In: Statistics and Probability
Using the simple random sample of weights of women from a data set, we obtain these sample statistics: n=50and xbar=140.23lb. Research from other sources suggests that the population of weights of women has a standard deviation given by σ=32.31 lb.
a. Find the best point estimate of the mean weight of all women.
b. Find a 99 % confidence interval estimate of the mean weight of all women.
Solution :
Given that,
(a)
Point estimate = sample mean =
= 140.23
Population standard deviation =
= 32.31
Sample size = n = 50
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2
= Z0.005 = 2.576
(b)
Margin of error = E = Z/2*
(
/
n)
= 2.576 * (32.31 /
50)
= 11.77
At 99% confidence interval estimate of the population mean is,
- E <
<
+ E
140.23 - 11.77 <
< 140.23 + 11.77
128.46 <
< 152.00
(128.46 , 152.00)