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 $000 for a sample of 8 Caribbean weddings. At the 0.025 significance level is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised? 12.4 8.5 10.1 12.6 10.5 9.1 9.8 11.2

a. State the null hypothesis and the alternate hypothesis. Use a .025 level of significance. (Enter your answers in thousands of dollars.) H0: μ ≥ H1: μ <

b. State the decision rule for 0.025 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Reject 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.) Value of the test statistic

d. What is the conclusion regarding the null hypothesis? H0. The cost is than $10,000.

Answer 1

(a)

Ho: ≥ 10000

Ha: < 10000

The Null hypothesis states that the mean wedding cost in Carribbean is greater than or equal to 10000

Alternative hypotheiss states that the mean wedding cost in Caribbean is less than 10000.

(b) Decision rule :

level of significance is 0.025

If p value is ≤ 0.025, we reject the Null hypothesis.

If p value is > 0.025, we fail to reject the Null hypothesis.

t critical at df = 7 , level of significance = 0.025

Reject Ho: If t stat < t critical (- 2.365)

Fail to reject Ho: If t stat > t ciritical ( -2.365)

(C) Test statistics

n = 8

sample mean = sum of all terms / no of terms = 84200 /8 = 10525

sample sd = s

data	data-mean	(data - mean)2
12400	1875	3515625
8500	-2025	4100625
10100	-425	180625
12600	2075	4305625
10500	-25	625
9100	-1425	2030625
9800	-725	525625
11200	675	455625

Assuming that the data is normally distributed and also as the population sd is not given, we will calculate t stat

t  = 1.011

(d)

p value = TDIST (t statistics, df, tails) = TDIST (1.011, 7,1) = 0.1729

As the p value (0.1729) is > 0.025, we fail to reject the Null hypothesis.

As the t stat (1.011) > t critical (-2.365), we fail to reject the Null hypothesis.

Hence, we do not have sufficient evidence to believe that the mean wedding cost is Carribean is less than 10000.