Question

A researcher wants to obtain an error of 30 minutes when estimating the hours of sleep a student gets at a confidence level of 90%. Use the sample standard deviation from the sleep data and determine the sample size the researcher should use to achieve the desired results.

sleep hours per person.

7
7
5
7
6
8
7
8
5
8
8
4
8
8
6
8
8
8
7
10
6
7
8
5
8
7
7
4
9
8
7
7
8
8
10

Answer 1

X	(X - X̄)²
7	0.040
7	0.040
5	4.840
7	0.040
6	1.440
8	0.640
7	0.040
8	0.640
5	4.840
8	0.640
8	0.640
4	10.240
8	0.640
8	0.640
6	1.440
8	0.640
8	0.640
8	0.640
7	0.040
10	7.840
6	1.440
7	0.040
8	0.640
5	4.840
8	0.640
7	0.040
7	0.040
4	10.24
9	3.24
8	0.64
7	0.040
7	0.040
8	0.640
8	0.640
10	7.840

	X	(X - X̄)²
total sum	252	67.600
n	35	35

sample variance =    Σ(X - X̄)²/(n-1)=   67.6000   /   34.0000   =   1.9882
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   1.9882   =       1.4100

--------------

Standard Deviation ,   σ =    1.4100                  
sampling error ,    E = 30 minutes= 0.5 hours
Confidence Level ,   CL=   90%                  
                          
alpha =   1-CL =   10%                  
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.645   *   1.4100 /   0.5   ) ² =   21.5
                          
                          
So,Sample Size needed=       22