The cost of weddings in the United States has skyrocketed in recent years. As a
ID: 2921377 • Letter: T
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 0.010 significance level is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised?
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State the null hypothesis and the alternate hypothesis. Use a .01 level of significance. (Enter your answers in thousands of dollars.)
State the decision rule for 0.010 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
Compute the value of the test statistic. (Negative amount should be indicated by a minus sign.Round your answer to 3 decimal places.)
What is the conclusion regarding the null hypothesis?
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 0.010 significance level is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised?
Explanation / Answer
Given that,
population mean(u)=10
sample mean, x =9.2625
standard deviation, s =0.7425
number (n)=8
null, Ho: >10
alternate, H1: <10
level of significance, = 0.01
from standard normal table,left tailed t /2 =2.998
since our test is left-tailed
reject Ho, if to < -2.998
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =9.2625-10/(0.7425/sqrt(8))
to =-2.809
| to | =2.809
critical value
the value of |t | with n-1 = 7 d.f is 2.998
we got |to| =2.809 & | t | =2.998
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -2.8094 ) = 0.01308
hence value of p0.01 < 0.01308,here we do not reject Ho
ANSWERS
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null, Ho: >=10
alternate, H1: <10
test statistic: -2.809
critical value: -2.998
decision: do not reject Ho
p-value: 0.01308
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