An inkjet cartridge manufacturer needs to determine the mean number of pages pri
ID: 3325606 • Letter: A
Question
An inkjet cartridge manufacturer needs to determine the mean number of pages printed per cartridge. A random sample of 16 cartridges printed the following number of pages:
2,650 2,480 2,550 2,820 2,370 2,420 2,450 2,620
2,770 2,280 2,680 2,750 2,420 2,360 2,670 2,710
(a) What assumptions must you make to obtain a confidence interval for the mean number of pages printed per cartridge?
(b) Find a 95% confidence interval estimate for the mean number of pages printed per cartridge. Present all your work.
(c) Interpret the confidence interval obtained in (b).
(d) The managers are not totally convinced with the confidence interval estimated in (b), because it is too wide. They would like to be 99% sure that the sample mean is within 50 pages of the true mean. What sample size is required to meet this requirement? Assume the population standard deviation equal to the one identified for the sample of 16 cartridges.
Explanation / Answer
a.
sample mean, x =2562.5
standard deviation, s =167.4714
sample size, n =16
b.
TRADITIONAL METHOD
given that,
sample mean, x =2562.5
standard deviation, s =167.4714
sample size, n =16
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 167.4714/ sqrt ( 16) )
= 41.868
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 15 d.f is 2.131
margin of error = 2.131 * 41.868
= 89.22
III.
CI = x ± margin of error
confidence interval = [ 2562.5 ± 89.22 ]
= [ 2473.28 , 2651.72 ]
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DIRECT METHOD
given that,
sample mean, x =2562.5
standard deviation, s =167.4714
sample size, n =16
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 15 d.f is 2.131
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 = [ 2562.5 ± t a/2 ( 167.4714/ Sqrt ( 16) ]
= [ 2562.5-(2.131 * 41.868) , 2562.5+(2.131 * 41.868) ]
= [ 2473.28 , 2651.72 ]
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c.
interpretations:
1) we are 95% sure that the interval [ 2473.28 , 2651.72 ] 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
d.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 167.4714
ME =50
n = ( 2.58*167.4714/50) ^2
= (432.08/50 ) ^2
= 74.68 ~ 75