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 0.05 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

Answer 1

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:

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

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

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{236100}{8} = 29512.5$

Sum of squares of differences = 3408750

$S.D = \sqrt{\frac{3408750}{7}} = 697.82$

Population mean, μ = $30,000

Sample mean, $\bar{x}$ = $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

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

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975$

Now,

$t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894$

Since,                  

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it.

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