Question

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?

29.1 28.5 28.8 29.4 29.8 29.8 30.1 30.6

Required:
a. State the null hypothesis and the alternate hypothesis. Use a 0.10 level of significance.
b. State the decision rule for 0.10 significance level.
c. Compute the value of the test statistic.
d. What is the conclusion regarding the null hypothesis?

Answer 1

a) Null hypothesis:
`\mu \geq 30`

Alternative hypothesis:
`\mu < 30`

b) Since we are conducting a left tailed test we need to find a quantile in the normal distribution who accumulates 0.1 of the area in the left and we got:

`z_(cric)= -1.28`

If the calculated value is lower than -1.28 we have enough evidence to reject the null hypothesis.

c)
`t=(29.51-30)/((0.698)/(√(8)))=-1.986`

d) Since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 10%. So then the true mean is significantly lower than 30000 for this case.

Explanation:

Data given and notation

For this case we have the following data:

29.1 28.5 28.8 29.4 29.8 29.8 30.1 30.6

We can calculate the mean and deviation with these formulas:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

`s = \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

`\bar X=29.51` represent the sample mean

`s=0.698` represent the sample standard deviation

`n=8` sample size

`\mu_o =30` represent the value that we want to test

`\alpha=0.1` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 30000, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 30`

Alternative hypothesis:
`\mu < 30`

Part b

Since we are conducting a left tailed test we need to find a quantile in the normal distribution who accumulates 0.1 of the area in the left and we got:

`z_(cric)= -1.28`

If the calculated value is lower than -1.28 we have enough evidence to reject the null hypothesis.

Part c

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

We can replace in formula (1) the info given like this:

`t=(29.51-30)/((0.698)/(√(8)))=-1.986`

Part d

Since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 10%. So then the true mean is significantly lower than 30000 for this case.