution:

the given data as follows:

sample size = n = 35

sample mean = $\bar X$ = 178.7

sample standard deviation = S = 14.8

a) point estimate of mean amount of time at take out window

$\bar X$ = 178.7

b) confidence level = 95% = 0.95

significence level = $\alpha$ = 1-0.95 = 0.05

degree of freedom df = n-1 = 35-1 = 34

since population standard deviation is not known so we will use t-distribution

critical value of t = $t_c=t_{0.05/2,34df}=2.032$

t critical value is found from te t table

margin of error = $E=t_c*\frac{S}{\sqrt{n}}=2.032*\frac{14.8}{\sqrt{35}}=5.08$

confidence interval is $\bar X \pm E$

upper limit = 178.7 + 5.08 = 183.78

lowe limit = 178.7 - 5.08 = 173.62

so 95% confidence interval for mean = ( 173.62 , 183.78)