Question

1. You have been hired as a consultant by a college foundation to conduct a survey of graduates. The foundation wants to estimate the mean amount of all charitable contributions their graduates make annually. What is the minimum number of graduates that should be surveyed if the foundation wants to be 98% confident that the sample mean is within $50 of the actual population mean? (Based on results of a pilot study, assume the true standard deviation of annual charitable contributions for all graduates is $337.)

2. Suppose that a random sample of 350 graduates is found for the problem above. The sample mean annual charitable donations was found to be $1154 with a sample standard deviation of $388.

1. Find a 98% interval estimate for the true population mean annual charitable contributions of graduates using the assumed population standard deviation of $337 (from problem #7 above).

2. Redo part (a) above: Find a 98% interval estimate for the true population mean annual charitable contributions of graduates, this time using the sample standard deviation of $388.

Answer 1

Solution :

Given that,

Population standard deviation = = 337

Margin of error = E = 50

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = Z0.01 = 2.326

sample size = n = [Z/2* / E] 2

n = [ 2.326 * 337 / 50 ]2

n = 245.78

Sample size = n = 246

( a )

Solution :

Given that,

Point estimate = sample mean = = 1154

Population standard deviation =    = 337

Sample size = n = 350

At 98% confidence level

= 1 - 98%  

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = Z0.01 = 2.326

Margin of error = E = Z/2 * ( /n)

= 2.326 * ( 337 /  350 )

= 41.899

At 98% confidence interval estimate of the population mean is,

- E < < + E

1154 - 41.899 <   < 1154 + 41.899  

1112.101 <   < 1195.899

( 1112.101 , 1195.899 )

The 98% confidence interval estimate of the population mean is : - ( 1112.101 ,1195.899 )

( b )

Given that,

Point estimate = sample mean = = 1154

sample standard deviation = s = 388

sample size = n = 350

Degrees of freedom = df = n - 1 = 1 - 350 = 349

At 98% confidence level

= 1 - 98%

=1 - 0.98 = 0.02

/2 = 0.01

t/2,df = t0.01 , 349 = 2.337

Margin of error = E = t/2,df * (s /n)

= 2.337 * ( 388 / 350 )

Margin of error = E = 48.468

The 98% confidence interval estimate of the population mean is,

- E < < + E

1154 - 48.468 < < 1154 + 48.468

1105.532 < < 1202.468

( 1105.532 , 1202.468 )

The 98% confidence interval estimate of the population mean is : - ( 1105.532 , 1202.468 )