# T-test - unpaired Testing for a difference. What does it do? Tests for a difference in means Compares two cases (eg areas of lichen found in two locations)

## Presentation on theme: "T-test - unpaired Testing for a difference. What does it do? Tests for a difference in means Compares two cases (eg areas of lichen found in two locations)"— Presentation transcript:

t-test - unpaired Testing for a difference

What does it do? Tests for a difference in means Compares two cases (eg areas of lichen found in two locations) Your two samples do not have to be the same size

If your data are not normal, use Mann-Whitney If your data are also paired, use paired t-test You have five or more values in each sample Your data are normally distributed  Only continuous data (lengths, weights etc) can be normally distributed  You can do a check by seeing if your data look roughly like this: Planning to use it? Make sure that…

How does it work? You assume (null hypothesis) there is no difference between the two means The test works by calculating the means and standard deviations of the two samples, and substituting into a formula.

Doing the test These are the stages in doing the test: 1.Write down your hypotheseshypotheses 2.Finding the means & standard deviationmeans & standard deviation 3.Use the formula to calculate the t-valueformula 4.Write down the degrees of freedomdegrees of freedom 5.Look at the tablestables 6.Make a decisiondecision Click here Click here for an example

Hypotheses H 0: mean1 = mean2 For H 1, you have a choice, depending on what alternative you were looking for. H 1: mean1 > mean2 eg: Area of lichen in unpolluted location is greater than area of lichen in polluted location orH 1: mean1  mean2 eg: Area of lichen in unpolluted location is different to area of lichen in polluted location Unless you have a good scientific reason for expecting one to be larger, you should choose “different” for H 1

Means & Standard Deviation For each sample, find the mean by adding up all the values and dividing by how many there are Find the estimate of the standard deviation s for the two samples, using the formula n 1 and n 2 = sizes of the two samples  and = means of the two samples  x 1 2 and  x 2 2 = sums of squares of sample values

Formula Substitute your values into the formula n 1 and n 2 = sizes of the two samples  and = means of the two samples s = estimate of standard deviation

Degrees of freedom The formula here for degrees of freedom is degrees of freedom = n 1 + n 2 – 2 Where n 1, n 2 are the sample sizes You do not need to worry about what this means –just make sure you know the formula! But in case you’re interested – the fewer values you have, the more likely you are to get a large t-value by chance – so the higher your value has to be to be significant.

Tables This is a t table Degrees of freedom Significance levels - note different values for 1 and 2-tailed

Make a decision If your value is bigger than the tables value then you can reject the null hypothesis. Otherwise you must accept it. Make sure you choose the right tables value – it depends whether your test is 1 or 2 tailed:  If you are using H 1 : mean1 > mean 2, you are doing a 1-tailed test  If you are using H 1: mean1  mean 2, you are doing a 2-tailed test

Example: Lichen areas in polluted & unpolluted locations A student obtained the following data on areas of lichen at a polluted and an unpolluted location Polluted: 300 400 400 330 370 Unpolluted 600 420 540 190 260 800 740 Hypotheses: H 0: Mean area at polluted location = mean area at unpolluted location. H 1 Mean area at polluted location  mean area at unpolluted location

Means & Standard Deviation Polluted sample (sample 1): = ( 300 + 400 + 400 + 330 + 370)  5 = 360  x 1 2 = 300 2 + 400 2 + 400 2 + 330 2 + 370 2 = 655800 n 1 = 5 Similarly, for unpolluted sample (sample 2): = 507.14  x 2 2 = 2119300 n 2 = 7

The test Degrees of freedom = 7 + 5  2 = 10 We are doing a 2-tailed test Tables value (5%) = 2.228 Since our value is smaller than the tables value, we accept H 0 – there’s no significant difference in lichen area content

Similar presentations
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}