Condé Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service. A sample of 37 ships that carry fewer than 500 passengers resulted in an average rating of 85.33, and a sample of 43 ships that carry 500 or more passengers provided an average rating of 81.9. Assume that the population standard deviation is 4.58 for ships that carry fewer than 500 passengers and 3.95 for ships that carry 500 or more passengers.

Round your all answers to two decimal places.

a. What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers?

b. At 95% confidence, what is the margin of error?

c. What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

[tex]\\[/tex](a) Let [tex]x_{1}[/tex] represent the average ratings of ships that carry fewer than 500 passengers and [tex]x_{2}[/tex] represent the ship that carry more than 500 passengers. [tex]\\[/tex]Then , the point of estimate of the difference between the population mean rating for the ships that carry fewer than 500 passengers and the population mean rating for ship that carry 500 or more passengers is given as :[tex]\\[/tex] [tex]x_{1}[/tex] – [tex]x_{2}[/tex] , which is

[tex]\\[/tex]85.33 – 81. 9

[tex]\\[/tex]= 3.43

[tex]\\[/tex](b) To find the margin error using 95% Confidence Interval and [tex]Z_{\frac{\alpha }{2} }[/tex] = 1.96

[tex]\\[/tex]Since we are dealing with two samples , the formula is given as

[tex]x_{1}[/tex] – [tex]x_{2}[/tex] ± [tex]Z_{\frac{\alpha }{2} }[/tex] [tex]\sqrt{\frac{sigma1^{2} }{n1}+\frac{sigma2}{n2} }[/tex]

[tex]\\[/tex]Where sigma is the standard deviation

[tex]\\[/tex]We have, 3.43 ± 1.96 [tex]\sqrt{\frac{4.58^{2} }{37}+\frac{3.95^{2} }{43} }[/tex]

[tex]\\[/tex]= 3.43 ± 1.96( 0.9643)

[tex]\\[/tex]= 3.43 ± 1.89

[tex]\\[/tex]Therefore the margin error is 1.89

[tex]\\[/tex](c) the 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships implies

[tex]\\[/tex]3.43 – 1.89 t0 3.43 + 1.89

[tex]\\[/tex]= 1.54 to 5.32