State the ‘null hypothesis’ State the ‘alternative hypothesis’ State either one-tailed or two-tailed test State the chosen statistical test with reasons State the level of significance (usually 5%) Present calculations Draw conclusions – accept or reject the ‘null hypothesis’

Nominal level data (categorical) Expected frequencies in any cell should not fall below 5 Testing Genetic Ratios The Chi-squared test is commonly used for comparing experimental genetic data with that expected from predicted ratios

In a breeding experiment in which tall, purple flowered pea plants were allowed to self pollinate, a Mendelian ratio of 9 : 3 : 3 : 1 was predicted among the progeny as follows: Tall/Purple : Tall/White : Dwarf/Purple : Dwarf/White = 9 : 3 : 3 : 1 The Chi-squared test was used to determine how well the observations fitted the predicted ratio; in this context, Chi-squared is used as a test of ‘goodness of fit’ Null Hypothesis: There is no difference between the results of the genetic cross and the predicted Mendelian ratio of 9 : 3 : 3 : 1 Alternative Hypothesis: The experimental results differ from the predicted 9 : 3 : 3 : 1 ratio

The basis of the Chi-squared test is the difference between observed results (O) and the expected results (E) predicted by the ‘null hypothesis’ Results of the Experimental Genetic Cross Tall with purple flowers245 Tall with white flowers75 Dwarf with purple flowers63 Dwarf with white flowers27 Calculate the expected results (according to a 9 : 3 : 3 : 1 ratio) Calculate the differences between the observed and expected results (O – E) and square the differences (O – E) 2 Divide each (O – E) 2 by the relevant expected result and sum the values obtained to obtain the  2 (Chi-squared) value Calculate the number of degrees of freedom (n – 1) = number of categories – 1 Reject the ‘null hypothesis’ and accept the ‘alternative hypothesis’ if the  2 value is greater than the critical value,  2 crit at the 5% significance level

Chi-squared result from a statistics package probability 2 (O – E) 2 /E(O – E) 2 Expected (E) Ratio Observed (O) 7.81Critical Value 3Degrees of Freedom The  2 value (3.52) is less than the critical value  2 crit (7.81), and thus we accept the ‘null hypothesis’; the experimental results do conform to a 9 : 3 : 3 : 1 ratio The value of p (probability) is > 0.05, so there is no significant difference between the experimental results and a 9 : 3 : 3 : 1 ratio

The probability value gives us a measure of the extent to which chance has caused any difference between the observed and expected results For a  2 value of 3.52 and 3 degrees of freedom there is between a 30% and 50% probability that chance alone has caused the difference between the observed and expected values, i.e. p lies between 0.30 and 0.50 (the actual value is 0.32); no significant difference between observed and expected results The  2 value (3.52) is less than the critical value,  2 crit (7.81) for a 5% (0.05) level of significance and thus we accept the ‘null hypothesis’ critical value

A group of students investigated the response of Daphnia to light by introducing approximately 12 organisms into a water filled tube, and counting the numbers present in the weakly illuminated and dark halves after a period of 100 seconds

Null Hypothesis: Lighting conditions have no effect on the distribution of Daphnia Alternative Hypothesis: Lighting conditions do have an effect on the distribution of Daphnia As nominal (categorical) data was obtained from the experimental procedure the Chi-squared test was used to determine whether the null hypothesis should be accepted The Yates’ correction is applied before squaring the differences between observed and expected results In this example there is only one degree of freedom and thus the Yates’ correction should be applied to enhance the accuracy of the  2 value

The Yates’ correction The Yates’ correction is applied to enhance the accuracy of the  2 value when there is only ONE degree of freedom The differences between the observed and expected results (O – E) are calculated in the normal way but any negative difference is converted to a positive value; this is sometimes written |O – E| and is called the absolute value or modulus of (O – E) When all positive values of (O – E) have been determined, the value 0.5 is subtracted from each of the positive differences before these quantities are squared The  2 value calculation therefore becomes:

Daphnia Distribution Results Dark area O Σ = Illuminated area [(O – E) – 0.5] 2 /E [(O – E) – 0.5] 2 (O – E) – 0.5(O – E)E Category Number of Daphnia Group Totals Illuminated area Dark area Calculate the totals and complete the table below Use the  2 value and critical values table to determine  2 crit, at the 5% significance level for one degree of freedom Use a statistic package to compare your values Interpret the findings

Category OE(O – E)(O – E) – 0.5[(O – E) – 0.5] 2 [(O – E) – 0.5] 2 /E Illuminated area Dark area Σ = 4.87 The  2 value is 4.87 Daphnia Distribution Results

For a  2 value of 4.87 and 1 degree of freedom, there is between a 1% and 5% probability that chance alone has caused the difference between the observed values and those expected if the ‘null hypothesis is true i.e. p lies between 0.01 and 0.05 ( the actual value is 0.02); there is a significant difference between observed and expected results The  2 value (4.87) is greater than the critical value,  2 crit (3.84) for a 5% (0.05) level of significance, and thus we reject the ‘null hypothesis’ and accept the ‘alternative hypothesis’ that lighting conditions do have an effect on the distribution of Daphnia with more organisms congregating in the weakly illuminated areas than in the darker areas

A student investigated the relationship between eye colour and hair colour for a biology project Null Hypothesis: There is no association between eye colour and hair colour Alternative Hypothesis: There is an association between eye colour and hair colour The student performed the Chi-squared test of association to determine if there was a significant relationship between eye colour and hair colour 23147brown green/grey 91753blue blackbrownfair/red Hair colour Eye colour

fair/redbrownblack Row totals blue53179 green/grey brown71423 Column totals Grand total Calculate the row and column totals Calculate the grand total for the data

fair/redbrownblack Row totals blue green/grey brown Column totals Grand total Calculate the expected frequency, E, for each data cell using the formula:

fair/redbrownblack Row totals blue green/grey brown Column totals Grand total Calculate the expected frequency, E, for each data cell using the formula: e.g. the expected value for brown hair & blue eyes = 79 x =

O E brown green / grey blue blackbrownfair/red Draw up a table and calculate (O – E) 2 ÷ E for each data cell Sum together the values of (O – E) 2 ÷ E to obtain the  2 test value

Observed frequency (O) Expected frequency (E) (O – E) 2 ÷ E Number of degrees of freedom df = (Number of rows – 1) x (Number of columns – 1) df = (3 – 1) x (3 – 1) df = 2 x 2 = 4 Use the critical values table to determine the critical value,  2 crit corresponding to four degrees of freedom at the 5% (0.05) significance level

Observed frequency (O) Expected frequency (E) (O – E) 2 ÷ E Reject the ‘null hypothesis’ and accept the ‘alternative hypothesis’ if the  2 value is greater than the critical value,  2 crit at the 5% significance level For 4df the  2 value (50.49) is greater than the critical value,  2 crit (9.49) for a 5% (0.05) level of significance and thus we reject the ‘null hypothesis’

Within the species Cepaea nemoralis (land snail), there is much variation in the colours and banding patterns of the shells A group of students investigated the distribution of banded and unbanded snails in two different localities - a deciduous woodland and open grassland Data was entered into a 2 x 2 contingency table Null Hypothesis: There is no relationship between banding pattern and locality Alternative Hypothesis: There is an association between banding pattern and locality

The students wanted to determine if there is an association between banding pattern and locality and performed the Chi-squared test of association 1852Grassland 10536Woodland UnbandedBanded Snail Type Location Snail Distribution Results

Location Snail Type Row totals BandedUnbanded Woodland36105 Grassland5218 Column totals Grand total Calculate the row and column totals Calculate the grand total for the snail data

Location Snail Type Row totals BandedUnbanded Woodland Grassland Column totals Grand total Calculate the expected frequency, E, for each data cell using the formula: e.g. the expected value for banded snails in the woodland = 141 x =

The Yates’ correction is applied before squaring the differences between observed and expected results In this example there is only one degree of freedom and thus the Yates’ correction should be applied to enhance the accuracy of the  2 value Table showing expected and observed results 1852 Grassland Woodland UnbandedBanded Snail Type Location O E

[(O – E) – 0.5] Calculate the absolute values of (O – E), i.e. convert negative values to positive, and then subtract 0.5 from each of the differences Draw up a table and calculate [(O – E) – 0.5] 2 ÷ E for each data cell Sum together the values of [(O – E) – 0.5] 2 ÷ E to obtain the  2 test value [(O – E) – 0.5] 2 Square the corrected differences Table showing expected and observed results 1852 Grassland Woodland UnbandedBanded Snail Type Location O E

Observed frequency (O) Expected frequency (E) [(O – E) – 0.5] 2 E Number of degrees of freedom, df = (Number of rows –1) x (Number of columns – 1) df = (2 – 1) x (2 –1) df = 1 x 1 = 1 Use the critical values table to determine the critical value,  2 crit corresponding to one degree of freedom at the 5% (0.05) significance level

Reject the ‘null hypothesis’ and accept the ‘alternative hypothesis’ if the  2 value is greater than the critical value,  2 crit at the 5% significance level Observed frequency (O) Expected frequency (E) [(O – E) – 0.5] 2 E

For 1df, the  2 value (43.74) is greater than the critical value,  2 crit (3.84) for a 5% (0.05) level of significance and thus we reject the ‘null hypothesis’ Observed frequency (O) Expected frequency (E) [(O – E) – 0.5] 2 E