DETERMINING THE SAMPLE SIZESource:EMRI (Educational Mentoring Resources, Inc., 2011)

FOR ONE GROUP WHERE OUTCOME IS EXPRESSED AS A DISCRETE VARIABLE (I.E., PROPORTION OR PERCENTAGE)

where Zά/2 = standard normal deviate

corresponding to the desired level of confidence

e = effect size or maximum tolerable error, or margin of error

p = estimate of the population proportion

q = 1 – p

 

2

22/ )(

e

pqzn

A researcher wants to do a survey to determine the prevalence of abusive behavior in children 6-12 years of age in a community in Manila. How many children should be included in the study if the prevalence of child abuse in the Philippines as reported in past studies is 10% (i.e., p=0.10), the desired level of confidence is 95% (i.e., α=0.05, Zά=1.96), and the desired precision of the estimate (tolerable error) is 5% (i.e., e=0.05)?

= (1.96²) (0.10) (0.90) = 138.29

(0.05)² 

This study will include at least 139 children aged 6-12 years.

 A drop-out or non-participation rate is factored in. In this study, we expect an attrition rate of 10%. Therefore, 139 + 13.9 = 152.9. Overall, we need at least 153 respondents for this study.

2

22/ )(

e

pqzn

FOR TWO OR MORE GROUPS AND OUTCOME IS EXPRESSED AS A DISCRETE VARIABLE (E.G.,PERCENTAGE OR PROPORTION)

n = [Z/2² (2pq) + Zβ² (p1q1+p2q2)]²

e² where

n = sample size per group p1, p2 = estimated population

proportions in groups 1 & 2, respectively p = p1 + p2 2 q = 1 - p

e = magnitude of difference to be detected, effect size

Zά/2 = standard, normal deviate corresponding to the desired level of confidence

Zβ = standard, normal deviate corresponding to β error rate

Suppose a researcher wishes to test a hypothesis comparing the proportion of passers in the two sections in a statistics class. She wants to detect a 15% difference (error) in the percentage of successful examinees between the two sections where her past experience shows that 95% of students passed in one section (p1), and only 80% passed in the other class (p2). What should be her sample size in each group? Given: α=0.05, power (ß)= 80%.

DETERMINE THE VALUE OF Zẞ ß = 80% = 0.80 Z ß 0.84 = 0.7995

z – 0.84 0.0005

z = 0.80000.01 0.0028

0.85 = 0.8023

8418.0

84.00028.0

)0005.0)(01.0(

0028.0

0005.0

01.0

84.0

z

zz

 n = [Zά/2² (2pq) + Zβ² (p1q1+p2q2)]²

e² n = [(1.96²)(2)(0.875)(0.125)+ 0.8418² ((0.95)(0.05) +(0.80)

(0.20))]

0.15²

= a minimum of 43.88 or 44 per group for a total of 88 subjects.

With 10% non-participation rate of 10%, at least 96.8 (or 97) respondents will taken for the two groups.

 

FOR ONE GROUP WHERE OUTCOME IS EXPRESSED AS A CONTINUOUS VARIABLE (E.G., MEANS)

where Zά/2 = standard normal deviate

corresponding to the desired level of confidence

e = maximum tolerable error; level of precision, effect size

s² = estimate of variance of observations

2

222/

e

szn

EXAMPLEA survey will be done to determine the

average number of times a selected group of adolescents have engaged in binge drinking in the past year. How many adolescents should be included in the study if past studies have shown that binge drinking in adolescents occurs about 8 times on the average per year (SD=0.40)? The desired level of confidence is 95% (i.e., α=0.05), and the desired precision of the estimate (e) is 5%.

= 1.96² (0.40) ² = 245.86

.05² A minimum number of 246

adolescents is needed for the study

plus 10% non-participation rate = at least 271 respondents

2

222/

e

szn

FOR TWO OR MORE GROUPS AND OUTCOME IS EXPRESSED AS CONTINUOUS VARIABLE (E.G., MEANS)

  2 s² (Zά/2+ Zβ)²

n = ------------------------ e² where n = sample size per groups² = estimate of variance of observationse = magnitude of difference to be detectedZά/2 = standard. normal deviate corresponding to the

desired level of confidenceZβ = std. normal deviate corresponding to β error rate

A group of acceptors and non acceptors of measles immunization were compared in terms of their beliefs in measles vaccination. Beliefs in measles vaccination was measured using a 5-point semantic differential scale. The pilot data indicated that a conservative value for the variance was 1.0. It was also decided that the smallest difference that the study should detect was 0.4 where ά=0.05, two-tailed, power=90%.

n = 2 s² (Zά/2+ Zβ)²

e² n = 2(1)² (1.96 + 1.2817) ² =131.35 = 132

0.4² Total sample size = 132 + 10%(132) = 146 When there is absolutely no prior knowledge about

the variance of the population, a maximum variance of 0.50 can be estimated.

It must be noted that the higher the variance, the larger will be the sample size.

SLOVIN’S FORMULA

21 Ne

Nn

An investigator wants to know the GPA of MMSU students. However, he does not have enough resources to survey the entire population of 3,000 students. If he wants to use a sample of this population, with a 5% margin of error, what should his sample size be?

Given: N = 3,000 e = 5% = 0.05 Required: n = ?

Solution: n = 3000 / ( 1 + [(3000)(0.05)(0.05)] = 352.9 or a

minimum of 353

The Slovin’s formula is a simple way of estimating sample size but is most commonly used.

Only applicable for one group surveys and when the population size is known.

REPRESENTATIVENESS

OF SAMPLES

As sample size increases, sample becomes more and more representative

of population.

Exercises:1. A student in public administration wants to determine the mean amount members of city councils in large cities earn per month as renumeration for being a council member. The error in estimating the mean is to be less than Php1,000 with a 95% level of confidence. The student found a report by the Department of Labor that estimated the standard deviation to be Php5,000. What is the required sample size?

2. The study in problem 1 also estimates the proportion of cities that have private collectors. The student wants the estimate to be within 0.10 of the population proportion, the desired level of confidence is 90%, and no estimate is available for the population proportion. What is the required sample size?

3. Will you assist the college registrar in determining how many transcripts to study? The registrar wants to estimate the arithmetic mean grade point average (GPA) of all graduating seniors during the past 10 years. GPA’s range between 2.0 and 4.0. The mean GPA is to be estimated within plus or minus 0.05 of the population mean. The standard deviation is estimated to be 0.279. Use the 99% level of confidence.

TEACHING BASIC STATISTICS ….

Session 4.21

If X has a distribution (not necessarily normal) with mean and variance 2,then the distribution of the sample mean approaches the normal distribution with mean and variance 2/n as the sample size increases.

CENTRAL LIMIT THEOREM

TEACHING BASIC STATISTICS ….

Session 4.22

Sampling Distributions of the Sample Mean

A. Uniform

B. Two right triangles

PARENT POPULATION n = 2 n = 5 n = 25

C. Exponential

TEACHING BASIC STATISTICS ….

Session 4.23

Large sample size, like n 25 only imply that “normality” of the sample mean may be assumed.

However, it does not imply that this is the “appropriate” sample size for inference.

REMARK