F distribution, equal variances test, ANOVA, mean squared error formula, two factor experiment, interaction effect + blocking variables

Topic 7

The F distribution

A continuous distribution

Cannot be negative

Asymmetric - POSITIVELY skewed

Is family of distributions with two degrees of freedom parameters, one in the numerator, one in the denominator

Asymptotic - the curve approaches the x-axis but doesn’t reach it

Equal variances test - two assumptions

Populations are normally distributed

The level of measurements is interval or ratio in order to calculate the variances

Also typically run two-tailed tests, as one tailed-tests not very useful

Equal variances test - steps

Step 1: set the hypotheses:

H_0:\sigma_1^2=\sigma_2^2

H_1:\sigma_1^2\ne\sigma_2^2

Step 2: Select the significance level

Step 3: Select the appropriate test statistic (F statistic) using:

F=\frac{\sigma_1^2}{\sigma_2^2} - where the larger sample variance goes in the numerator - meaning only large values of the F statistic reject the null

Step 4: Formulate the decision rule - we reject the null hypothesis if the test statistic is in the critical region (F > F_crit)

Step 5: Make your decision - find F and the critical value from the tables, and make conclusion within context

ANOVA - analysis of variance

Enables us to test the equality of several means simultaneously

Previously z/t tests only allowed us to compare two means simultaneously

Assume the sampled populations must follow a normal distribution, have equal standard deviations, and the samples are randomly selected and independent

ANOVA - steps

Step 1: The null hypothesis is that the population means are all the same, and the alternate hypothesis is that at least one of the means is different

H₀: =\mu_1=\mu_2=\mu_3=\mu_4

H₁: the means are not all equal

Step 2: Choose significance level which is usually \alpha = 0.01

Step 3: Use the F distribution (table) to find F_{critical value}

Step 4: The decision rule is to reject the null hypothesis if the F statistic is in the critical region of the F distribution (if the computed F value is LARGER than F_{critical value} we can then REJECT null hypothesis)

The degrees of freedom in the numerator = the number of populations sampled (k) minus 1

The degrees of freedom in the denominator are the total number of observations (n) minus k

Step 5: Compute the test statistic and make a decision using

F=\frac{\frac{SST}{\left(k-1\right)}}{\frac{SSE}{\left(n-k\right)}} - SST: treatment sum of squares, and SSE: error sum of squares

Calculate SSTotal (each individual xvalue - overall mean) and SSE (each individual xvalue - mean for that particular sample the value is in)- then you can get SST by doing SSTotal - SSE (easier)

Analysis of variance - sum of squares

We need to calculate 3 suANOVA, m of squares terms:

Total sum of squares: SSTotal = SST + SSE - Sum of squares of all deviations from the overall averageSSTotal=\Sigma\left(x_{i}-\overline{x}_{G}\right)^2

Treatment sum of squares: SST - Sum of squares of all deviations of factory means from the overall average - add/multiply n (sample) number to the front of each ()² brackets which is each individual mean - the overall mean

Random/Error sum of squares: SSE - This is the sum of squares of deviations within each factory from the factory average - sum of all ()² which includes each observation minus the mean for that particular sample

Mean squared error formula

MSE = SSE/n-k

WHERE Random/Error sum of squares: SSE - This is the sum of squares of deviations within each factory from the factory average - sum of all ()² which includes each observation minus the mean for that particular sample

Confidence interval for finding the DIFFERENCE in treatment means

We may know there is a difference but we don’t know which ‘treatment’/sample is different

However ONLY do this test if the null hypothesis is REJECTED

Use formula:

\left(\overline{x}_1-\overline{x}_2\right)\pm t_{\alpha,n-k}\sqrt{MSE\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}

x₁ and x₂ = means of first and second samples

t_{\alpha,n-k} - critical value obtained from t-distribution tables for the required alpha with n-k degrees of freedom - USING entire sample size for n

Then if the confidence interval INCLUDES 0, the difference between means is NOT statistically significant

Therefore if both ends of the confidence interval have the same sign e.g. both negative then we can conclude these treatment means differ

Blocking variable

A second treatment variable that when included in the ANOVA analysis will have the effect of reducing the SSE term

Formula:

SSB=k\!\sum_{i=1}^{b}\left(\overline{x}_{i}-\overline{x}_{G}\right)^2 - k = number of treatments, b = number of blocks, x_G = overall/grand mean

e.g. A test to compare the mean travel time on 4 different routes from point A to B - we can assume differences in travel time are either due to the routes or random

But if we then get 5 different drivers to drive each route - the drivers would be the blocking variable

Then find SSE = SSTotal - SST - SSB

Should be lower than SSE before adding blocking variable

Two-factor experiment

Conducting a hypothesis test for the difference in BLOCK means

First we conduct test for treatment means first:

H₀ = block means are all equal (write with mew)

H₁= the block means are not all equal

use k-1 for degrees of freedom in numerator

(b-1)(k-1) for denominator

Find critical value, and F statistic (using F = MST/MSE)

MST = SST/(k-1) AND MSE = SSE/[(k-1)(b-1)]

Decide if reject if F > F_critical value

Then conduct test for block means

use b-1 for degrees of freedom in numerator

and (b-1)(k-1) for denominator

Find new critical value and F statistic (using F = MSB/MSE)

MSB = SSB/(b-1)

Decide if reject if F>F_critical value - then conclude

Interaction effect

The effect of one factor on a response variable differs depending on the value of another factor

E.g. differences in mean travel time may depend on the COMBINED effect of driver and route

If means are plotted with NO interaction effect the lines would be parallel

Testing for interaction effect

First find and plot the means for each driver/route (the two factors) combination

If lines are NOT parallel there are likely interaction effects

Need a hypothesis test to see if observed interactions are SIGNFICANT

H₀: There is no interaction between drivers and routes

H₁: There is interaction between drivers and routes

Use ANOVA With Interaction Table - and use Excel to find p-value - to know if we reject null hypothesis

If there is interaction present test for differences in factor means using a one-way ANOVA for each level of the other factor

Then conduct a one-way ANOVA for each route