The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have thelr 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? 29.7 29.4 31.7 29.0 29.1 30.5 29.1 29.8 Excel Data a. State the null hypothesis and the alternate hypothesis. (Enter your answers in thousands of dollars.) b. State the decision rule for 0.05 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) H0 if t c. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) of the test statistic d. What is the conclusion regarding the null hypothesis? o not reject cost is not less n $30 Reference links for lation Mean ulation Sta

Step-by-step explanation:

The mean of the set of data given is

Mean = (29.7 + 29.4 + 31.7 + 29.0 + 29.1 + 30.5 + 29.1 + 29.8)/8 = 29.788 = $29788

Standard deviation = √(summation(x - mean)/n

n = 8

Summation(x - mean) = (29700 - 29788)^2 + (29400 - 29788)^2 + (31700 - 29788)^2 + (29000 - 29788)^2 + (29100 - 29788)^2 + (30500 - 29788)^2 + (29100 - 29788)^2 + (29800 - 29788)^2 = 5888752

Standard deviation = √(5888752/8) = 857.96

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ < 30000

For the alternative hypothesis,

µ > 30000

It is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8,

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 29788

µ = population mean = 30000

s = samples standard deviation = 857.96

t = (29788 - 30000)/(857.96/√8) = - 0.7

We would determine the p value using the t test calculator. It becomes

p = 0.253

Since alpha, 0.05 < than the p value, 0.253, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, the sample data did not show evidence that the mean wedding cost is not less than $30,000 as advertised

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step-by-step explanation:

i think asking google will you get the answer to that question