7. The warranty for batteries for mobile phones is set at 200 operating hours, w

Question

7. The warranty for batteries for mobile phones is set at 200 operating hours, with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 200 hours despite proper charging procedures. (a) (8) Do these experimental results go against the claim that less than 0.2 percent of the company's batteries will fail during the warranty period, with proper charging procedures? Test at a level of significance of 0.01. (b) (2) What is the p-value for the test in (a)? (c) (4) If the company wishes to be 95% confident that their estimate of the proportion of batteries that fail during the warranty period, with proper charging procedures, is within 0.1% of the true value, how many batteries must be tested?

Explanation / Answer

7.

Given that,

possibile chances (x)=15

sample size(n)=5000

success rate ( p )= x/n = 0.003

success probability,( po )=0.002

failure probability,( qo) = 0.998

null, Ho:p=0.002  

alternate, H1: p<0.002

level of significance, = 0.01

from standard normal table,left tailed z /2 =2.33

since our test is left-tailed

reject Ho, if zo < -2.33

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.003-0.002/(sqrt(0.001996)/5000)

zo =1.5827

| zo | =1.5827

critical value

the value of |z | at los 0.01% is 2.33

we got |zo| =1.583 & | z | =2.33

make decision

hence value of |zo | < | z | and here we do not reject Ho

p-value: left tail - Ha : ( p < 1.58272 ) = 0.94326

hence value of p0.01 < 0.94326,here we do not reject Ho

ANSWERS

---------------

null, Ho:p=0.002

alternate, H1: p<0.002

test statistic: 1.5827

critical value: -2.33

decision: do not reject Ho

p-value: 0.94326

a.

we do not have enough evidence to support the claim that less than 0.2% of company batteries will fail during the warranty period

b.

p-value = 0.94326

c.

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.05 is = 1.96

Sample Proportion = 0.003

ME = 0.001

n = ( 1.96 / 0.001 )^2 * 0.003*0.997

= 11490.226 ~ 11491

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