Question

Name : The owner of a specialty coffee shop wants to study coffee purchasing habits of customers of her shop. She selects a random sample of 60 customers during a certain week, with the following results. Amount spent: 7.25. S - $1.75 31 customers say they "definitely will" recommend the specialty coffee shop to family and friends. a) At the .05 level of significance is there evidence that the population mean amount spent by all of her customers is different from $6.50? STEPS Tasks Your Solution Develop the null and alternative hypotheses L. and H.) Specify the level of significance (a). sample size Determine what stat test to use Establish the decision/rejection rule Calculate the test statistic Make a decision in statistical terms and in practical business terms

Answer 1

1.

Null and Alternative hypothesis:

Ho : µ = 6.5

H1 : µ ≠ 6.5

2.

x̅ = 7.25, s = 1.75

n = 31

α = 0.05

3.

Test used : one sample t test

4.

df = n-1 = 30

Critical value :

Two tailed critical value, t-crit = T.INV.2T(0.05, 30) = 2.042

Reject Ho if t < -2.042 or if t > 2.042

5.

Test statistic:

t = (x̅ - µ)/(s/√n) = (7.25 - 6.5)/(1.75/√31) = 2.3862

6.

Decision:

t = 2.3862 < 2.042, Reject the null hypothesis.

Conclusion:

There is enough evidence to conclude that the population mean amount spent by all of her customers is different from $6.50 at 0.05 significance level.