Question

Because of safety considerations, the FAA changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 pounds as a typical passenger weight (including carry-on luggage). Frontier Airlines conducted a study to estimate the mean passenger weight (including carry-on luggage). They took a random sample of 100 passengers and found the mean weight of 190 pounds with a standard deviation of 23 pounds. Is there evidence that the FAA was correct in changing its guidelines because the typical passenger weight (including carry-on luggage) is now different from 180 pounds?

Answer 1

Answer:

$t=\frac{190-180}{\frac{23}{\sqrt{100}}}=4.348$    

$df=n-1=100-1=99$  

$p_v =2*P(t_{(99)}>4.348)=0.000033$  

Since the p value obtained for this case is a very low value we have enough evidence to reject the null hypothesis that the true mean is equal to 180 at many of the possible significance levels commonly used.  So then makes sense the claim that the true mean for the weigth is different from 180

Step-by-step explanation:

Information provided

$\bar X=190$ represent the mean weight

$s=23$ represent the sample standard deviation for the weight

$n=100$ sample size  

$\mu_o =180$ represent the value to compare

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to determine if the true mean weight is different from 180 pounds, the system of hypothesis would be:  

Null hypothesis:$\mu = 180$  

Alternative hypothesis:$\mu \neq 180$  

We don't know the population deviation for the variable of interest so then the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

Replacing the data given we got:

$t=\frac{190-180}{\frac{23}{\sqrt{100}}}=4.348$    

Now we can find the p value but first we need to find the degrees of freedom given by:

$df=n-1=100-1=99$  

Since we are conducting a two tailed test the p value can be calculated on this way:

$p_v =2*P(t_{(99)}>4.348)=0.000033$  

Since the p value obtained for this case is a very low value we have enough evidence to reject the null hypothesis that the true mean is equal to 180 at many of the possible significance levels commonly used.  So then makes sense the claim that the true mean for the weigth is different from 180