Question

# Consider a normal population with an unknown population standard deviation. A random sample results in x−x−...

Consider a normal population with an unknown population standard deviation. A random sample results in x−x− = 62.88 and s2 = 16.81. [You may find it useful to reference the t table.]

a. Compute the 90% confidence interval for μ if x−x− and s2 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 s2 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 s2 = 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.