A travel agency reported that a random sample of 30 couples vacationing in Jamai
ID: 3226022 • Letter: A
Question
A travel agency reported that a random sample of 30 couples vacationing in Jamaica for a week spent an average of $4,750 with a sample standard deviation
of $100. A random sample of 30 couples vacationing in Bermuda spent an average of $5050 with a standard deviation of $125. The agency has been interested to see if there is a difference in the two means.
What is an appropriate null hypothesis for the difference in two means? (1 pt)
What assumption about the population variances must you make in order to test these two means? (1pt)
At 90% confidence level, what is the test statistic? (1pt)
What is the critical value? (1pt)
What is your conclusion concerning the null hypothesis? (1pt)
Explanation / Answer
Null HYpothesis : There is no difference between weekly amount spent by couples in Jamaica is equal to weekly amount by couples in Bermuda. Jamaica - Bermuda = 0
ALternative Hypothesis : There is significant difference between weekly amount spent by couples in Jamaica is equal to weekly amount by couples in Bermuda. Jamaica - Bermuda 0
Assumption : we will take assumption that variances of amount spent by couples on Jamaica and bermuda are equal.
Test Statistic:
xJamacia= $ 4750 and s1 = $ 100
XBermuda = $ 5050 and s2= $ 125
pooled standard Deviation sp= sqrt [ ((n1 -1)s12 + (n2-1)s22) /(n1+ n2-2)]
sp= sqrt [ (29 * 100 * 100 + 29 * 125 * 125 )/ 58] = 113.1923
t - test
t = (XBermuda- xJamacia)/ sp * sqrt [ 1/n1+ 1/n2] = ( 5050 - 4750)/ 113.1923 * sqrt [ 1/30 + 1/30 ]
t = 300/ 29.226 = 10.265
Critical value at dF = 58 and alpha = 0.90
tcritical= t0.10, 58 = +/- 1.67
so t > tcritical so we can reject the null hypothesis and conclude their is significant difference in mean value of spent in Jamaica in comparison to Bermuda by couples.
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