91Ó°ÊÓ

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 \(85.36,\) and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40 (Condé Nast Traveler, February 2008 ). 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? b. \(\quad\) 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?

Step by step solution

01

Define the Problem Variables

Let \( \mu_1 \) be the mean rating for ships carrying fewer than 500 passengers, and \( \mu_2 \) for ships carrying 500 or more passengers. We have the sample sizes: \( n_1 = 37 \) and \( n_2 = 44 \). The sample means are \( \bar{x}_1 = 85.36 \) and \( \bar{x}_2 = 81.40 \). The standard deviations are \( \sigma_1 = 4.55 \) and \( \sigma_2 = 3.97 \). We are asked to find the point estimate and confidence interval for \( \mu_1 - \mu_2 \).

02

Calculate the Point Estimate

The point estimate for the difference between the population means \( \mu_1 - \mu_2 \) is the difference between the sample means: \( \bar{x}_1 - \bar{x}_2 \). Calculate it as follows:\[ \bar{x}_1 - \bar{x}_2 = 85.36 - 81.40 = 3.96 \].

03

Determine the Standard Error

To find the standard error (SE) of the difference between sample means, use the formula:\[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{4.55^2}{37} + \frac{3.97^2}{44}} \].Calculate each term step by step:- \( \frac{4.55^2}{37} = \frac{20.7025}{37} \approx 0.5595 \),- \( \frac{3.97^2}{44} = \frac{15.7609}{44} \approx 0.3582 \).Therefore, \[ SE = \sqrt{0.5595 + 0.3582} \approx \sqrt{0.9177} \approx 0.958 \].

04

Find the Margin of Error

For a 95% confidence level, use the z-value of 1.96. The margin of error (ME) is calculated as:\[ ME = z \times SE = 1.96 \times 0.958 \approx 1.877 \].

05

Construct the Confidence Interval

The 95% confidence interval for the difference in population means \( \mu_1 - \mu_2 \) is given by:\[ (\bar{x}_1 - \bar{x}_2) \pm ME = 3.96 \pm 1.877 \].Calculate the values:- Lower bound: \( 3.96 - 1.877 \approx 2.083 \),- Upper bound: \( 3.96 + 1.877 \approx 5.837 \).Therefore, the confidence interval is \( (2.083, 5.837) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate

The point estimate is a statistical way to give us the best guess of a population parameter based on a sample from that population. When you hear about point estimates, you're essentially dealing with a single value that represents the characteristic you're estimating. In our case, we want to estimate the difference between the average ratings of two types of cruise ships. Here, the point estimate for the difference between the population mean ratings (\( \mu_1 - \mu_2 \)) of the two groups is calculated by subtracting the average rating of larger ships (500 or more passengers) from that of smaller ones (fewer than 500 passengers). The formula for that is: \[ \bar{x}_1 - \bar{x}_2 \approx 3.96 \] This number, 3.96, suggests that on average, smaller ships have a higher rating by this amount in the sample.

Standard Error

Standard error (SE) helps to quantify the amount of variability or dispersion of a sample statistic. It's especially useful for understanding how precise our point estimate is. Simply put, standard error helps to gauge the accuracy of our sample mean as a representation of the population mean.To find the SE of the difference between the averages (means) from our two samples, we use the formula:\[SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]Where:

• \( \sigma_1 \) and \( \sigma_2 \) are the standard deviations of the two groups.
• \( n_1 \) and \( n_2 \) are the sample sizes of the two groups.

In our problem, the standard error calculates to approximately 0.958, which indicates the extent of variability within the difference in the mean ratings.With a smaller SE, our point estimate becomes more reliable.

Margin of Error

The margin of error (ME) tells us about the range within which we can be reasonably certain the true population parameter lies. It gives a "safety buffer" around our point estimate, reflecting the maximum expected difference between our point estimate of a mean and the actual population mean.To calculate the margin of error, we multiply the standard error by the critical value (z-value) from the standard normal distribution that corresponds to our desired confidence level.In a 95% confidence interval, the critical z-value is 1.96.Hence, the margin of error is:\[ ME = z \times SE = 1.96 \times 0.958 \approx 1.877\]This tells us that the range of values within which we can confidently say the true population mean difference lies, extends roughly 1.877 units to either side of the point estimate.

Population Mean Difference

The population mean difference (\( \mu_1 - \mu_2 \)) is what we're trying to estimate using our sample data. This difference helps us understand if there's a significant disparity in the ratings given to different types of cruise ships.After calculating the point estimate (3.96) and the margin of error (1.877), we can construct a confidence interval to give a range for this difference:\[(\bar{x}_1 - \bar{x}_2) \pm ME\]The result is a confidence interval of around (2.083, 5.837).This range indicates that we're 95% confident that the actual mean difference in ratings between smaller and larger cruise ships lies between 2.083 and 5.837. It suggests that smaller ships may indeed have a higher average rating but also underscores the uncertainty inherent in statistical sampling.