An online travel agent that specializes in cruises wanted to compare the cost of

Question

An online travel agent that specializes in cruises wanted to compare the cost of cruising to different destinations on various cruise lines. The following random sample data show the average cost of a seven-day cruise in a standard room to two locations. Assume the population variances for the cruise fares for these destinations are equal. Complete parts a through c. Location 1 $999 $132 Location 2 $930 $125 Sample mean Sample standard deviation Sample size a. Construct a 95% confidence interval to estimate the difference between the average rates of cruising to location 1 vs. cruising to ocation 2 The 95% confidence interval is (TI). Round to two decimal places as needed.) b. What conclusions can be made about the difference in rates between these destinations? With 95% confidence, it can be said that the difference in rates between these destinations the endpoints of the confidence interval. The fact that this interval zero provides evidence that there | a significant difference between these two rates.

Explanation / Answer

Population variances are equal so,

poolsed sample variance sp2 = ((n-1)s12 + (m-1)s22)/(n+m-2) = ((7)(132)2 +(10)(125)2)/(8+11-2)

sp = 127.9287

(a) 95% confidence interval : (sample mean 1- sample mean 2) +- t0.025,17 ((sp)((1/n)+(1/m))0.5)

= (999-930)+-2.110((127.9287)((1/8)+(1/11))0.5) = (-56.43, 194.43)

(b) The above mentioned interval contains 0 so there is no significant difference between these two rates

(c) assumption - option (c) If both sample sizes are not atleast 30, then one must assume that the underlying poulations are normally distributed to perform this procedure. Also, one must assume that the samples are independent.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.