I need the solutions in detils Question 8 The range of a new type of mortar shel

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I need the solutions in detils

Question 8 The range of a new type of mortar shell is being investigated. The observed ranges, in meters, of 20 such shells are as follows: 2,100 1,950 2,043 2,210 2,018 1,984 1,992 2,218 2,152 2,106 2,072 2,096 2,244 1,962 1,938 1,898 2,103 2,206 2,007 1,956 Assuming that a shell's range is normally distributed, construct a) a 95 percent two-sided confidence interval for the mean range of a shell; b) a 99 percent two-sided confidence interval for the mean range of a shell c) Determine the largest value v that, "with 95 percent confidence," will be less than the mean range.

Explanation / Answer

8.

a.
TRADITIONAL METHOD
given that,
sample mean, x =2062.75
standard deviation, s =104.3435
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 104.3435/ sqrt ( 20) )
= 23.332
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 = 19 d.f is 2.093
margin of error = 2.093 * 23.332
= 48.834
III.
CI = x ± margin of error
confidence interval = [ 2062.75 ± 48.834 ]
= [ 2013.916 , 2111.584 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =2062.75
standard deviation, s =104.3435
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
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 = [ 2062.75 ± t a/2 ( 104.3435/ Sqrt ( 20) ]
= [ 2062.75-(2.093 * 23.332) , 2062.75+(2.093 * 23.332) ]
= [ 2013.916 , 2111.584 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2013.916 , 2111.584 ] 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
b.
TRADITIONAL METHOD
given that,
sample mean, x =2062.75
standard deviation, s =104.3435
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 104.3435/ sqrt ( 20) )
= 23.332
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 = 19 d.f is 2.861
margin of error = 2.861 * 23.332
= 66.753
III.
CI = x ± margin of error
confidence interval = [ 2062.75 ± 66.753 ]
= [ 1995.997 , 2129.503 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =2062.75
standard deviation, s =104.3435
sample size, n =20
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.861
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 = [ 2062.75 ± t a/2 ( 104.3435/ Sqrt ( 20) ]
= [ 2062.75-(2.861 * 23.332) , 2062.75+(2.861 * 23.332) ]
= [ 1995.997 , 2129.503 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 1995.997 , 2129.503 ] 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
c.
TRADITIONAL METHOD
given that,
sample mean, x =2062.75
standard deviation, s =104.3435
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 104.3435/ sqrt ( 20) )
= 23.332
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table,left tailed value of |t /2| with n-1 = 19 d.f is 1.729
margin of error = 1.729 * 23.332
= 40.341
III.
CI = x ± margin of error
confidence interval = [ 2062.75 ± 40.341 ]
= [ 2022.409 , 2103.091 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =2062.75
standard deviation, s =104.3435
sample size, n =20
level of significance, = 0.05
from standard normal table,left tailed value of |t /2| with n-1 = 19 d.f is 1.729
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 = [ 2062.75 ± t a/2 ( 104.3435/ Sqrt ( 20) ]
= [ 2062.75-(1.729 * 23.332) , 2062.75+(1.729 * 23.332) ]
= [ 2022.409 , 2103.091 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2022.409 , 2103.091 ] 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

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