A random sample of eight customers was interviewed in order to find the number o

Question

A random sample of eight customers was interviewed in order to find the number of computers they planned to order next year. The results were 22, 18, 24, 47, 64, 32, 45, and 35. You are interested in knowing about the larger population that these customers represent.

a. Find the usual summary measure of the variability of individuals.
b. Approximately how far is the sample average from the population mean?
c. Find the 95% confidence interval for the population mean.

d.Find the 99% confidence interval for the population mean.

Explanation / Answer

a.
given that data,we find that
sample mean, x =35.875
standard deviation, s =15.4313
sample size, n =8

b.
the sample average from the population mean is depends on confidence intervals(95%,99%) values
c.
TRADITIONAL METHOD
given that,
sample mean, x =35.875
standard deviation, s =15.4313
sample size, n =8
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 15.4313/ sqrt ( 8) )
= 5.456
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 = 7 d.f is 2.365
margin of error = 2.365 * 5.456
= 12.903
III.
CI = x ± margin of error
confidence interval = [ 35.875 ± 12.903 ]
= [ 22.972 , 48.778 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =35.875
standard deviation, s =15.4313
sample size, n =8
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 2.365
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 = [ 35.875 ± t a/2 ( 15.4313/ Sqrt ( 8) ]
= [ 35.875-(2.365 * 5.456) , 35.875+(2.365 * 5.456) ]
= [ 22.972 , 48.778 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 22.972 , 48.778 ] 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.
TRADITIONAL METHOD
given that,
sample mean, x =35.875
standard deviation, s =15.4313
sample size, n =8
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 15.4313/ sqrt ( 8) )
= 5.456
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 3.499
margin of error = 3.499 * 5.456
= 19.09
III.
CI = x ± margin of error
confidence interval = [ 35.875 ± 19.09 ]
= [ 16.785 , 54.965 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =35.875
standard deviation, s =15.4313
sample size, n =8
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 3.499
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 = [ 35.875 ± t a/2 ( 15.4313/ Sqrt ( 8) ]
= [ 35.875-(3.499 * 5.456) , 35.875+(3.499 * 5.456) ]
= [ 16.785 , 54.965 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 16.785 , 54.965 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean

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