Question

A chemical manufacturer claims that its​ ice-melting product will melt snow and ice at temperatures as low as 0◦ Fahrenheit. A representative for a large chain of hardware stores is interested in testing this claim. The chain purchases a large shipment of​ 5-pound bags for distribution. The representative wants to​ know, with 90​% confidence and within ±0.03,

what proportion of bags of the chemical perform the job as claimed by the manufacturer. Complete parts​ (a) through​ (c) below.

a. How many bags does the representative need to​ test? What assumption should be made concerning the population​ proportion? (This is called destructive​ testing; that​ is, the product being tested is destroyed by the test and is then unavailable to be​ sold.)

The population proportion should be assumed to be equal to

​(Round to four decimal places as​ needed.)

The representative needs to test ___ bags.

​(Type a whole​ number.)

b. Suppose that the representative tests 50 ​bags, and 37 of them do the job as claimed. Construct a 90​%

confidence interval estimate for the population proportion that will do the job as claimed.

___ ≤ π ≤ _____

​(Round to four decimal places as​ needed.)

c. How can the representative use the results of​ (b) to determine whether to sell the​ manufacturer's ​ice-melting product?

The representative can conclude

(with 90% confidence OR that there is a 90% probability)

that the proportion of all bags that will do the job as claimed is between the lower and upper limits of the confidence interval estimate. If the​ representative's minimum acceptable proportion is

(greater OR less)

than the upper​ limit, the representative should not sell the product.

Answer 1

a)

The population proportion should be assumed to be equal to 0.5

here margin of error E =	0.03
for90% CI crtiical Z          =	1.645
estimated prop.=p=	0.5000
reqd. sample size n=	p*(1-p)*(z/E)2=	752

The representative needs to test 752 bags

b)

sample success x =		37
sample size          n=		50
sample proportion p̂ =x/n=		0.7400
std error se= √(p*(1-p)/n) =		0.0620
for 90 % CI value of z=		1.645
margin of error E=z*std error   =		0.1020
lower bound=p̂ -E                       =		0.6380
Upper bound=p̂ +E                     =		0.8420

0.6380  ≤ π ≤ 0.8420

c)

The representative can conclude with 90% confidence that the proportion of all bags that will do the job as claimed is between the lower and upper limits of the confidence interval estimate If the​ representative's minimum acceptable proportion is greater than the upper​ limit, the representative should not sell the product.