請各大大 , 幫幫手....好難的maths

2009-12-17 11:45 pm
The customer services manager of a bank wishes to estimate the true
average time it takes for a teller to serve a customer. A random sample of
the service time for 14 individual customers is collected. The results are
listed as follows (in minutes):

1.7 1.9 1.7 1.5 1.6 1.4 1.6 1.0 1.0 2.0 1.3 1.2 0.8 2.3

a With respect to the above given information:
i Explain why a Student’s t distribution should be used in
constructing a confidence interval for the true average amount of
time it takes to serve a customer. What assumption(s) must be
made initially?
ii State the condition(s) of when a normal distribution can be used
instead.

b Find a 90% confidence interval estimate for the true average time it
takes to serve a customer. Keep your final answers to two decimal
places.

c Given that the service times are normally distributed and the
population standard deviation is known to be 0.45 minutes, what
sample size is required if the manager wishes to be 98% confident
that the sample average is correct to within ±0.2 minutes of the true
average service time?

回答 (1)

2009-12-18 4:14 am
✔ 最佳答案
(a)(i) Since the population standard deviation is not known, and the sample size of 14 is small, Student's distribution has to be used. The assumption is that the data should be normally distributed.
(ii) A normal distribution should be used when the population standard deviation is known. A normal distribution can also be used if the sample size is big enough, for example > 30.
(b) From the provided data, the mean of the sample = 1.5
The standard deviation of the sample = 0.4036
t(0.05) = 1.771
90% confidence intervall = 1.5 +/- 1.771*0.4036 = [0.79, 2.21]
(c) Since the data is normally distributed with a population standard deviation of 0.45 known, we can use a normal distribution for estimation.
Standard deviation of the sample mean = 0.45/√n
z-score for 98% confidence interval is 2.326
0.2 / (0.45/√n) = 2.326
√n = 2.326*0.45/0.2
n = 27.4
The sample size should be at least 28


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