Question

The Following are prices of a random sample of 18 smart TVs sold at a particular department store with screen sizes between 46’’ and 50’’. Find the 90% confidence interval for the true mean price of a smart TV purchased at this store. Assume the prices are normally distributed.

448 498 597 599 649 699 529 749 628 950 798 369 498 669 499 599 355 469

Answer 1

Solution:

x	x2
448	200704
498	248004
597	356409
599	358801
649	421201
699	488601
529	279841
749	561001
628	394384
950	902500
798	636804
369	136161
498	248004
669	447561
499	249001
599	358801
355	126025
469	219961
∑x=10602	∑x2=6633764

Mean ˉx=∑xn

=448+498+597+599+649+699+529+749+628+950+798+369+498+669+499+599+355+46918

=1060218

=589

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√6633764-(10602)21817

=√6633764-624457817

=√38918617

=√22893.2941

=151.305

Degrees of freedom = df = n - 1 = 18 - 1 = 17

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,17 =1.740

Margin of error = E = t/2,df * (s /n)

= 1.740 * (151.30 / 18)

= 62.04

Margin of error = 62.04

The 90% confidence interval estimate of the population mean is,

- E < < + E

589 - 62.04< < 589 + 62.04

526.96 < < 651.04

(526.96, 651.04 )