Question

The cost of a wedding 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 that the cost of a Caribbean wedding was less than $10 000. Listed below is the total cost (in $ thousands) for a sample of eight Caribbean weddings:

$9.7

$10.2

$9.1

$10.6

$9.3

$10.1

$11.0

$9.8

a. State the null hypothesis and the alternate hypothesis. (Enter the final answers in thousands of dollars.)

H₀: μ ≥           10 10 Correct

H₁: μ <           10 10 Correct

b. State the decision rule for 0.10 significance level. (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Reject H₀ if t <            Not attempted .

c. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Value of the test statistic           -0.133 -0.133 Incorrect

d. At the 0.10 significance level, is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised?

Do not reject  CorrectH₀. There is not enough  Correctevidence to conclude that the cost is less than $10,000

Answer 1

Solution:

(in $ thousands)

x	x2
9.7	94.09
10.2	104.04
9.1	82.81
10.6	112.36
9.3	86.49
10.1	102.01
11	121
9.8	96.04
---	---
∑x=79.8	∑x2=798.84

Mean ˉx=∑xn

=9.7+10.2+9.1+10.6+9.3+10.1+11+9.88

=79.88

=9.975

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√798.84-(79.8)28/7

=√798.84-796.005/7

=√2.835/7

=√0.405

=0.6364

This is the left tailed test .

The null and alternative hypothesis is ,

H0 :    ≥ 10000

Ha : < 10000

Test statistic = t

= ( - ) / S / n

= (9975-10000) / 636 / 8

= -0.111

Test statistic = t = -0.111

P-value =0.4573

= 0.10  

P-value >

0.4573 > 0.10

Do not reject the null hypothesis .

. There is not enough  Correctevidence to conclude that the cost is less than $10,000