# 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 d

_{0}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 n_{x}, n_{y} 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 p_{0} 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 O
_{i}), 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; O_{ij} 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, E_{ij} is

where R_{i} is the row i sum of counts; C_{j} 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

n_{i} is the number of observations in group i

is the sample mean in group i

is the jth observation in group i