5 steps please for A and plot Do B please too Clear handwriting PROBLEMS: KINDI

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5 steps please for A and plot Do B please too
Clear handwriting PROBLEMS: KINDI y CoMPI.ETE ALL PARTs oFALL PROBLEMS. PARTIAL CREDIT wILL BE GIVEN ON THE PARTS OF THE PROBLEMS THAT ARE COMPLETED. ENSURE THAT YOU ANSWER EACH QUESTION ON THE SPACE PROVIDED AND SHow ALL YOUR WORK 1. Sony manufactures Color Televisions at two plants and wanted to compare the proportions of filled orders of Color Televisions on time from the two plants. The first plant is located in Tai Pi in Taiwan and the second one is located in Seoul, South Korea, Sony selected 1.000 orders at random from each plant and found that 798 orders were completed on time from the Tai Pi plant and only 631 orders were completed ontime from the Seoul plant. A. Does the data in this sample support the claim of that the proportion of all filled orders from the Tai Pi plant is greater than the proportion of all filled orders from the Seoul plant? Perform a Hypothesis test: list all the required 5 steps in order to answer the question. Do not use the PNALUE or the Confidence Interval as rejection rules here. (17 Points) Repeat only the three required steps to test the same Hypothesis as in part A but use ONLY the P-VALUE as a rejection rule here. (6 Points)

Explanation / Answer

5 A)

let p1 be the filled in orders from Tai Pi plant and p2 be the same from Seoul plant.

the claim is p1>p2

so the null hypothesis is H0:p1=p2 and the alternative hypothesis is H1:p1>p2

now to test these hypothesis we have sample values with level of significance=alpha=0.05

we have a sample of size n1=1000 from Tai Pi plant out of which f1=798 are complete orders and n2=1000 from Seoul plant out o which f2=631 are complete orders.

let p1hat=f1/n1 be the sample proportion of complete orders from Tai Pi plant and p2hat=f2/n2 be the sample proportions of complete orders from Seoul plant

then E[p1hat]=p1 E[p2hat]=p2 V[p1hat]=p1(1-p1)/n1 V[p2hat]=p2(1-p2)/n2

since n1 and n2 are very large hence the distribution of p1hat and p2hat can be assumed to be normal

p1hat~N(p1,p1(1-p1)/n1) p2hat~N(p2,p2(1-p2)/n2)

so p1hat-p2hat~N(p1-p2,p1(1-p1)/n1+p2(1-p2)/n2)

under H0 p1=p2=p say

so under H0 p1hat-p2hat~N(0,p(1-p)(1/n1+1/n2))

where p is unknown and estimated by phat=(f1+f2)/(n1+n2)

hence the test statistic is given by Z=(p1hat-p2hat)/sqrt[phat(1-phat)(1/n1+1/n2)] which under H0 follows N(0,1)

since the alternative hypothesis is right tailed hence H0 is rejected iff Zobs>z0.05   where z0.05 is the upper 0.05 point of a N(0,1) distribution. and Zobs is the observed value of Z

now p1hat=798/1000=0.798 p2hat=631/1000=0.631 phat=(798+631)/(1000+1000)=0.7145 n1=n2=1000

so Zobs=(0.798-0.631)/sqrt[0.7145*(1-0.7145)(1/1000+1/1000)]=8.27

z0.05=1.64

hence Zobs>z0.05   hence H0 is rejected and the conclusion is that p1>p2 that is proportions of all filled orders from TaiPi plant is greater than that of Seoul plant

B) now we have to use p value

then as previous

p1hat~N(p1,p1(1-p1)/n1) p2hat~N(p2,p2(1-p2)/n2)

so p1hat-p2hat~N(p1-p2,p1(1-p1)/n1+p2(1-p2)/n2)

under H0 p1=p2=p say

so under H0 p1hat-p2hat~N(0,p(1-p)(1/n1+1/n2))

where p is unknown and estimated by phat=(f1+f2)/(n1+n2)

hence the test statistic is given by Z=(p1hat-p2hat)/sqrt[phat(1-phat)(1/n1+1/n2)] which under H0 follows N(0,1)

H0 is rejected iff pvalue<alpha=0.05

where p value=P[Z>Zobs] where Z~N(0,1) Zobs is the observed value of Z

now from part A Zobs=8.27

so p value=P[Z>8.27]=1.110223e-16

so p value<0.05

hence H0 is rejected and the conclusion is that p1>p2 that is proportions of all filled orders from TaiPi plant is greater than that of Seoul plant

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