10.15 ur ol uecreasing the confidence level. 10.13 a. The mean of a sample of 25

Question

10.15

ur ol uecreasing the confidence level. 10.13 a. The mean of a sample of 25 was calculated as x = 500, The sample was randomly drawn from a population with a standard deviation of 15.Estimate the population mean with 99% confidence. Repeat part (a) changing the population standard deviation to 30. b. c. Repeat part (a) changing the population stan- d. dard deviation to 60 Describe what happens to the confidence interval estimate when the standard deviation is increased. 10.14 a. A statistics practitioner randomly sampled 100 observations from a population with a standard deviation of 5 and found that x is 10. Estimate the population mean with 90% confidence. b. Repeat part (a) w b a sample size of 25 c. Repeat part (a) vin a smmple size of 10. d. Describe what happens to the confidence inter- val estimate when the sample size decreases 10.15 a. From the information given here determine the 95% confidence interval estimate of the popula- tion mean. = 100 = 20 n=25 b. Repeat part (a) with@ = 200. c. Repeat part (a) with x = 500. d. Describe what happens to the width of the con- fidence interval estimate when the sample mean increases. 10.16 a. A random sample of 100 observations was ran

Explanation / Answer

10.15
a.

TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =100
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 25) )
= 4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 4
= 7.84
III.
CI = x ± margin of error
confidence interval = [ 100 ± 7.84 ]
= [ 92.16,107.84 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =100
population size (n)=25
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 100 ± Z a/2 ( 20/ Sqrt ( 25) ) ]
= [ 100 - 1.96 * (4) , 100 + 1.96 * (4) ]
= [ 92.16,107.84 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [92.16 , 107.84 ] 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
[ANSWERS]
best point of estimate = mean = 100
standard error =4
z table value = 1.96
margin of error = 7.84
confidence interval = [ 92.16 , 107.84 ]

b.
sample mean = 200
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =200
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 25) )
= 4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 4
= 7.84
III.
CI = x ± margin of error
confidence interval = [ 200 ± 7.84 ]
= [ 192.16,207.84 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =200
population size (n)=25
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 200 ± Z a/2 ( 20/ Sqrt ( 25) ) ]
= [ 200 - 1.96 * (4) , 200 + 1.96 * (4) ]
= [ 192.16,207.84 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [192.16 , 207.84 ] 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
[ANSWERS]
best point of estimate = mean = 200
standard error =4
z table value = 1.96
margin of error = 7.84
confidence interval = [ 192.16 , 207.84 ]

c.
sample mean =500
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =500
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 25) )
= 4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 4
= 7.84
III.
CI = x ± margin of error
confidence interval = [ 500 ± 7.84 ]
= [ 492.16,507.84 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =500
population size (n)=25
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 500 ± Z a/2 ( 20/ Sqrt ( 25) ) ]
= [ 500 - 1.96 * (4) , 500 + 1.96 * (4) ]
= [ 492.16,507.84 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [492.16 , 507.84 ] 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
[ANSWERS]
best point of estimate = mean = 500
standard error =4
z table value = 1.96
margin of error = 7.84
confidence interval = [ 492.16 , 507.84 ]

d.
sample mean increases then respected confidence interval should be increased in each case

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