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?

a) Null hypothesis:  

Alternative hypothesis:  

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:

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

c)    

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.

Step-by-step 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:

represent the sample mean

represent the sample standard deviation

sample size  

represent the value that we want to test

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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:  

Alternative hypothesis:  

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:

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:

 (1)  

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

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.

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