Hypothesis testing formula sheet
Test of mean of a normal distribution
a) case population variance known
Critical value(s) from Z table
b) case population variance is unknown
Critical value(s) from t table with df = n-1
Test for the difference between 2 means
a) case of matched pairs
and d0 is the hypothesized difference.
Critical value(s) from t table with df = n-1.
Independent samples with population variances known or large samples
For large nx, ny and unknown population variances, use sample variances.
c) Independent samples, small samples with population variances equal but unknown
where
Test of the population proportion in large samples
where p0 is the value of the hypothesized value of the proportion. Use Z table.
Testing the difference between 2 proportions in large samples
Use Z table.
Goodness-of-fit test formula sheet
Consider a random sample such that:
- there are n observations
- observation i (let’s denote it Oi), i=1,..n, falls into only one of k categories
The null hypothesis specifies the probabilities for an observation falling into each into each category.The test statistic that the data do follow the given probabilities is:
where the expected count for cell i if we there null were true, is
Under the null, the test follows a chi-square distribution with df = number of categories (K) – number of any parameters estimated to calculate the probabilities (m) -1
Reject the null if the test statistic exceeds the critical value for a given significance level. The goodness-of-fit test is a one-tailed test.
Crosstabs formula sheet
Crosstabs is short for cross tabulation (and is also known as “contingency table”)
Consider a random sample of size n that are cross classified according to 2 attributes in a table with r rows and c columns.
The null hypothesis is: there is no relationship between the 2 attributes in the population. The test statistic is:
where the summation is over all cells; Oij is the actual observed count in cell row i, column j. The estimated expected count for cell in row i and columnn j if the null were true, Eij is
where Ri is the row i sum of counts; Cj is column j sum of counts.
Under the null, the test follows a chi-square distribution with df = (r-1)*(c-1).
This test is a one-tailed test, so reject the null if the test statistic exceeds the critical value for a given significance level.
One way between groups ANOVA
One-Way ANOVA (ANOVA stands for Analysis of Variance) is used where we wish to test the equality of population means of 3 or more groups. (So it is like an extension of a t-test).
Conditions needed to run a one way ANOVA
- data are normally distributed
- each group has unknown but common variance
One way ANOVA table
| Source | Degree of freedom | Sum of square(SS) | Mean square(MS) | F |
| Betwen | K-1 | B | MSB = B / (K-1) | MSB / MSW |
| Within | n-K | W | MSW = W / (n-K) | |
| Total | n-1 | SST |
The null hypothesis is that population means in each group is equal. Reject the null at a given significance level if the test statistic exceeds the critical value from the F table with degree of freedom K-1(numerator), n-K (denominator):
The quantities in the table are:
K is number of group means being compared
ni is the number of observations in group i
is the sample mean in group i
is the jth observation in group i
Tags: crosstabs formula, hypothesis tests formula, t-test formula, z-test
No comments
Comments feed for this article
Trackback link: http://www.statisticsmentor.com/2012/07/05/hypothesis-tests-formula-for-basic-tests/trackback/