Statistical Tests
A statistical test is a technique of statistical inference. The aim is to conclude whether there is sufficient evidence
to reject a claim or hypothesis.
Important terms
Interpretation of Result
Problem:
I have taken the test scores of ten seventh grade students of Bishop Westcort girls school. We want to compare these students with city score , which is 100. The null hypothesis is that Bishop students are similar to the city average and the alternative hypothesis is that they are different from the city average, being either higher or lower.
Step 1: State Null and Alternate Hypothesis
H0 : M = 100
H1: M ≠ 100
Step 2: Set the Significance Level (α)
Confidence Interval percentage = 95%=0.95
α=1-0.95=0.05
Step 3:Calculate the Test Statistic and Corresponding P-Value
M= sample mean=100.40
μ = population mean=100
S / √n = sample error mean=5.46/√10=1.73
t = (M — μ) / (s / √n)=(100.40-100)/1.73=.23
p=.822
Step 4: Calculating Confidence Interval
(a): Subtract 1 from your sample size. (In the above example n=10). So degrees of freedom (df) = 10 – 1 = 9.
(b): Subtract the confidence level from 1, then divide by two for two tailed test.
(1 – 0.95) / 2 = 0.025
(c): Look up the answers to (a) and (b) in the t-distribution table. For 9 degrees of freedom (df) and α = 0.025, result is 2.262.
df α = 0.1 0.05 0.025 0.01 0.005 0.001 0.0005
∞ tα=1.282 1.645 1.960 2.326 2.576 3.091 3.291
1 3.078 6.314 12.706 31.821 63.656 318.289 636.578
2 1.886 2.920 4.303 6.965 9.925 22.328 31.600
3 1.638 2.353 3.182 4.541 5.841 10.214 12.924
4 1.533 2.132 2.776 3.747 4.604 7.173 8.610
5 1.476 2.015 2.571 3.365 4.032 5.894 6.869
6 1.440 1.943 2.447 3.143 3.707 5.208 5.959
7 1.415 1.895 2.365 2.998 3.499 4.785 5.408
8 1.397 1.860 2.306 2.896 3.355 4.501 5.041
9 1.383 1.833 2.262
(d): Multiply t-socre (2.262) by sample error mean (1.73)
2.262 × 1.73 = 3.913
(e): For lower range, subtract step {d) from the mean difference
0.4 - 3.913 = -3.513
(f): For the upper range, add step {d) from the mean difference
0.4 + 3.913 = 4.313
Step 5: Drawing a Conclusion
if P-value <= significance level (α) => Reject null hypothesis because result is statistically significant.
P-value > significance level (α) => Fail to reject null hypothesis because result is not statistically significant.
in our example p (.822) value is greater than .05, so we fail to reject the null hypothesis. It means Bishop students are similar to city average.
Important terms
Types of Test
- Parametric tests – involve certain parameters of the population and sample such as mean, standard deviation etc. It includes Student's T-tests, Z-tests and ANOVA tests and believe data is from a normal distribution.
- Non-parametric tests – These tests are distribution-free tests because having fewer assumptions and no parameters. In non parametric test we have several tests such as Chi-square test, ann-Whitney test, Kruskal-Wallis test etc.
t-test
There are three major types of t-test:- One sample t-test: Used to tests the mean of a single group against a known mean.
- Independent Samples t-test: used to compare the means of two groups.
- Paired sample t-test: It compares means from the same group but at different times. Used in mainly experimental research design
One sample t-test in PSPP
Steps
1. Create a data set
My Sample (Math Score) : 99.00, 101.00, 98.00, 102.00, 89.00, 110.00, 105.00, 101.00, 102.00, 97.00
2. Analyze-> Compare Means-> One Sample T Tests...
3. Assign test varibles and test malue (in our ex. test variable is Math Score and Test Value is 100)
4. Click options and specify Confidence Interval percentage
4. Press ok
Output
Interpretation of Result
Problem:
I have taken the test scores of ten seventh grade students of Bishop Westcort girls school. We want to compare these students with city score , which is 100. The null hypothesis is that Bishop students are similar to the city average and the alternative hypothesis is that they are different from the city average, being either higher or lower.
Step 1: State Null and Alternate Hypothesis
H0 : M = 100
H1: M ≠ 100
Confidence Interval percentage = 95%=0.95
α=1-0.95=0.05
Step 3:Calculate the Test Statistic and Corresponding P-Value
M= sample mean=100.40
μ = population mean=100
S / √n = sample error mean=5.46/√10=1.73
t = (M — μ) / (s / √n)=(100.40-100)/1.73=.23
p=.822
Step 4: Calculating Confidence Interval
(a): Subtract 1 from your sample size. (In the above example n=10). So degrees of freedom (df) = 10 – 1 = 9.
(b): Subtract the confidence level from 1, then divide by two for two tailed test.
(1 – 0.95) / 2 = 0.025
(c): Look up the answers to (a) and (b) in the t-distribution table. For 9 degrees of freedom (df) and α = 0.025, result is 2.262.
df α = 0.1 0.05 0.025 0.01 0.005 0.001 0.0005
∞ tα=1.282 1.645 1.960 2.326 2.576 3.091 3.291
1 3.078 6.314 12.706 31.821 63.656 318.289 636.578
2 1.886 2.920 4.303 6.965 9.925 22.328 31.600
3 1.638 2.353 3.182 4.541 5.841 10.214 12.924
4 1.533 2.132 2.776 3.747 4.604 7.173 8.610
5 1.476 2.015 2.571 3.365 4.032 5.894 6.869
6 1.440 1.943 2.447 3.143 3.707 5.208 5.959
7 1.415 1.895 2.365 2.998 3.499 4.785 5.408
8 1.397 1.860 2.306 2.896 3.355 4.501 5.041
9 1.383 1.833 2.262
(d): Multiply t-socre (2.262) by sample error mean (1.73)
2.262 × 1.73 = 3.913
(e): For lower range, subtract step {d) from the mean difference
0.4 - 3.913 = -3.513
(f): For the upper range, add step {d) from the mean difference
0.4 + 3.913 = 4.313
Step 5: Drawing a Conclusion
P-value > significance level (α) => Fail to reject null hypothesis because result is not statistically significant.
in our example p (.822) value is greater than .05, so we fail to reject the null hypothesis. It means Bishop students are similar to city average.
