A sport utility vehicle design is suspected to have propensity for rollover. In

Question

A sport utility vehicle design is suspected to have propensity for rollover. In a test, out of 60 vehicles, 17 are observed to rollover. At a significance level of 0.05, test the hypothesis that proportion of vehicles with rollover problem is 0.2.

a. Using P-value approach

b. Using z-test

c. Confidence Interval method.

d. If the true fraction of defects is found to be 25%, find the type II error.

e. What is the minimum sample size needed to detect type II error within 5% accuracy if the true fraction of defects is 25%?

Explanation / Answer

a.

Given that,

possibile chances (x)=17

sample size(n)=60

success rate ( p )= x/n = 0.2833

success probability,( po )=0.2

failure probability,( qo) = 0.8

null, Ho:p<0.2

alternate, H1: p>0.2

level of significance, = 0.05

from standard normal table,right tailed z /2 =1.64

since our test is right-tailed

reject Ho, if zo > 1.64

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

zo=0.28333-0.2/(sqrt(0.16)/60)

zo =1.6137

| zo | =1.6137

p-value: right tail - Ha : ( p > 1.61374 ) = 0.05329

hence value of p0.05 < 0.05329,here we do not reject Ho

b.

critical value

the value of |z | at los 0.05% is 1.64

we got |zo| =1.614 & | z | =1.64

make decision

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

c.

given that,

possibile chances (x)=17

sample size(n)=60

success rate ( p )= x/n = 0.283

CI = confidence interval

confidence interval = [ 0.283 ± 1.96 * Sqrt ( (0.283*0.717) /60) ) ]

= [0.283 - 1.96 * Sqrt ( (0.283*0.717) /60) , 0.283 + 1.96 * Sqrt ( (0.283*0.717) /60) ]

= [0.169 , 0.397]

ANSWERS

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

null, Ho:p=0.2

alternate, H1: p>0.2

test statistic: 1.6137

critical value: 1.64

decision: do not reject Ho

p-value: 0.05329

d.

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

Z a/2 at 0.5 is = 0.674

Sample Proportion = 0.25

ME = 0.05

n = ( 0.674 / 0.05 )^2 * 0.25*0.75

= 34.071 ~ 35

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