2. The owner of an appliance factory has established that no more than 30 percen

Question

2. The owner of an appliance factory has established that no more than 30 percent of all consumers would purc director convinced the owner to produce solar-powered machines IF he can prove that more than 30% of potential buyers indicate they will purchase the newly designed products. The marketing director conducts a test. In a sample of 500 potential buyers, 160 indicated that they would buy such a product. Assume alpha 0.05 hase solar-powered appliances. However, the marketing department State the null and the alternative hypotheses Calculate the critical value (to 3 decimals). Calculate the test statistic (to 2 decimals). Using the Critical Value Approach, what decision should be made and WHY? What is the p-value (to 4 decimals)? Using the pvalue Approach, what decision should be made and WHY?

Explanation / Answer

Given that,
possibile chances (x)=160
sample size(n)=500
success rate ( p )= x/n = 0.32
success probability,( po )=0.3
failure probability,( qo) = 0.7
null, Ho:p<0.3  
alternate, H1: p>0.3
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=0.32-0.3/(sqrt(0.21)/500)
zo =0.9759
| zo | =0.9759
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =0.976 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 0.9759 ) = 0.16456
hence value of p0.05 < 0.16456,here we do not reject Ho
ANSWERS
---------------
null, Ho:p<=0.3
alternate, H1: p>0.3
test statistic: 0.9759
critical value: 1.64
decision: do not reject Ho
p-value: 0.16456

no evidence that it is more than 30%

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