In: Statistics and Probability
Company A is trying to sell its website to Company B. As part of the sale, Company A claims that the average user of their site stays on the site for 10 minutes. To test this claim Company B collects the times (in minutes) below for a sample of 20 users. Assume normality.
Time |
0.5 |
1.7 |
17.7 |
8.5 |
1.1 |
0 |
4 |
10.1 |
8 |
4.6 |
13.5 |
1.3 |
20.6 |
2 |
14.2 |
1.4 |
1.2 |
6.9 |
16.3 |
1 |
Construct a 98% confidence interval for the true mean time spent on the web site.
a) What is the lower limit of the 98% interval? Give your answer to three decimal places.
b) What is the upper limit of the 98% interval? Give your answer to three decimal places.
TRADITIONAL METHOD
given that,
sample mean, x =6.73
standard deviation, s =6.5648
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6.5648/ sqrt ( 20) )
= 1.468
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 =
19 d.f is 2.539
margin of error = 2.539 * 1.468
= 3.727
III.
CI = x ± margin of error
confidence interval = [ 6.73 ± 3.727 ]
= [ 3.003 , 10.457 ]
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DIRECT METHOD
given that,
sample mean, x =6.73
standard deviation, s =6.5648
sample size, n =20
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 =
19 d.f is 2.539
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 6.73 ± t a/2 ( 6.5648/ Sqrt ( 20) ]
= [ 6.73-(2.539 * 1.468) , 6.73+(2.539 * 1.468) ]
= [ 3.003 , 10.457 ]
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interpretations:
1) we are 98% sure that the interval [ 3.003 , 10.457 ] contains
the true population mean
2) If a large number of samples are collected, and a confidence
interval is created
for each sample, 98% of these intervals will contains the true
population mean
a.
the lower limit of the 98% interval is 3.003
b.
the upper limit of the 98% interval is 10.457