Question

Presto Pizza delivers pizzas throughout its local market area at no charge to the customer. Mr. Presto wants to estimate the average delivery distance. A random sample of 11 deliveries revealed the following round trip distances (in km.):
2.2     1.8     5.2     6.0     5.8     3.6     4.0     4.2      3.6     4.2     3.9
a) What is the point estimate of the population mean (in 1 decimal places)?
b) At an 90% level of confidence, what is the average delivery distance (km., in 1 decimal place)?
c) Is it reasonable to conclude that, generally, the delivery distance is 4.5 km.? Very briefly support your conclusion.

Answer 1

a)

Point estimate = sample mean, xbar = 4

b)

sample standard deviation, s = 1.3095
sample size, n = 11
degrees of freedom, df = n - 1 = 10

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.812

ME = tc * s/sqrt(n)
ME = 1.812 * 1.3095/sqrt(11)
ME = 0.715

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (4 - 1.812 * 1.3095/sqrt(11) , 4 + 1.812 * 1.3095/sqrt(11))
CI = (3.3 , 4.7)

c)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 4.5
Alternative Hypothesis, Ha: μ ≠ 4.5

Rejection Region
This is two tailed test, for α = 0.05 and df = 10
Critical value of t are -2.228 and 2.228.
Hence reject H0 if t < -2.228 or t > 2.228

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (4 - 4.5)/(1.3095/sqrt(11))
t = -1.266

P-value Approach
P-value = 0.2342
As P-value >= 0.05, fail to reject null hypothesis.

There is not sufficient evidence to conclude that the delivery distance is 4.5 km