A travel agency reported that a random sample of 30 couples vacationing in Jamai

Question

A travel agency reported that a random sample of 30 couples vacationing in Jamaica spent an average of $4500 with a sample standard deviation of $300. A random sample of 30 couples vacationing in Bermuda spent an average of $4900 with a sample standard deviation of $400. A travel agency would like to know if there is a difference in these two means at a 95% confidence level.

My calculations using excel were

p value= 0.0001

upper critical value= 2.0017

t statistic= -4.3818

g. Using the Critical Value Test (test statistic v., critical value), explain your conclusion concerning the null hypothesis? .

h. Using the p value test (p value vs. value), explain your conclusion concerning the null hypothesis?

i. If the Xbar values actually closely estimated the mean value spent for each vacation site, which site is likely to be more expensive?

Explanation / Answer

where we use t-test with

null hypothesis H0:mean(Jamaica)=mean(Bermuda) and

alternate hypothesis H1:mean(Jamaica) not = mean(Bermuda) and

t=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2) =-4.3818

and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2

(g) absolute t = 4.3818 is more than critical t(0.05,58)=2.0017 so we reject H0 and conclude that there is a difference in these two means

(h) p-value=0.0001 is less than alpha=0.05 ( 1-confidence=1-0.95=0.05), so we reject H0, and conclude that there is a difference in these two means

(i) Bermuda, since mean of Bermuda is more than Jamiaca

following calculation is done using ms-excel

sample mean s s2 n (n-1)s2 Jamaica 4500 300 90000 30 2610000 Bermuda 4900 400 160000 30 4640000 total= 9400 250000 60 7250000 sp2= 125000 sp= 353.5533906 t= -4.38178046 one tailed p-value= 0.00003 two tailed p-value= 0.00007

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