Question

2. 16 alkaline and 16 heavy-duty Radio shack batteries were placed individually in a cir- cuit consisting of two flashlight bulbs wired in parallel, a switch, a battery holder, and a Hewlet Packard 427-A analog DC voltmetre. Each battery was drained to a reference failure voltage of 0.9 volts and the time to failure (min) was measured. Alkaline Heavy-duty 15 29 31 17 22 22 18 27 30 26 33 18 17 22 15 16 23 23 29 27 26 23 24 24 30 19 22 20 22 25 22 26 (a) Test to see if the mean time to failure are equal for Alkaline land Heavy-duty at 5% significance level. i. State the null and alternative hypotheses. ii. Compute the test statistic value. iii. Find the corresponding p-value. iv. Make the decision according to (iii) (p-value approach). v. Give the conclusion by the decision made from (iv). (b) Find a 95% confidence interval for the difference in the mean time to failure, and win the interpretation of the result.

Answer 1

2 a)

i) Hypothesis

let $\small \mu_{1}$ be the population mean failure time of Alkaline battery.

let $\small \mu_{2}$ be the population mean failure time of Heavy duty battery.

H0:

11 = 12

H1 :

11 + 12

ii) test statistic

	Alkaline	Heavy duty
Mean	$\small \bar{x_{1}}$ = 22.69	$\small \bar{x_{2}}$ = 23.75
Std dev	s1 = 6.30	s2 = 2.86
sample size	n1 = 16	n2 = 16

assuming variances unknown and not equal

test statistic t = = -0.6141

11-12 V + + 5

iii) degrees of freedom df =

eleulat au ale +

by substituting all values we get df = 21

corresponding p-value = 0.5457

iv) for a given alpha = 0.05

since p-value is more than alpha we fail to reject null hypothesis and there is no significant evidence to conclude that the mean failure times differ for alkaline and heavy duty batteries.

v) the mean failure times differ for alkaline and heavy duty batteries are equal.

b) 95% CI

CI =

1 - 2 Eta/2.0 * 1

t0.025,21 = 2.080

95% CI = (-4.658 , 2.538)

We are 95% confident that the mean difference between the failure times of alkaline and heavy duty batteries will lie in between 4.658 and 2.538.

4)

a)

let $\small \mu_{1}$ be the population mean failure time of fertilizer A.

let $\small \mu_{2}$ be the population mean failure time of fertilizer B.

H0:

11 = 12

H1 :

μι μ2

b)

	A	B
Mean	40.71	39.14
Std dev	1.98	2.34
sample size	7	7

assuming variances unknown and equal

Sp = pooled std dev =

(ni - 1)s + (n2 - 1)s ni + n2 - 2 V

by substituting all values we get Sp = 2.17

test statistic t =

| X1 - V2 Says +

by substituting all values we get t = 1.357

c) df = n1+n2 - 2 = 7+7-2 = 12

corresponding degrees of freedom p-value = 0.0998

d)

for a given alpha = 0.05

since p-value is more than alpha we fail to reject null hypothesis.

f) Fertilizer A is not more effective than fertilizer B.