Question

The cost of weddings in the United States has skyrocketed in recent years. As a result, 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 $10,000. Listed below is a total cost in $1000s for a sample of 8 Caribbean weddings.

9.1, 9.5, 10.1, 8.8, 8.5, 10.1, 9.8, 9.2
At the .005 significance level, is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised?

Answer 1

`t=(9.39-10)/((0.591)/(√(8)))=-2.917`

`p_v =P(t_((7))<-2.917)=0.0112`

If we compare the p value and the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 10000 at 5% of signficance.

Explanation:

Data given and notation

We have the following data: 9.1, 9.5, 10.1, 8.8, 8.5, 10.1, 9.8, 9.2

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=9.39` represent the mean height for the sample

`s=0.591` represent the sample standard deviation

`n=8` sample size

`\mu_o =10` represent the value that we want to test

`\alpha=0.01` 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is lower than 10000 (10), the system of hypothesis would be:

Null hypothesis:
`\mu \geq 10`

Alternative hypothesis:
`\mu < 10`

If we analyze the size for the sample is < 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

`t=(9.39-10)/((0.591)/(√(8)))=-2.917`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=8-1=7`

Since is a one side test the p value would be:

`p_v =P(t_((7))<-2.917)=0.0112`

Conclusion

If we compare the p value and the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 10000 at 5% of signficance.