Temperatures Listed below are temperatures (\'F) in San Francisco, Los Angeles,

Question

Temperatures Listed below are temperatures ('F) in San Francisco, Los Angeles, and ramento from January of a recent year. Construct the 95% confidence interval estimates the mean January temperature in each of the three cities, and then compare the results. of Sac San Francisco 47.5 450 46.0 480 51.0 52.0 52.5 54.5 57.0 55.0 51.5 54.5 55.0 50.5 52.0 53.0 545 570 56.0 55.0 55.0 52.0 56.0 565 56.0 56.0 61.0 565 575 57.0 61.0 Los Angeles 47 47 51 52 65 69 66 61 64 57 58 58 59 58 62 62 62 60 58 56 59 61 62 62 64 66 62 64 66 63 61 Sacramento 47 41 41 46 49 51 53 50 51 49 49 54 58 47 46 49 53 51 53 52 56 52 57 6356 53 51 56 56 59 59

Explanation / Answer

3.

a.
confidence inteval 95% for san francisco
TRADITIONAL METHOD
given that,
sample mean, x =53.9354
standard deviation, s =3.8138
sample size, n =31
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 3.8138/ sqrt ( 31) )
= 0.685
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 = 30 d.f is 2.042
margin of error = 2.042 * 0.685
= 1.399
III.
CI = x ± margin of error
confidence interval = [ 53.9354 ± 1.399 ]
= [ 52.537 , 55.334 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =53.9354
standard deviation, s =3.8138
sample size, n =31
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 30 d.f is 2.042
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 = [ 53.9354 ± t a/2 ( 3.8138/ Sqrt ( 31) ]
= [ 53.9354-(2.042 * 0.685) , 53.9354+(2.042 * 0.685) ]
= [ 52.537 , 55.334 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 52.537 , 55.334 ] 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.
confidence inteval 95% for Los angles
TRADITIONAL METHOD
given that,
sample mean, x =60.0645
standard deviation, s =5.2404
sample size, n =31
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 5.2404/ sqrt ( 31) )
= 0.941
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 = 30 d.f is 2.042
margin of error = 2.042 * 0.941
= 1.922
III.
CI = x ± margin of error
confidence interval = [ 60.0645 ± 1.922 ]
= [ 58.143 , 61.986 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =60.0645
standard deviation, s =5.2404
sample size, n =31
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 30 d.f is 2.042
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 = [ 60.0645 ± t a/2 ( 5.2404/ Sqrt ( 31) ]
= [ 60.0645-(2.042 * 0.941) , 60.0645+(2.042 * 0.941) ]
= [ 58.143 , 61.986 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 58.143 , 61.986 ] 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.
confidence inteval 95% for sacramento

TRADITIONAL METHOD
given that,
sample mean, x =51.8709
standard deviation, s =5.03151
sample size, n =31
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 5.03151/ sqrt ( 31) )
= 0.904
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 = 30 d.f is 2.042
margin of error = 2.042 * 0.904
= 1.845
III.
CI = x ± margin of error
confidence interval = [ 51.8709 ± 1.845 ]
= [ 50.026 , 53.716 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =51.8709
standard deviation, s =5.03151
sample size, n =31
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 30 d.f is 2.042
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 = [ 51.8709 ± t a/2 ( 5.03151/ Sqrt ( 31) ]
= [ 51.8709-(2.042 * 0.904) , 51.8709+(2.042 * 0.904) ]
= [ 50.026 , 53.716 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 50.026 , 53.716 ] 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:
a.confidence inteval 95% for san francisco = [ 52.537 , 55.334 ]
b.confidence inteval 95% for Los angles = [ 58.143 , 61.986 ]
c.confidence inteval 95% for sacramento = [ 50.026 , 53.716 ]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.