Small Sample Data Analysis

1
Small Sample Data Analysis
2
Overview

• To frame our discussion, consider
• Justification
• t test

3
Outline
4
Sample Size

• For large samples (n30) the sampling
  distribution is approximately normal. Hence, we
  can utilize the z-statistic to determine
  significance.
• What of small(er) samples?

5
Techniques

• Students t distribution
• Differences in means
• F-distribution
• Differences in variances.
• Chi-square distribution
• Differences in frequencies.

6
Students t distribution

• Applied when working with normal or nearly normal
  populations.
• Distribution is similar to normal. Shape is a
  function of the number of degrees of freedom.
• Degrees of freedom (v) is n-k, where k is the
  number of population parameters that must be
  estimated.

7
t statistic
8
Example

• Research group develops estimation technique for
  software projects which they claim provides an
  accuracy of 15. The technique is applied to 10
  projects, yielding a sample mean of 15.9 and
  standard deviation of 0.9.

9
Hypotheses?

• H0 m 15
• H1 m ? 15

10
Compute t

• Sample Mean 15.9
• s 0.9
• n 10
• Population mean?, use 15 as estimate
• t3

11
Use t test

• a .05
• To accept H0, t must fall in the .95 range.
• Two tailed test, where 2.5 falls on each side.
• Annex III, p. 385
• Degrees of freedom, v10-1 (had to estimate
  population mean)
• 2.26 provides the t values for which 95 fall
  between.
• Since t3, we can reject the null hypothesis with
  a 95 confidence level.

12
Example

• Suppose two different code inspection techniques
  are applied to 24 programs of similar size with
  each technique applied to 12 programs. The first
  technique results in a mean of 4.8 errors per
  unit of time with a standard deviation of 0.4
  second. The second technique results in a mean of
  5.1 errors per unit of time with a standard
  deviation of 0.36 second.
• Is the observed value of the response variable
  significant?

13
Hypotheses?

• H0 m1 m2
• H1 m1 lt m2

14
Difference of Means

• Based on the assumptions, we have

15
Solution

• n112 n212x15.1x24.8s1.36s2.4
• s 0.38
• t1.93
• a .01
• One tailed test, would reject H0 if tgt t.99
• Degrees of freedom 22 (n1 n2-2).
• tgt t.99 is false, hence we do not reject the null
  hypothesis.