Problem 2 (40 points total) Atlanta temperatures. According to weather.com https

Question

Problem 2 (40 points total) Atlanta temperatures. According to weather.com https://weather.com/weather/tenday//USGA0028:1:US the maximum temperatures in the next 10 days will be 63, 69, 67, 68, 64, 65, 65, 67, 64, 54 a. (20 points) Construct a 95% confidence interval for the maximum daily temperatures at the beginning of December in Atlanta. b. (10 points) Suppose you want to reduce the width of the 95% confidence interval to half the size obtained in part a. How many more data points are required in the sample in order to obtain the desired confidence interval width? Is this possible? c. (10 points) According to historical data https://en.wikipedia.org/wiki/Atlanta#Climate the mean maximum temperature in December in Atlanta is 70.8 F. Based on the current data, are we experiencing a cold spell, that is, is the temperature on the next 10 days significantly lower than the historical average? Draw the appropriate conclusions

Explanation / Answer

PART A.

TRADITIONAL METHOD

given that,

sample mean, x =64.6

standard deviation, s =4.1952

sample size, n =10

I.

stanadard error = sd/ sqrt(n)

where,

sd = standard deviation

n = sample size

standard error = ( 4.1952/ sqrt ( 10) )

= 1.327

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 = 9 d.f is 2.262

margin of error = 2.262 * 1.327

= 3.001

III.

CI = x ± margin of error

confidence interval = [ 64.6 ± 3.001 ]

= [ 61.599 , 67.601 ]

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DIRECT METHOD

given that,

sample mean, x =64.6

standard deviation, s =4.1952

sample size, n =10

level of significance, = 0.05

from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262

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 = [ 64.6 ± t a/2 ( 4.1952/ Sqrt ( 10) ]

= [ 64.6-(2.262 * 1.327) , 64.6+(2.262 * 1.327) ]

= [ 61.599 , 67.601 ]

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interpretations:

1) we are 95% sure that the interval [ 61.599 , 67.601 ] 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

PART B.

margin of error in part A is = 3.001

and now consider this to be reduced to half a size = 3.001 / 2 = 1.5005

To reduce the width of the confidence to half with ME = 1.5005, standard deviation, s =4.1952 , CI = 95%

Compute Sample Size  

n = (Z a/2 * S.D / ME ) ^2

Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )

Standard Deviation ( S.D) = 3.001

ME =1.5005

n = ( 1.96*3.001/1.5005) ^2

= (5.88/1.5005 ) ^2

= 15.37 ~ 16

PART C.

Given that,
population mean(u)=70.8
sample mean, x =64.6
standard deviation, s =4.1952
number (n)=10
null, Ho: =70.8
alternate, H1: <70.8
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =64.6-70.8/(4.1952/sqrt(10))
to =-4.6735
| to | =4.6735
critical value
the value of |t | with n-1 = 9 d.f is 1.833
we got |to| =4.6735 & | t | =1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -4.6735 ) = 0.00058
hence value of p0.05 > 0.00058,here we reject Ho
ANSWERS
---------------
null, Ho: =70.8
alternate, H1: <70.8
test statistic: -4.6735
critical value: -1.833
decision: reject Ho
p-value: 0.00058
have evidence that is lower than historical average

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