Question

Your factory's inventory level was determined at 12 randomly selected times last year with the following results: 313, 891,153, 387, 584,162,742, 684, 277, 271,285,845.

a) find the typical inventory level throughout the whole year using the standard statistical summary.

b) identify the population.

c) find the 95% confidence interval for the population mean inventory level.

d) is the average of the measured inventory levels significantly different from 500 which is the number used for management budgeting purposes? Justify your answer.

Answer 1

TRADITIONAL METHOD
a.
the typical inventory level throughout the whole year using the standard statistical summary.
b.
the population mean is 500
given that,
sample mean, x =466.166
standard deviation, s =256.325
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 256.325/ sqrt ( 12) )
= 73.995
II.
margin of error = t α/2 * (standard 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 = 11 d.f is 2.201
margin of error = 2.201 * 73.995
= 162.862
III.
CI = x ± margin of error
confidence interval = [ 466.166 ± 162.862 ]
= [ 303.304 , 629.028 ]
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DIRECT METHOD
given that,
sample mean, x =466.166
standard deviation, s =256.325
sample size, n =12
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
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 = [ 466.166 ± t a/2 ( 256.325/ Sqrt ( 12) ]
= [ 466.166-(2.201 * 73.995) , 466.166+(2.201 * 73.995) ]
= [ 303.304 , 629.028 ]
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interpretations:
1) we are 95% sure that the interval [ 303.304 , 629.028 ] 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
c.
the 95% confidence interval for the population mean inventory level
95% sure that the interval [ 303.304 , 629.028 ]
d.
Given that,
population mean(u)=500
sample mean, x =466.166
standard deviation, s =256.325
number (n)=12
null, Ho: μ=500
alternate, H1: μ!=500
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =466.166-500/(256.325/sqrt(12))
to =-0.4572
| to | =0.4572
critical value
the value of |t α| with n-1 = 11 d.f is 2.201
we got |to| =0.4572 & | t α | =2.201
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.4572 ) = 0.6564
hence value of p0.05 < 0.6564,here we do not reject Ho
ANSWERS
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null, Ho: μ=500
alternate, H1: μ!=500
test statistic: -0.4572
critical value: -2.201 , 2.201
decision: do not reject Ho
p-value: 0.6564
we do not have enough evidence to support the claim that the average of the measured inventory levels significantly different from 500.