A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard deviation of $50.30. The 95% confidence interval around this sample mean is .

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $393.50

Standard deviation r = $50.30

Number of samples n = 25

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

$393.50+/-1.96($50.30/√25)

$393.50+/-1.96($10.06)

$393.50+/-$19.7176

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

imma say 16%. sorry if its wrong. i'm doing the same stuff right now

step-by-step explanation: