24. The university data center whether computer 1 is receiving tasks which requi

Question

24. The university data center whether computer 1 is receiving tasks which require compa of computer 2. A random sample of 10 processing times from computer e of 66 seconds with a standard deviation of 20 seconds, while a random sample ot g rocsing times from computer 2 (chosen independently of those for computer 1) has two main computers. The center wants to examine rable processing time to those wed a mean of 56 seconds with a standard deviation of 19 seconds. Assume that the ns of processing times are normally distributed for each of the two computers sho populatio and that the variances are equal. (4 pt. ea.) a) Find a 90% confidence interval for the difference in processing times between the two computers. b) Interpret your results. Do you have reason to believe with 90% confidence that e processing time for computer 2 is faster than that of computer 1? Explain. uppose you are told that a 99% confidence interval for the average price of a gall oline in your state is from $2.60 to $3.88. Compute the sample mean and the al margin of error E. (4 pt.)

Explanation / Answer

Q24.
TRADITIONAL METHOD
given that,
mean(x)=66
standard deviation , s.d1=20
number(n1)=10
y(mean)=56
standard deviation, s.d2 =19
number(n2)=9
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*400 + 8*361) / (19- 2 )
s^2 = 381.647
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 381.647 * (1/10+1/9) )
=8.976
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, ? = 0.1
from standard normal table, two tailed and value of |t ?| with (n1+n2-2) i.e 17 d.f is 1.74
margin of error = 1.74 * 8.976
= 15.618
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (66-56) ± 15.618 ]
= [-5.618 , 25.618]
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DIRECT METHOD
given that,
mean(x)=66
standard deviation , s.d1=20
sample size, n1=10
y(mean)=56
standard deviation, s.d2 =19
sample size,n2 =9
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 66-56) ± t a/2 * sqrt( 381.647 * (1/10+1/9) ]
= [ (10) ± 15.618 ]
= [-5.618 , 25.618]
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interpretations:
1. we are 90% sure that the interval [-5.618 , 25.618]contains the true population proportion
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 proportion

we don't have evidence that computer 2 is faster than that of copmuter 1

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