Question

s 261518.1 QUESTION points sample of 36 ems resulted in a sample mean of 125 and a standand devation of 20. Develop a A simple popuanion mean. Whuat is the lower imit of the intervalt (NOTE: USE 4 DECI random sample of 36 tem confidee nerval for the QUESTION 2 1 peints if the populasion variance of the lienme of a lightbub s 19600 hours, how lange of a samole munt be taken in ander to be 9% confhdent margin of error will not exceed 25 hours? QUESTION 3 The mean ife-ome of a cell phone battery has been 3000 minutes. The quality conerol division of the factory suspects defects in the production o the battery whe h resutinai , we for the product Atamp e of 40 baneres was vspected and-was fund Nie-age u" ษ" was 2930 minutes with a standard deiation of 300 you mant to beut the claim of the quality conerol division at so levei of confidence Coude value tor this test (NOTE USE 4 DECIMAL DIGITS Clek Save end Subovi to save and subi. Cck Save Al Aaers to ve all anvers

Answer 1

QUESTION 1:

SE=s/$\sqrt{n}$

= 20/$\sqrt{36}$ = 3.3333

$\alpha$ = 0.15

ndf = 36 - 1 = 35

From Table, critical values of t = $\pm$ 1.4718

Confidence interval:

125 $\pm$ (1.4718 X 3.3333)

= 125 $\pm$4.9060

= ( 120.0940 ,129.9060)

So,

Lower limit = 120.0940

QUESTION 2:

$n=\frac{Z^{2}_{\alpha /2}\times \sigma ^{2}}{e^{2}}$

Given:

$\alpha$= 0.01

From Table, critical values of Z = $\pm$ 2.576

$\sigma ^{2}$ = 19600

e = 25

Substituting, we get:

$n=\frac{2.576^{2}\times 19600}{25^{2}}=209$

So,

Answer is:

209

QUESTION 3:

SE=s/$\sqrt{n}$

= 300/$\sqrt{40}$ = 47.4342

Test statistic is:

t = (2930 - 3000)/47.4342 = - 1.4757

ndf = 40 - 1 = 39

By Technology, p - value = 0.0740

So,

Answer is:

0.0740