Question

1.You are interested in estimating the the mean weight of the local adult population of female white-tailed deer (doe). From past data, you estimate that the standard deviation of all adult female white-tailed deer in this region to be 21 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 98% confidence level, to within 9 pounds of the actual weight?

2.You measure 30 backpacks' weights, and find they have a mean weight of 72 ounces. Assume the population standard deviation is 4.9 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean backpack weight.

3.You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be 80% confident that you estimate is within 2% of the true population proportion. How large of a sample size is required?

Answer 1

Solution :

1) Given that,

Population standard deviation = = 21

Margin of error = E = 9

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z_/2 = 2.326

sample size = n = [Z_/2* / E] ²

n = [2.326 * 21 / 9 ]²

n = 29.45

Sample size = n = 30

2) Given that,

Point estimate = sample mean = = 72

Population standard deviation =    = 4.9

Sample size = n = 30

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

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

E = 1.96 * ( 4.9 /  30 )

E = 1.75

3) Given that,

= 1 - = 0.5

margin of error = E = 0.02

At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

Z/2 = 1.282

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.282 / 0.02)² * 0.5 * 0.5

= 1027.20

sample size = n = 1028