Question

The sales manager for a hardware wholesaler finds that 229 of the previous 500 calls to hardware store owners resulted in new product placements. Assume that the 500 calls represent a random sample.

a) Find a 95% confidence interval for the long-run proportion of new product placements.
b) Give a careful verbal interpretation of the 95% confidence interval for the long-run proportion of new product placements.
c) Find the confidence interval for estimating the population proportion for 90% confidence level; sample size n = 675; and sample proportion p = 0.10.

Answer 1

->Suppose, p is the ppopontion of calls to hand wane store owners resulted in new product placements. of calls is considen ed A pandom. sample of size n m calls being mutually fndependent and ff oe den ote by f the number of calls to handwane stane owners Desulted in new product placements, then. the sampling distribution of failL be of the binomial foom, with P-m-f. -49-T However 2as n is sufficiently Jange, normal distriburtion, instead of binomial distpibution. Je can use A binomial d istpibution tends to the nonma4 foem ifnis darge, provided p is not close to 1 on 0. As a may should Jie betaeen p in sist that working rule 9/n and Cn-9)/n.. one has mean mp and vaniance p(1-P Since one may use- (du-f VFL(- P(1-) approximately standand normal vaniable as np(4-P) ahene p= s ample proportion 11

In onden to get a pain of confidence Jrmits fon p with cofidence coeffieient 1- we see that, as nis large in the express ion. n (- be replaced by P although. the exact vaniance of is P(1- hus, we have, approeimaiely = 1-c or P[-Ta/2E (1-FAS + Ta/2E (1-F)m - 1 -d Hence for the given sample the obsenved values of p-tol/a(1-F/n +Ta/2 F(a-F)/n ll be confidence limits to b -aith. confidence coefficient Opproximately equal to 1-. Therefore :- a 95% confidence fnterval :- Tn this case, (1-d)= 95/o= o.95 d= o.05 Td12 To.05/2 From the standand normal table we get To.025 To-o25= 1-96 239 11 =0.458 500 E(1-E) 0-02228 = 0.4143312 0. 414 To.a 25 X O.02228 O.501 p+ To.025 XO.022 28 = 0.50 1 6688 The requined 95% confrdence intenval is (o.114,0.501)

c 900/0 con fidence intenval :- (1-a)= 900/0 1-a= o. Th this case = 0.1 sample propontion.: =0-1 (atven) mow Co/2 To-1/2 = To.05 From standand nor Mal table , we get, To.o5 = 1- 645 VG (1-6)/n = .011 54 To.o5x O.01154 = 0.08101 G7 o.081 . P+T0.05 X0. 01154 - O.11 8 9833 O.1 18 Thenefone, the pequined 90 % confrdence interval is:- (0.081,0.118) Inter pretation prodortion of new peoduct place ment wilL be beta een We have 95°% confidence that the (0.414, 0.50 1) in other words the Intenval wilL Enclude the Jongeun pPoportion. of new ppoduct pl ace ment, while In the pemaining 5 lo ttaflL fail to do so.