Question

Question 1

The lengths of metal rods produced by an industrial process are normally distributed with a standard deviation of 1.8 millimeters. Based on a random sample of nine observations from this population, the 99% confidence interval was found for the population mean length to extend from 194.65 to 197.75. Suppose that a production manager believes that the interval is too wide for practical use and, instead, requires a 99% confidence interval extending no further than 0.50 mm on each side of the sample mean. How large a sample is needed to achieve such an interval?

a) 86

b) 97

c) 30

d) 52

Question 2

How large a sample is needed to estimate the population proportion for the following? ME = 0.03; α=0.05

a) 962

b) 1068

c) 1297

d) 1242

Question 3

A research group wants to estimate the proportion of consumers who plan to buy a scanner for their PC during the next 3 months. How many people should be sampled so that the sampling error is at most 0.04 with a 95% confidence interval?

a) 601

b) 390

c) 423

d) 541

Question 4

A country club wants to poll a random sample of its 320 members to estimate the proportion likely to attend a early-season function. The number of sample observations should be sufficiently large to ensure that a 99% confidence interval for the population extends to most 0.05 on each side of the sample proportion. How large of a sample is necessary?

a) 204

b) 217

c) 304

d) 317

Question 5

Suppose that in a city last year, 1,118 mortgages were taken out and that a simple random sample is to be taken in order to estimate the mean amount of these mortgages. From previous experience of such populations, it is estimated that the population standard deviation is approximately $20,000. A 95% confidence interval for the population mean must extend $4,000 on each side of the sample mean. How many sample observations are needed to achieve this objective?

a) 96

b) 91

c) 89

d) 85

Answer 1

1)

	for99% CI crtiical Z          =	2.58
	standard deviation σ=	1.800
	margin of error E =	0.5
required sample size n=(zσ/E)2                  =		86.0

2)

here margin of error E =	0.0300
for95% CI crtiical Z          =	1.96
estimated proportion=p=	0.5000
required sample size n =	p*(1-p)*(z/E)2=	1068.00

3)

here margin of error E =	0.0400
for95% CI crtiical Z          =	1.96
estimated proportion=p=	0.5000
required sample size n =	p*(1-p)*(z/E)2=	601.00

4)

N=	320
for99% CI crtiical Z          =	2.576
estimated proportion p=	0.5
margin of error E =	0.05
required sample size n=Np(1-p)z2/(p(1-p)z2+E2(N-1)) =	217

5)

N=	1118
for95% CI crtiical Z          =	1.960
σ=	20000
margin of error E =	4000
required sample size n=Nz2σ2/(z2σ2+E2(N-1)) =	89