Please only answer if you will answer all parts to the question. Thank you Probl
ID: 3180601 • Letter: P
Question
Please only answer if you will answer all parts to the question. Thank you
Problem 2 (10 points, 2 points each) The Port Authority is interested in the amount of time commuters wait for a trolley to get out of town during rush hour. A random sample of 18 people who were leaving work during rush hour was taken. The amount of time they waited was recorded in minutes and stored in the Minitab dataset. Assume waiting times are normally distributed. a. Construct a 98% confidence interval for the mean amount of time commuters spend waiting for a trolley during rush hour. (You may use Minitab, but make sure you can do the calculations by hand.) b. Interpret the confidence interval from part (a)Explanation / Answer
(a)
SE = std. dev. / sqrt(n)
Here n = 18
Lower limit = mean - z*SE
upper limit = mean + z*SE
(b)
This means that for a sample size of 18 we are sure 98% sure that mean will fall into CI (4.0435, 6.7121).
(c)
H0: mu = 7
H1: mu not equal to 7
(d)
Reject null hypothesis as value 7 does not fall in the CI.
(e)
With std. dev. = 2.5, below are the computation
As we see CI shifted to right by the value 0.0387.
Mean 5.3778 std. dev. 2.4296 SE 0.5727 z-value for 98% CI 2.3300 Lower Limit 4.0435 Upper Limit 6.7121