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 35 ships that carry fewer than 500 passengers resulted in an average rating of 85.82, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.90. 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.)

Answer 1

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:  

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$  

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$  

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$.

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)  

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:  

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$  

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$  

The standard error is given by the following formula:  

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$  

And replacing we have:  

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$  

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

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

Confidence interval  

Now we have everything in order to replace into formula (1):  

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$  

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$  

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$.