Answer :
Answer:
(40, 41.537)
Step-by-step explanation:
Given that a random sample of 30 airline flights during a storm had a mean delay of 40 minutes with a standard deviation of 5 minutes.
n =30: Sample mean = 40 and sample std dev s = 5
a) The population mean delay time we want to estimate
b) Point estimate = sample mean=[tex]\bar x=40[/tex]
c) Std error of sample = [tex]\frac{s}{\sqrt{n} } \\= 0.913[/tex]
df = 39
t critical value for 95% one side = 1.684
Margin of error = [tex]1.684*SE=1.537[/tex]
Confidence interval one tailed = (40, 41.537)
We are 95% confidence that at random for samples of large size, the mean delay time would be within 40 and 41.537