8.4. Choosing the sample size for a study
Key results for finding sample size:
- Margin of error depends on standard error for sampling distribution of point estimate
- Standard error depends on sample size
Once we specify margin of error with particular confidence level, we can determine the value of n
that has a standard error giving that margin of error
Choosing the sample size for estimating a population proportion
- Decide desired margin of error
- Choose confidence level for that margin of error
If we specify a margin of error of 0.04, this means a 95% confidence interval should equal the sample
proportion plus and minus 0.04
Selecting a sample size without guessing a value for ^p
- The largest possible value for ^p (1− ^p ) is 0.25 , which occurs when ^p=0.50
- This is the safe approach to guarantee a large enough sample
General sample size formula for estimating a population proportion
^p ( 1− ^p ) z 2
- n=
m2
Reducing the margin of error by a factor of one-half requires quadrupling n
- ME=0.04 with n=600 ME=0.02 with 2400
Revisiting the approximation 1/ √ n for the margin of error
- For a 95% confidence interval, the margin of error is approximately 1/ √ n when ^p is near 0.5
Choosing the sample size for estimating a population mean
- To derive the sample size for estimating a population mean, you set the desired margin of
error and solve for n
- Confidence interval for population mean x ± t α / 2 (se )
- se=s / √ n & s = sample standard deviation
- If you don’t know n, you also don’t know the degrees of freedom and the t-score. However,
when df >30 , the t-score is very similar to the z-score from a normal distribution (1.96 for
95% confidence)
- Also, we don’t know the sample standard deviation s before collecting the data
- When we use a z-score instead of a t-score, supply and educated guess for the standard
deviation of the sample mean σ / √n and then set z (σ √ n)equal to a desired margin of error
m and solve for n
- The random sample size n for which a confidence interval for a population mean μ has
σ 2 z2
margin of error approximately equal to m is: n= 2
m
o To use this, you need to guess the value for the population standard deviation σ
,
Key results for finding sample size:
- Margin of error depends on standard error for sampling distribution of point estimate
- Standard error depends on sample size
Once we specify margin of error with particular confidence level, we can determine the value of n
that has a standard error giving that margin of error
Choosing the sample size for estimating a population proportion
- Decide desired margin of error
- Choose confidence level for that margin of error
If we specify a margin of error of 0.04, this means a 95% confidence interval should equal the sample
proportion plus and minus 0.04
Selecting a sample size without guessing a value for ^p
- The largest possible value for ^p (1− ^p ) is 0.25 , which occurs when ^p=0.50
- This is the safe approach to guarantee a large enough sample
General sample size formula for estimating a population proportion
^p ( 1− ^p ) z 2
- n=
m2
Reducing the margin of error by a factor of one-half requires quadrupling n
- ME=0.04 with n=600 ME=0.02 with 2400
Revisiting the approximation 1/ √ n for the margin of error
- For a 95% confidence interval, the margin of error is approximately 1/ √ n when ^p is near 0.5
Choosing the sample size for estimating a population mean
- To derive the sample size for estimating a population mean, you set the desired margin of
error and solve for n
- Confidence interval for population mean x ± t α / 2 (se )
- se=s / √ n & s = sample standard deviation
- If you don’t know n, you also don’t know the degrees of freedom and the t-score. However,
when df >30 , the t-score is very similar to the z-score from a normal distribution (1.96 for
95% confidence)
- Also, we don’t know the sample standard deviation s before collecting the data
- When we use a z-score instead of a t-score, supply and educated guess for the standard
deviation of the sample mean σ / √n and then set z (σ √ n)equal to a desired margin of error
m and solve for n
- The random sample size n for which a confidence interval for a population mean μ has
σ 2 z2
margin of error approximately equal to m is: n= 2
m
o To use this, you need to guess the value for the population standard deviation σ
,
