Question

	x
	120,130,100,205,185,220
10	The variance of the sample mean is,
a	449.839
b	436.737
c	424.017
d	411.667
11	The upper end of the 95% interval estimate is,
a	220.7
b	212.2
c	203.7
d	199.6
12	In the previous question to build a 95% interval estimate for the population mean number of customers with a margin of error of ±15 days, determine the required minimum sample size using a planning value of 50 days.
a	64
b	52
c	43
d	35

Answer 1

Given sample is ,

x : 120, 130 , 100 , 205, 185 , 220

Q10 )

Formula for variance of the sample mean is ,

where , s² is sample variance.

So, s² = 2470    { Using Excel function ,   =VAR()   }

n is sample size = 6

So, variance of the sample mean is ,

Option d is correct !!

Q11)

Formula for upper end of confidence interval is ,

Where , is sample mean.

= 160                        { using Excel function ,    =AVERAGE( )     )

s² = 2470 , s = sqrt(2470 ) = 49.6991

n = 6

is t critical value at given confidence level.

Confidence level = 95% = 0.95 , = 1-0.95 = 0.05 , /2 = 0.025

degrees of freedom ( df ) = n - 1 = 6 - 1 = 5

So, =    = 2.5706          

{ Using Excel function , =TINV( , df ) , This function returns two tailed inverse of t distribution

=TINV( 0.05 , 5 ) = 2.5706 }

So Upper end of the 95% interval estimate is ,

Option b is correct !!

12 )

Formula for sample size is,

Where , is z critical value for given confidence level.

Confidence level = 95% = 0.95 , = 1-0.95 = 0.05 , /2 = 0.025

So, = = 1.96            { Using excel function , =NORMSINV(0.025 ) = 1.96 }

E is margin of error = 15

s is sample standard deviation = 49.6991

So sample size is ,

     ( Rounded to next integer )

Option c is correct !!