Answer :
Final answer:
Using statistical methods, we have constructed 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, finding that the new FAA recommendations align with our calculated intervals but are positioned conservatively towards the upper limits.
Explanation:
To answer the student's question about constructing and interpreting 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, and then commenting on the new FAA recommendations, we'll follow statistical methods.
Part A: Summer Weight
For the summer weight, the mean is 183 lb, the standard deviation is 20 lb, and the sample size (n) is 100 passengers. The formula for a 95% confidence interval is: mean ± (z*standard error), where the standard error is (standard deviation/√n). Using z=1.96 for 95% confidence, we calculate the interval as 183 ± (1.96*(20/√100)), which simplifies to 183 ± 3.92. Thus, the 95% confidence interval for summer is ±183 lb ± 3.92 lb.
Part B: Winter Weight
Similar calculations for the winter weight (mean = 190 lb, standard deviation = 23 lb, n = 100) yield a 95% confidence interval of 190 ± (1.96*(23/√100)), or 190 ± 4.51. Therefore, the 95% confidence interval for winter is ±190 lb ± 4.51 lb.
Part C: Comment on FAA Recommendations
The new FAA recommendations are 190 lb for summer and 195 lb for winter. These fall within our confidence intervals, but lean towards the upper limit, suggesting a conservative approach by the FAA, possibly to ensure safety by accommodating potential underestimations of passenger weights.
Answer:
a) [tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]
[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]
So on this case the 95% confidence interval would be given by (179.032;186.968)
b) [tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]
[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]
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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] 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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=100-1=99[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]
[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]
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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=100-1=99[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]
[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]
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
