Question

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 10 users. Assume normality.

Time 1.6

25.9

7

23.3

7.3

8.8

18.5

8.6

10.8

12

Construct a 95% confidence interval for the true mean time spent on the web site.

a) What is the lower limit of the 95% interval? Give your answer to three decimal places. Enter 0 if your lower limit is less than 0.

b) What is the upper limit of the 95% interval? Give your answer to three decimal places.

c) Based on this data, do you believe the claim made by Company A?

Yes because 10 is not inside the interval.

No because 10 is not inside the interval.

Yes because 10 is inside the interval.

No because 10 is inside the interval.

d) Which of the following assumptions should be checked before constructing the above confidence interval?

the data need to have small variance

the data need to follow a t distribution

the data need to be skewed

the data need to follow a normal distribution

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =12.38
standard deviation, s =7.7474
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 7.7474/ sqrt ( 10) )
= 2.45
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 2.262
margin of error = 2.262 * 2.45
= 5.542
III.
CI = x ± margin of error
confidence interval = [ 12.38 ± 5.542 ]
= [ 6.838 , 17.922 ]
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DIRECT METHOD
given that,
sample mean, x =12.38
standard deviation, s =7.7474
sample size, n =10
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 2.262
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 = [ 12.38 ± t a/2 ( 7.7474/ Sqrt ( 10) ]
= [ 12.38-(2.262 * 2.45) , 12.38+(2.262 * 2.45) ]
= [ 6.838 , 17.922 ]
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interpretations:
1) we are 95% sure that the interval [ 6.838 , 17.922 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

a.
the lower limit of the 95% interval = 6.838
b.
the upper limit of the 95% interval = 17.922
c.
yes,
believe the claim made by Company A
Yes because 10 is inside the interval
d.
the following assumptions should be checked before constructing the above confidence interval.
the data need to follow a normal distribution