The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than $30,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings. At the 0.10 significance level, is it reasonable to conclude the mean wedding cost is less than $30,000 as advertised?

29.7 29.4 31.7 29.0 29.1 30.5 29.1 29.8

Required:
a. State the null hypothesis and the alternate hypothesis.
b. State the decision rule for 0.05 significance level.
c. Compute the value of the test statistic.
d. What is the conclusion regarding the null hypothesis?

Summation(x - mean)² = (29.7 - 29.7875)^2 + (29.4 - 29.7875)^2 + (31.7 - 29.7875)^2 + (29.0 - 29.7875)^2 + (29.1 - 29.7875)^2 + (30.5 - 29.7875)^2 + (29.1 - 29.7875)^2 + (29.8 - 29.7875)^2 = 5.88875

Standard deviation = √(5.88875/8

s = 0.88

s = 0.88 × 1000 = $880

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0:µ ≤ 30000

For the alternative hypothesis,

H1:µ > 30000

This is a right tailed test.

b) The decision rule is to reject the null hypothesis if the significance level of 0.05 is greater than the probability value. If it is otherwise, we would fail to reject the null hypothesis.

c) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 29787.5

µ = population mean = 30000

s = samples standard deviation = 880

t = (29787.5 - 30000)/(880/√8) = - 0.68

We would determine the p value using the t test calculator. It becomes

p = 0.26

d) Since alpha, 0.1 < the p value, 0.26, then we would fail to reject the null hypothesis.