Chapter 9

question #34

A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95% confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.

Robert Koeppel

Step 1: identify problem

Is it reasonable that the population

mean is 28 weeks?

Robert Koeppel

step 2: Formula

Confidence interval for the population Mean

_

s

n

X + t

-

X = sample mean

t = t-distribution

s = sample standard deviation

n = the number of observations in the sample

To determine the confidence limits when the population standard deviation is known, we use the z distribution.

Robert Koeppel

step 3: arrange data

_

_

_

X = 26 + (2.010) 6.2

_

50

n = 50

X = 26

Standard deviation = 6.2

Degree of freedom: 50-1= 49

Confidence interval: 95%

t-table: 2.010

Mashaam Latif

step 4: solve

=

_

26 + 1.762

=

27.762

26 - 1.762

=

24.238

Mashaam Latif

step 5: Comment on result

26 + 1.762

27.762

26 + 1.762

=

26 - 1.762

=

24.238

It is not reasonable that population mean is 28 weeks since 28 does not lie within the confidence interval

question #36

The American Restaurant Association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meal per week. Construct a 99% confidence interval for the population mean.

step 1: identify problem

Construct a 99% confidence interval for the population mean

step 2: Formula

Confidence Interval for the Population Mean Formula

X = sample mean

t = t-value for a particular confidence level

s = sample standard deviation

n = number of observations in sample

FIND T-VALUE

step 3: find t-value

With the Confidence Interval and the degree of freedom we are able to find the T-Value:

the Degree of freedom

"df=n-1"

n=60

60-1=59

Confidence Interval=99%

t=2.662

step 4: solve

_

0.75

2.76 + 2.662

=

2.76 + 0.258

X = 2.76

t = 2.662

s = .75

n = 60

√60

3.018

2.76 + 0.258 =

2.76 - 0.258 =

2.502

The 99% confidence interval for the mean is

3.018 and 2.502

question #48

During a national debate on changes to health care, a cable news service performs an opinion poll of 500 small-business owners. It shows that 65% of small-business owners do not approve of the changes. Develop a 95% confidence interval for the proportion opposing health care changes. Comment on the result.

step 1: identify problem

Develop a 95% confidence interval for the proportion opposing health care changes.

step 2: formula

Confidence interval for a population proportion formula

P(1-P)

P + Z n

P = Sample proportion (p=x/n)

z = the z value for particular confidence level

n = number of observations in sample

step 3: plug & play

Level of significance = 0.95

P = x/n

x=500/.65=325

325/500= 0.65

z = 0.9500/2= 0.475 = 1.96

N = 500

= 0.6082

95% CL =

0.65(1-0.65)

0.65 + 1.96 500

= 0.6918

0.65 (1-0.65)

95% CL = 0.65 - 1.96 500

step 4: comment on result

Population proportion is between 0.6082 and 0.6918.

This means that the sample probability 95% of opposing

health care changes is in between 61% to 69%

question #55

You are to conduct a sample survey to determine the mean family income in a rural area of central Florida. The question is, how many families should be sampled? In a pilot sample of 10 families, the standard deviation of the sample was $500. The sponsor of the survey wants you to use the 95% confidence level. The estimate is to be within $100. How many families should be interviewed?

step 1: identify problem

How many families should be interviewed?

step 2: formula

Sample size for estimating population mean

n = Zs 2

E

E= Margin of Error

S= Standard Deviation

Z= Z-Distribution

step 3: arrange data

n = 1.96 x 500 2

100

Z= 0.9500/2 =0.475

Find: 0.475 on table= 1.96

s= 500

E= 100

step 4: solve

2

n = (9.8)

n = 1.96 x 500 2

100

n = 96.04

Round up to whole number

97 families should be interviewed

question #62

You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 95% confidence level and state that the estimated proportion must be within 2% of the population proportion. A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements?

step 1: identify problem

How many in the workforce should be interviewed to meet your requirements?

step 2: formula

Sample size for the population proportion formula

z

2

n=p(1-p)( -- )

E

step 3: arrange data

Z = 0.9500/2= 0.475= 1.96 (CL = 95%) (standard normal value)

E = 0.02 (estimated proportion)

p = 5/50 = 0.10 (sample proportion)

step 4: solve

2

= 865

= 864.36

(1.96)

(0.10)(1-0.10) 0.02

step 5: comment on answer

Out of the workforce population, the data shows that 865 people should be interviewed to meet my requirements