Question

PrintTech, Inc. is introducing a new line of ink-jet printers and would like to promote the number of pages a user can expect from a print cartridge. A sample of 10 cartridges revealed the following number of pages printed. (Use Distibution Table) 3,245 2,529 2,484 2,558 2,536 2,558 2,537 2,811 3,649 3,144 nts picture Click hers for the Excel Data File a. What is the point estimate of the population mean? (Round your answer to 2 decimal places.) Answer is complete but not entirely correct 2848 33 b. Develop a 90% confidence interval for the population mean. (Round your answers to 2 decimal places.) 3 Answer is complete but not entirely correct 2,539 22 3157 44

Answer 1

Solution :

a) Point Estimate of population mean is :

(3245+2529 +2484 +2558 +2536 +2550+2537+2811 + 3649 + 3144) / 10

=2804.3

So, Point Estimate of Population mean is 2804.3

Now we will find out S.d

=[ (3245 - 2804.3)² + (2529 - 2804.3 )² + (2484 - 2804.3)² + (2558 - 2804.3 )² +(2536 - 2804.3)² +(2550 - 2804.3)² +(2537 - 2804.3)² +(2811 - 2804.3)² + (3649 -2804.3)² +(3144 -2804.3 )2 ]   / 10

=147032.41

= $\sqrt{}147032.41$

=383.44

b ) Develop a 90 % Confidence Interval :

So D.o.f = 10 -1 = 9

t = 1.833

M..E = 1.833 * (383.44) / $\sqrt{}9$

=234.28

So ,X1 = 2804.3 - 234.28 = 2570.02

X2 = 2804.3+ 234.28 = 3038.58

So, 90 % Confidence Interval for Population Mean = (2570.02 , 3038.58)