Question

QUESTION 5:   A certain type of aircraft battery is known to have a lifetime, which is normally distributed.

1. Suppose that an operator selects 16 batteries from that population. The average lifetime of batteries is found as 1200 days in the sample. Assuming population standard deviation is known to be 200 days, find the 98% confidence interval for the population average. (Calculate the margin of error, lower and upper limits for the 98% confidence interval)*ANSWER USING EXCEL FUNCTIONS *                                                                                         
2. Suppose that an operator selects 16 batteries from that population. The average lifetime of batteries is found as 1200 days in the sample. Assuming population standard deviation is known to be 500 days, find the 98% confidence interval for the population average. (Calculate the margin of error, lower and upper limits for the 98% confidence interval) * ANSWER USING EXCEL FUNCTIONS *
3. Compare the width of the confidence intervals in (a) and (b). If they are different, give reasons why they are different.          
4. Suppose that an operator selects 16 batteries from that population. In that sample, average lifetime of batteries is found as 1100 days with a standard deviation of 80 days. Find the 90% confidence interval for the population average. (Calculate the margin of error, lower and upper limits for the 90% confidence interval) * ANSWER USING EXCEL FUNCTIONS *
5. Suppose that an operator selects 25 batteries from that population. In that sample, average lifetime of batteries is found as 1100 days with a standard deviation of 80 days. Find the 90% confidence interval for the population average. (Calculate the margin of error, lower and upper limits for the 90% confidence interval) * ANSWER USING EXCEL FUNCTIONS *
6. Compare the width of the confidence intervals in (d) and (e). If they are different, give reasons why they are different.

- Thank you in advance

Answer 1

5.

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =200
sample mean, x =1200
population size (n)=16
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 200/ sqrt ( 16) )
= 50
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
margin of error = 2.326 * 50
= 116.3
III.
CI = x ± margin of error
confidence interval = [ 1200 ± 116.3 ]
= [ 1083.7,1316.3 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =200
sample mean, x =1200
population size (n)=16
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 1200 ± Z a/2 ( 200/ Sqrt ( 16) ) ]
= [ 1200 - 2.326 * (50) , 1200 + 2.326 * (50) ]
= [ 1083.7,1316.3 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [1083.7 , 1316.3 ] 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
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =500
sample mean, x =1200
population size (n)=16
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 500/ sqrt ( 16) )
= 125
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
margin of error = 2.326 * 125
= 290.75
III.
CI = x ± margin of error
confidence interval = [ 1200 ± 290.75 ]
= [ 909.25,1490.75 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =500
sample mean, x =1200
population size (n)=16
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 1200 ± Z a/2 ( 500/ Sqrt ( 16) ) ]
= [ 1200 - 2.326 * (125) , 1200 + 2.326 * (125) ]
= [ 909.25,1490.75 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [909.25 , 1490.75 ] 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
Compare the width of the confidence intervals in (a) and (b)
a.
98% sure that the interval [1083.7 , 1316.3 ]
b.
98% sure that the interval [909.25 , 1490.75 ]

c.
TRADITIONAL METHOD
given that,
sample mean, x =1100
standard deviation, s =80
sample size, n =16
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 80/ sqrt ( 16) )
= 20
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 1.753
margin of error = 1.753 * 20
= 35.06
III.
CI = x ± margin of error
confidence interval = [ 1100 ± 35.06 ]
= [ 1064.94 , 1135.06 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1100
standard deviation, s =80
sample size, n =16
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 1.753
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 = [ 1100 ± t a/2 ( 80/ Sqrt ( 16) ]
= [ 1100-(1.753 * 20) , 1100+(1.753 * 20) ]
= [ 1064.94 , 1135.06 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 1064.94 , 1135.06 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
d.
TRADITIONAL METHOD
given that,
sample mean, x =1100
standard deviation, s =80
sample size, n =25
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 80/ sqrt ( 25) )
= 16
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 1.711
margin of error = 1.711 * 16
= 27.376
III.
CI = x ± margin of error
confidence interval = [ 1100 ± 27.376 ]
= [ 1072.624 , 1127.376 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1100
standard deviation, s =80
sample size, n =25
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 1.711
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 = [ 1100 ± t a/2 ( 80/ Sqrt ( 25) ]
= [ 1100-(1.711 * 16) , 1100+(1.711 * 16) ]
= [ 1072.624 , 1127.376 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 1072.624 , 1127.376 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
Compare the width of the confidence intervals in (d) and (c),
c.
90% sure that the interval [ 1064.94 , 1135.06 ]
d.
90% sure that the interval [ 1072.624 , 1127.376 ]