Question

A travel magazine conducts an annual survey where readers rate their favorite cruise ship. Ships are rated on a 10 point scale, with higher values indicating better service. A sample of 20 ships that carry fewer than 500 passengers resulted in a average rating of 6.93 with standard deviation 0.31. A sample of 55 ships that carry more than 500 passengers resulted in an average rating of 7.07 with standard deviation 0.6. statcrunch. 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?

Answer 1

Answer:

a) The point estimate of the difference between the populations is Md=-0.14.

b) The margin of error at 95% confidence is 0.212.

c) The 95% confidence interval for the difference between means is (-0.352, 0.072).

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1 (ships under 500 passengers), of size n1=20 has a mean of 6.93 and a standard deviation of 0.31.

The sample 2 (ships over 500 passengers), of size n2=55 has a mean of 7.07 and a standard deviation of 0.6.

The difference between sample means is Md=-0.14.

$M_d=M_1-M_2=6.93-7.07=-0.14$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.31^2}{20}+\dfrac{0.6^2}{55}}\\\\\\s_{M_d}=\sqrt{0.005+0.007}=\sqrt{0.011}=0.11$

The critical t-value for a 95% confidence interval is t=1.993.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.993 \cdot 0.11=0.212$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -0.14-0.212=-0.352\\\\UL=M_d+t \cdot s_{M_d} = -0.14+0.212=0.072$

The 95% confidence interval for the difference between means is (-0.352, 0.072).