Question

Here are summary statistics for randomly selected weights of newborn​ girls:

nequals=189189​,

x overbarxequals=30.530.5

​hg,

sequals=7.47.4

hg. Construct a confidence interval estimate of the mean. Use a

9090​%

confidence level. Are these results very different from the confidence interval

28.528.5

hgless than<muμless than<31.731.7

hg with only

1212

sample​ values,

x overbarxequals=30.130.1

​hg, and

sequals=3.13.1

​hg?

What is the confidence interval for the population mean

muμ​?

hgless than<muμless than<nothing

hg ​(Round to one decimal place as​ needed.)

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =30.5
standard deviation, s =7.4
sample size, n =189
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 7.4/ sqrt ( 189) )
= 0.538
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 = 188 d.f is 1.653
margin of error = 1.653 * 0.538
= 0.89
III.
CI = x ± margin of error
confidence interval = [ 30.5 ± 0.89 ]
= [ 29.61 , 31.39 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =30.5
standard deviation, s =7.4
sample size, n =189
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 188 d.f is 1.653
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 = [ 30.5 ± t a/2 ( 7.4/ Sqrt ( 189) ]
= [ 30.5-(1.653 * 0.538) , 30.5+(1.653 * 0.538) ]
= [ 29.61 , 31.39 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 29.6 , 31.4 ] 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
b.
Given that,
population mean(u)=28.5
sample mean, x =30.5
standard deviation, s =7.4
number (n)=189
null, Ho: μ=28.5
alternate, H1: μ!=28.5
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.653
since our test is two-tailed
reject Ho, if to < -1.653 OR if to > 1.653
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =30.5-28.5/(7.4/sqrt(189))
to =3.716
| to | =3.716
critical value
the value of |t α| with n-1 = 188 d.f is 1.653
we got |to| =3.716 & | t α | =1.653
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 3.7156 ) = 0.0003
hence value of p0.1 > 0.0003,here we reject Ho
ANSWERS
---------------
null, Ho: μ=28.5
alternate, H1: μ!=28.5
test statistic: 3.716
critical value: -1.653 , 1.653
decision: reject Ho
p-value: 0.0003
we have enough evidence to support the claim that these results very different from the confidence interval 28.5
c.
TRADITIONAL METHOD
given that,
sample mean, x =30.1
standard deviation, s =3.1
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 3.1/ sqrt ( 12) )
= 0.9
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 = 11 d.f is 1.796
margin of error = 1.796 * 0.9
= 1.6
III.
CI = x ± margin of error
confidence interval = [ 30.1 ± 1.6 ]
= [ 28.5 , 31.7 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =30.1
standard deviation, s =3.1
sample size, n =12
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 1.796
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 = [ 30.1 ± t a/2 ( 3.1/ Sqrt ( 12) ]
= [ 30.1-(1.796 * 0.9) , 30.1+(1.796 * 0.9) ]
= [ 28.5 , 31.7 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 28.5 , 31.7 ] 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