Download Constructing Confidence Interval with Unknown Population

Constructing Confidence Interval with
Unknown Population Standard Deviation
Suppose we randomly select 40 students and ask them how many hours do they watch TC per
week. The sample data are as follows:
6
14
18
22
7
15
18
22
8
15
19
22
10
16
19
23
11
16
19
23
11
17
20
23
12
17
20
24
12
18
21
24
13
18
21
25
13
18
22
27
Find a point estimate of the population mean and a 95% confidence interval for the population
mean.
Solution:
The sample mean for this set of data is 17.48. Hence, a point estimate of the population mean
is 17.48.
From the population of ACT scores we can form many, many samples of size 40. One these
many, many samples is:
6
14
18
22
7
15
18
22
8
15
19
22
10
16
19
23
11
16
19
23
11
17
20
23
12
17
20
24
12
18
21
24
13
18
21
25
13
18
22
27
Note: Sample mean = x = 17.48 and Sample Standard Deviation = s = 5.193623
Sample Size = 40 and Degree of Freedom = 4 - 1 = 39
For each sample of size 40, a sample mean and sample standard deviation can be calculated.
Hence there are many, many sample means and sample standard deviations. For each pair of
sample mean and corresponding sample standard deviation, a confidence interval can be
formed. Consequently, there are many, many confidence intervals.
Since our level of confidence is set 95%, about 95% of all confidence intervals will contain the
population mean and 5% of the confidence intervals do not contain the population mean.
From earlier discussion, a confidence interval has the form:

 s , x  t  s  
 /2 
 x  t / 2 
n 
n  



where x is the sample mean;
t / 2
is the number of standard error from the population mean;
s is the standard deviation of the population of ACT scores;
n is the sample size
Since the sample size is greater than 30, 95% of the t-scores will lie between - t / 2 and t / 2 .
t / 2
Left Area = 2.5%
Middle Area = 95%
Right Area = 2.5%
t / 2 is the t-score corresponding to a right area of 0.025. Hence, t / 2 = 2.019.
Standard Error = s
n
=
5.193623
 0.8211839 .
40

, x  t  s

Confidence Interval =  x  t / 2  s


 /2 
n
n  



= 17.48  2.019  0.8211839  , 17.48+2.019  0.8211839  
= 15.82, 19.14 
Comments:
We do not know if 15.82, 19.14  contains the population mean or not since this interval is
one of many, many confidence intervals. However, since we know that 95% of the confidence
intervals do contain the population mean, we can be 95% confident that the interval
15.82, 19.14  does contain the population mean.
Also, the interval 15.82, 19.14  is an interval estimate of the population mean.