Answer to Question #268770 in Statistics and Probability for jopert

Question

Suppose we would like to determine if the typical amount spent per customer for dinner is more than 20. A sample of 49 customers was randomly selected and the average amount spent was 22.60. Assume that the SD is known to be 2.50. Using a 0.01 level of significance, would we conclude the typical amount spent per customer is more than 20?

Answer 1

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

=NORM.S.DIST(7.28,0)

P-value $= 1.23 \times 10^{-12}$

P-value < 0.02

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.