Question

You may need to use the appropriate appendix table or technology to answer this question. A magazine 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 39 ships that carry fewer than 500 passengers resulted in an average rating of 85.67, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.70. Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers. (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? (Use smaller cruise ships – larger cruise ships.) (b) At 95% confidence, what is the margin of error? (Round your answer to two decimal places.) (c) What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships? (Use smaller cruise ships – larger cruise ships. Round your answers to two decimal places.) to

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=85.67
standard deviation , σ1 =4.55
population size(n1)=39
y(mean)=81.7
standard deviation, σ2 =3.97
population size(n2)=44
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((20.7025/39)+(15.7609/44))
= 0.94
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.94
= 1.85
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (85.67-81.7) ± 1.85 ]
= [2.12 , 5.82]
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DIRECT METHOD
given that,
mean(x)=85.67
standard deviation , σ1 =4.55
number(n1)=39
y(mean)=81.7
standard deviation, σ2 =3.97
number(n2)=44
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 85.67-81.7) ±Z a/2 * Sqrt( 20.7025/39+15.7609/44)]
= [ (3.97) ± Z a/2 * Sqrt( 0.89) ]
= [ (3.97) ± 1.96 * Sqrt( 0.89) ]
= [2.12 , 5.82]
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interpretations:
1. we are 95% sure that the interval [2.12 , 5.82] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero
a.
point estimate of the difference between the population mean rating for ship that carry fewer than 500 passengers and the population mean rating for ship that carry
more 500 passengers is 3.97
b.
margin of error = 1.85
c.
95% sure that the interval [2.12 , 5.82]