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III Question Help 4.80 5.19 5.70 The following data represent the pH of rain for a random sample of 12 rain dates. A normal p
The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data
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Answer #1

a.
point estimate for population mean is 5.0583 = 5.06
b.
TRADITIONAL METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.3948/ sqrt ( 12) )
= 0.11
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 * 0.11
= 0.25
III.
CI = x ± margin of error
confidence interval = [ 5.0583 ± 0.25 ]
= [ 4.81 , 5.31 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
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 = [ 5.0583 ± t a/2 ( 0.3948/ Sqrt ( 12) ]
= [ 5.0583-(2.201 * 0.11) , 5.0583+(2.201 * 0.11) ]
= [ 4.81 , 5.31 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 4.81 , 5.31 ] 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.
TRADITIONAL METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.3948/ sqrt ( 12) )
= 0.11
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 3.106
margin of error = 3.106 * 0.11
= 0.35
III.
CI = x ± margin of error
confidence interval = [ 5.0583 ± 0.35 ]
= [ 4.7 , 5.41 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 3.106
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 = [ 5.0583 ± t a/2 ( 0.3948/ Sqrt ( 12) ]
= [ 5.0583-(3.106 * 0.11) , 5.0583+(3.106 * 0.11) ]
= [ 4.7 , 5.41 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 4.7 , 5.41 ] 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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