Ch8 Confidence Interval Estimation Part2 - Suggested Problems Solutions

STATISTICS FOR SOCIAL SCIENCES BUS 152  ___________________________________________________________________________  CONFIDENCE INTERVAL ESTIMATION 8.3 Confidence Interval Estimation for the Proportion

Problem 8 (8.35) If you want to be 99% confident of estimating the population mean to within a sampling error of  ± 20 and the standard deviation is assumed to be 100, what sample size is required?

SOLUTION: 2

n

 Z  =

2 σ  

e

2.582 100 2 ⋅

=

2

= 166.41

2

20

Use n  = 167

 _________________________________________________________________________________  Problem 9 (8.39) If the manager of a paint supply store wants to estimate the mean amount of paint in a 1-gallon can to within ± 0.004 gallon with 95% confidence and also assumes that the standard deviation is 0.02 gallon, what sample size is needed?

SOLUTION: 2

n

 Z  =

2 σ  

e

1.96 2 (0.02)2 ⋅

=

2

= 96.04

2

(0.004)

Use n  = 97

 _________________________________________________________________________________  Problem 10 (8.41) If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottles to within ± 0.01 liter with 95% confidence and also assumes that the standard deviation is 0.05 liter, what sample size is needed?

SOLUTION: 2

n

 Z  =

2

1.96 2 (0.05) 2

σ  

e

2

⋅

=

(0.01)

= 96.04

2

Use n  = 97

 _________________________________________________________________________________  Problem 11 (8.43) An advertising agency that serves a major radio station wants to estimate the mean amount of time that that the station’s station’s audience audience spends spends listening listening to the radio daily. daily. From From past studied, studied, the standard standard deviation is estimated as 45 minutes. a. What sample size is needed if the agency wants to be 90% confident of being correct to within ± 5 minutes? b. If 99% confidence is desired, what sample size is necessary?

SOLUTION: 2

a)

n

 Z  =

e 2

b)

n

 Z  =

2 σ   2

1.6452 45 2 ⋅

=

2 σ  

e

2

2

5

2.582 4 52 ⋅

=

2

5

= 219.19 = 539.17

Use

n 

= 220

Use

n 

= 540

 _________________________________________________________________________________