Question

Suppose we would like to determine the typical amount spent per customer for dinner. A

sample of 49 customers was randomly selected and the average amount spent was $22.60.

Assume that the amount spent is normally distributed and the population SD is known to be

$2.50.

a. Construct a 95% confidence interval for μ, the average amount spent per customer for

dinner.

b. Using a 0.02 level of significance, would we conclude the typical amount spent per

customer is more than $20.00?

Answer 1

a. We have to use standard normal distribution because population standard deviation is known.

$N = 49 \\ \bar{x} = 22.60 \\ σ = 2.5$

Confidence level = 0.95

Level of significance = 0.05

Critical value $Z_{α/2} = Z_{0.025}=1.96$

Using Excel

= NORMSINV( 0.025 ) = 1.96

Formula for confidence interval for mean:

$CI = \bar{x}±Z_{α/2} \times \frac{σ}{\sqrt{n}} \\ = 22.60±1.96 \times \frac{2.5}{\sqrt{49}} \\ = 22.60±0.7$

95% confidence interval for the average amount spent per customer for dinner at this new restaurant is

Lower limit = 22.60 - 0.7 = 21.9

Upper limit = 22.60 + 0.7 = 23.3

b. H₀: μ = 20

H₁: μ > 20

Level of significance α=0.02

Test-statistic

$X = \frac{bar{x}-μ}{s/ \sqrt{n}} \\ = \frac{22.6-20}{2.50/ \sqrt{49}} \\ = 7.28$

For α=0.02 and right-tailed, using the NORM.S.DIST function of Excel

P-value < 0.00001

Since P-value < α

Reject the null hypothesis.

There is sufficient evidence to conclude the typical amount spent per customer is more than $20.00.