Consider a normal population with an unknown population standard
deviation. A random sample results in x−x− = 62.88 and
s^{2} = 16.81. [You may find it useful to
reference the t table.]
a. Compute the 90% confidence interval for
μ if x−x− and s^{2} were obtained from a
sample of 24 observations. (Round intermediate calculations
to at least 4 decimal places. Round "t" value to 3 decimal
places and final answers to 2 decimal places.)
b. Compute the 90% confidence interval for μ if x−x− and s^{2} were obtained from a sample of 29 observations. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)
c. Use your answers to discuss the impact of the
sample size on the width of the interval.
The bigger sample size will lead to a larger interval width and therefore a more precise interval.
The bigger sample size will lead to a smaller interval width and therefore a more precise interval.
Given that x bar = 62.88 and s^{2} = 16.81
So , s = 4.1
( a ) 90% confidence interval : with n= 24
to find C. I we have formula
x bar t critical ( s.d / sqrt ( n)) ----------- ( * )
at alpha = 0.10 l.o.s , d.f =24 -1 = 23 , t critical value = 1.714
from ( * )
62.88 1.714 ( 4.1 / sqrt ( 24 ))
62 .88 1.4345
61.4455 , 64.3145
90% confidence interval : ( 61. 45 , 64 .31 )
( b )
90% confidence interval : with n = 29
at alpha = 0.10 l.o.s , d.f =29 -1 = 28 , t critical value = 1.701
from ( * )
62.88 1.701 ( 4.1 / sqrt ( 29))
62.88 1.2951
( 61.5849 , 64.1751 )
90% confidence interval ( 61.58 , 64.18 )
( c )
Answer : The bigger sample size will lead to a smaller interval width and therefore a more precise interval.
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