Question

In: Statistics and Probability

Company A is trying to sell its website to Company B. As part of the sale,...

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.

Solutions

Expert Solution

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 ]
-----------------------------------------------------------------------------------------------
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 ]
-----------------------------------------------------------------------------------------------
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


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