A brewery company claims that the time to cool an aluminum bottle from 75 F to 4
ID: 3362930 • Letter: A
Question
A brewery company claims that the time to cool an aluminum bottle from 75 F to 45 F is less than it takes to cool a glass bottle. Test to a confidence level of 0.05 if the claim s supported by the following data: for the aluminum bottle n = 35, sample mean is 98.4 and the sample standard deviation is 7.3 ; for the glass bottle n 42, sample mean is 101.8 and the sample standard deviation is 9.9 The Null Hypothesis is The Alternate Hypothesis is The value of the test statistic is The Critical value is The P-value is The Confidence Interval is: The results of the test indicate Confidence Interval The final conclusion is the Null Hypothesis because (what?) is or is not (specify) within theExplanation / Answer
Given that,
mean(x)=98.4
standard deviation , s.d1=7.3
number(n1)=35
y(mean)=101.8
standard deviation, s.d2 =9.9
number(n2)=42
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.032
since our test is two-tailed
reject Ho, if to < -2.032 OR if to > 2.032
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =98.4-101.8/sqrt((53.29/35)+(98.01/42))
to =-1.731
| to | =1.731
critical value
the value of |t | with min (n1-1, n2-1) i.e 34 d.f is 2.032
we got |to| = 1.73142 & | t | = 2.032
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.7314 ) = 0.092
hence value of p0.05 < 0.092,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.731
critical value: -2.032 , 2.032
decision: do not reject Ho
p-value: 0.092
we do not have enough evidence to support the claim
TRADITIONAL METHOD
given that,
mean(x)=98.4
standard deviation , s.d1=7.3
number(n1)=35
y(mean)=101.8
standard deviation, s.d2 =9.9
number(n2)=42
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((53.29/35)+(98.01/42))
= 1.964
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 34 d.f is 2.032
margin of error = 2.03 * 1.964
= 3.986
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (98.4-101.8) ± 3.986 ]
= [-7.386 , 0.586]
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DIRECT METHOD
given that,
mean(x)=98.4
standard deviation , s.d1=7.3
sample size, n1=35
y(mean)=101.8
standard deviation, s.d2 =9.9
sample size,n2 =42
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 98.4-101.8) ± t a/2 * sqrt((53.29/35)+(98.01/42)]
= [ (-3.4) ± t a/2 * 1.964]
= [-7.386 , 0.586]
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interpretations:
1. we are 95% sure that the interval [-7.386 , 0.586] 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
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