Question

X 6.6.86 A random sample of size na 20 was drawn from a population of size N= 300. The measurements shown in the table to the right were obtained a. Estimate with an approximate 95% confidence interval b. Estimate p, the proportion of measurements in the population that are greater than 30, with an approximate 95% confidence interval a. An approximate 95% confidence interval estimate for pis 38.4 (Round to three decimal places as needed.) 4.873

Answer 1

6.6.86

a.

TRADITIONAL METHOD
given that,
sample mean, x =36.35
standard deviation, s =10.9076
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 10.9076/ sqrt ( 20) )
= 2.439
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 = 19 d.f is 2.093
margin of error = 2.093 * 2.439
= 5.105
III.
CI = x ± margin of error
confidence interval = [ 36.35 ± 5.105 ]
= [ 31.245 , 41.455 ]
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DIRECT METHOD
given that,
sample mean, x =36.35
standard deviation, s =10.9076
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/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 = [ 36.35 ± t a/2 ( 10.9076/ Sqrt ( 20) ]
= [ 36.35-(2.093 * 2.439) , 36.35+(2.093 * 2.439) ]
= [ 31.245 , 41.455 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 31.245 , 41.455 ] 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.

given data,
measurements in population that are greater than 30
variables 13 are in the given data ,
sample size is 20
possible chances (x)=13
sample size(n)=20
TRADITIONAL METHOD
given that,
possible chances (x)=13
sample size(n)=20
success rate ( p )= x/n = 0.65
I.
sample proportion = 0.65
standard error = Sqrt ( (0.65*0.35) /20) )
= 0.107
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.107
= 0.209
III.
CI = [ p ± margin of error ]
confidence interval = [0.65 ± 0.209]
= [ 0.441 , 0.859]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=13
sample size(n)=20
success rate ( p )= x/n = 0.65
CI = confidence interval
confidence interval = [ 0.65 ± 1.96 * Sqrt ( (0.65*0.35) /20) ) ]
= [0.65 - 1.96 * Sqrt ( (0.65*0.35) /20) , 0.65 + 1.96 * Sqrt ( (0.65*0.35) /20) ]
= [0.441 , 0.859]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.441 , 0.859] contains the true population proportion
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 proportion