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 0.05 significance level, is it reasonable to conclude the mean wedding cost is less than $30,000 as advertised? 29.128.528.829.429.829.830.130.6

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Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{236100}{8} = 29512.5[/tex]

Sum of squares of differences = 3408750

[tex]S.D = \sqrt{\frac{3408750}{7}} = 697.82[/tex]

Population mean, μ = $30,000

Sample mean, [tex]\bar{x}[/tex] = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars}[/tex] We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894[/tex]

Since,                  

[tex]t_{stat} < t_{critical}[/tex]

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

At the significance level, it is reasonable to conclude the mean wedding cost is less than $30,000 as advertised.

How to explain the significance level?

From the information, the null hypothesis is that the Caribbean wedding was less than $30,000.

The test statistic is -3.813. In this case, since the test statistic is more than -1.895 which is the decision rule, then we won't reject the null hypothesis as the cost is not less than $30000.

Therefore, it can be concluded that the mean wedding cost is less than $30,000 as advertised.

Learn more about significance level on:

https://brainly.com/question/15712887

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