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

Answer:

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) $t=\frac{29.51-30}{\frac{0.698}{\sqrt{8}}}=-1.986$    

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.

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:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

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

$\bar X=29.51$ 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

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

$t=\frac{29.51-30}{\frac{0.698}{\sqrt{8}}}=-1.986$    

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.