Question

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 lb as a typical passenger weight (including carry-on luggage) in warm months and 185 lb as a typical weight in cold months. The Alaska Journal of Commerce (May 25, 2003) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 lb and a winter average of 190 lb. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 lb for the summer weights and 23 lb for the winter weights. A. Construct and interpret a 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers.B. Construct and interpret a 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers.C. The new FAA recommendations are 190 lb for sum- mer and 195 lb for winter. Comment on these recommen- dations in light of the confidence interval estimates from Parts (a) and (b).

Answer 1

Answer:

a) $183-1.984\frac{20}{\sqrt{100}}=179.032$    

$183+1.984\frac{20}{\sqrt{100}}=186.968$    

So on this case the 95% confidence interval would be given by (179.032;186.968)    

b) $190-1.984\frac{23}{\sqrt{100}}=185.437$    

$190+1.984\frac{23}{\sqrt{100}}=194.563$    

So on this case the 95% confidence interval would be given by (185.437;194.563)    

c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Part a : Summer

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=100-1=99$

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: "=-T.INV(0.025,99)".And we see that $t_{\alpha/2}=1.984$

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

$183-1.984\frac{20}{\sqrt{100}}=179.032$    

$183+1.984\frac{20}{\sqrt{100}}=186.968$    

So on this case the 95% confidence interval would be given by (179.032;186.968)    

Part b: Winter

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=100-1=99$

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: "=-T.INV(0.025,99)".And we see that $t_{\alpha/2}=1.984$

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

$190-1.984\frac{23}{\sqrt{100}}=185.437$    

$190+1.984\frac{23}{\sqrt{100}}=194.563$    

So on this case the 95% confidence interval would be given by (185.437;194.563)    

Part c

For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance