CHAPTER 8: HYPOTHESIS TESTING: DO MY SAMPLES COME FROM THE SAME POPULATION

Table of Contents

1. CHAPTER 8: HYPOTHESIS TESTING: DO MY SAMPLES COME FROM THE SAME POPULATION? NON-PARAMETRIC DATA
   1. Exercise 1
   2. Exercise 2

Exercise 2

This example was also used in interactive exercise 2 in Chapter 7. The aim of this exercise (exercise 2, Chapter 8) is to allow you to test your understanding of sections 4.1.2, 8.5 and 8.7 and illustrates why you may obtain negative values in a non-parametric ANOVA. .

Example W8.3: The inhibitory effect of oregano oil and selected antibiotics on certain bacteria

As part of a larger investigation an undergraduate examined the effectiveness of the essential oil from oregano in comparison with commercial antibiotics. Each substance (oregano oil etc) was added to a small disc of paper and placed on to agar plates inoculated either with Escherichia coli or Pseudomonas aeruginosa. After incubating the plates for 24 hours at 37^°c the zone of inhibition was recorded. The zone of inhibition is the mean diameter (mm) of the area of clear agar around the discs containing the substance under test (Table W8.3). The size of the zone of inhibition indicates the area over which the bacteria have been killed or their growth inhibited. Eight replicate plates were used for each treatment.

Table W8.3: The mean diameter of the zone of inhibition (mm) produced on agar plates inoculated either with Escherichia coli or Pseudomonas aeruginosa as a result of treatment with either oregano essential oil or streptomycin

	Mean diameter of	zone of inhibition (mm)
	E. coli	P. aeruginosa
Streptomycin	17.0	12.0
	16.0	11.0
	16.0	11.0
	15.0	10.0
	16.0	10.0
	16.0	11.0
	16.0	11.0
	17.0	11.0
Oregano oil	21.0	10.0
	22.0	9.0
	21.0	10.0
	21.0	10.0
	24.0	10.0
	22.0	10.0
	20.0	10.0
	22.0	10.0

The scale of measurement used (mm) is an interval scale but there are only 8 replicates for each treatment so it is difficult to confirm that the data are parametric. If the data were parametric it could be analysed using a parametric two-way ANOVA. One criteria for using this parametric test is that the variances have to be homogeneous (7. 7.1) which can be confirmed using an F_max test (BOX 7.5). When the F_max test was carried out on this data it was clear that F_{max calculated} (11.29) was greater than F_{max critical} (8.44) at p = 0.05, so the variances are not homogeneous and this criterion is not met. Therefore the data will be analysed using non-parametric tests.

1

Q W2.1 The undergraduate decided to use a two-way non-parametric ANOVA to test general hypotheses. Write out in full the general hypotheses she was testing. [If you would like to save a record of your answer, please type it into this Word document instead of the text box below] Since this is a two-way ANOVA we are examining three pairs of hypotheses. The first relate to variable one (species of bacteria), the second to variable two (potential bactericide treatments) and the third to the possible interaction between the two variables. (Columns) H0: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides. H1: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides. (Rows) H0: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria. H1: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria. (Interaction) H0: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide. H1: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide.

2

Q W2.2 Test the first and second of these pairs of hypotheses. What is the outcome? In the answer boxes below record Fcalculated to 2 decimal places. [If you would like to save a record of your answer, please type it into this Word document] Columns: Difference between bacteria species; Fcalcuated = Rows: Difference between bactericides; Fcalcuated = Columns: Difference between bacteria species; Fcalcuated = 22.72 Rows: Difference between bactericides; Fcalcuated = -0.3

Full calculation for the Columns hypotheses

1. Hypotheses to be tested

H₀: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides.

H₁: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides.

2. To work out K calculated.

i, ii and iii Table W8.4.

There are two columns E. coli and P. aeruginosa and 16 observations in each column.

Table W8.4: Calculation table for two-way non-parametric ANOVA test: General hypotheses (columns) examining the mean diameter of the zone of inhibition (mm) produced on agar plates inoculated either with E. coli or P. aeruginosa as a result of treatment with either oregano essential oil or streptomycin

	E.	coli	P. aeruginosa
	Mean diameter of zone of inhibition (mm)	Rank	Mean diameter of zone of inhibition (mm)	Rank
Streptomycin	17.0 16 .0 16.0 15.0 16.0 16.0 16.0 17.0	23 20 20 17 20 20 20 23	12.0 11.0 11.0 10.0 10.0 11.0 11.0 11.0	16 13 13 6 6 13 13 13
Oregano oil	21.0 22.0 21.0 21.0 24.0 22.0 20.0 22.0	27 30 27 27 32 30 25 30	10.0 9.0 10.0 10.0 10.0 10.0 10.0 10.0	6 1 6 6 6 6 6 6
Totals for the columns	n R R2	16 391 152881 9555.0625	n R R2	16 136 18496 1156

iv.

Σ (R²/n) = 9555.0625 + 1156 = 10711.0625

N = 16 + 16 = 32

v. K_{calculated columns}

= =

= (10711.0625 × 0.01136) - 99

= 121.71662 - 99 = 22.71662

3. To find K critical

Using a table of critical values for and p = 0.05. The degrees of freedom () = number of columns - 1. This is a two-tailed test.

= 2 - 1 = 1

K_{critical columns} = 3.84 at p = 0.05

4. The rule

K _{calculated columns} (22.72) is more than K_critical (3.84). You may therefore reject the null hypothesis (H₀).

In fact at p = 0.001, K_critical = 10.83. Therefore, we can reject the null hypothesis at this higher level of significance.

5. What does this mean in real terms?

There is a highly significant difference (K = 22.72, p = 0.001) medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides.

Full calculation for the Rows hypotheses

1. Hypotheses to be tested

H₀: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria.

H₁: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria.

2. To work out K calculated.

i, ii. Table W8.5.

There are two rows 'streptomycin' and 'oregano oil' each with 16 observations in the row.

Table W8.5: Calculation table for two-way non-parametric ANOVA test: General hypotheses (columns) examining the mean diameter of the zone of inhibition (mm) produced on agar plates inoculated either with E. coli or P. aeruginosa as a result of treatment with either oregano essential oil or streptomycin

		Mean diameter of zone of inhibition (mm)	Rank	Terms for calculation
Streptomycin	E.coli	17.0 16.0 16.0 15.0 16.0 16.0 16.0 17.0	23 20 20 17 20 20 20 23	n R R2	16 256 65536 4096
	P. aeruginosa	12.0 11.0 11.0 10.0 10.0 11.0 11.0 11.0	16 13 13 6 6 13 13 13
Oregano oil	E.coli	21.0 22.0 21.0 21.0 24.0 22.0 20.0 22.0	27 30 27 27 32 30 25 30	n R R2	16 271 73441 4590.0625
	P. aeruginosa	10.0 9.0 10.0 10.0 10.0 10.0 10.0 10.0	6 1 6 6 6 6 6 6

iii.

ΣR²/n = 4096 + 4590.0625 = 8686.0625

Σn = N = 16 + 16 = 32

iv. K_{calculated rows}

= =

= (8686.0625 × 0.01136) - 99 = 98.70526 - 99 = -0.2947

3. To find K critical

Using a table of critical values for (² and p = 0.05. The degrees of freedom () = number of columns - 1. This is a two-tailed test.

Since = 2 - 1 = 1, then K_{critical rows} = 3.84 at p = 0.05

4. The rule

K_{calculated rows} (-0.2947) is less than K_critical (3.84). You may not reject the null hypothesis (H₀).

5. What does this mean in real terms?

There is no significant difference (K = -0.3, p = 0.05) between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria.

Full calculation for the Interaction hypotheses

1. Hypotheses to be tested

H₀: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide.

H₁: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide.

2. To work out K calculated

i. Table W8.6. There are four samples A - D, each with 8 observations.

Table W8.6: Calculation table for two-way non-parametric ANOVA test: General hypotheses (interaction) examining the mean diameter of the zone of inhibition (mm) produced on agar plates inoculated either with E. coli or P. aeruginosa as a result of treatment with either oregano essential oil or streptomycin

	E.	coli	P. aeruginosa
	Mean diameter of zone of inhibition (mm)	Rank	Mean diameter of zone of inhibition (mm)	Rank
Streptomycin	17.0 16 .0 16.0 15.0 16.0 16.0 16.0 17.0	23 20 20 17 20 20 20 23	12.0 11.0 11.0 10.0 10.0 11.0 11.0 11.0	16 13 13 6 6 13 13 13
Totals for the sample	A n R R2 RA2/nA	A 8 163 26569 3321.125	B n R R2 RB2/nB	B 8 93 8649 1081.125
Oregano oil	21.0 22.0 21.0 21.0 24.0 22.0 20.0 22.0	27 30 27 27 32 30 25 30	10.0 9.0 10.0 10.0 10.0 10.0 10.0 10.0	6 1 6 6 6 6 6 6
Totals for the sample	C n R R2 RC2/nC	C 8 228 51984 6498	D n R R2 RD2/nD	D 8 43 1849 231.125

ii. ΣR²/n = 3321.125 + 1081.125 + 6498 + 231.125 = 11131.375

iii. N = 8 + 8 + 8 + 8 = 32

iv. K_{calculated total} =

= =

= 126.4929 - 99 = 27.4929

v.

K_interaction

K_columns = 22.71662

K_rows = -0.2947

K_total = 27.4929

K_interaction = 27.4929 - 22.71662 - (-0.2947) = 5.07098

3. To find K critical

for K_total= 4 - 1 = 3

for K_bacteria= 4 - 1 = 3

for K_bactericides = 4 - 1 = 3

for K_interaction= 4 - 1 = 3

Therefore, the critical value for K_interaction at p = 0.05 and × = 1 is 3.84.

4. The rule

K_{calculated interaction} (5.07) is more than K_critical (3.84) so you may reject the null hypothesis.

5. What does this mean in real terms?

There is a significant difference (K_{calculated interaction} 5.07, p = 0.05) between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide.

3

Q W2.3 When testing the hypotheses relating to the rows the result is a negative number. Negative numbers can arise in this calculation if you have a number of tied ranks. If you examine table W8.4 you will see that only 4 of the 32 values do not have a tied rank. It is not therefore possible to continue with this test. As a result the student decided to use the Scheirer-Ray-Hare test to examine these data. Are the criteria for using this test met? [If you would like to save a record of your answer, please type it into this Word document] a) Yes b) No Correct. Criterion Is it met? 1. Wish to test for differences in population medians. Yes. 2. Have two treatment variables each with at least two categories. Yes. The two variables are 'species of bacteria' and 'possible bactericides' 3. The design is orthogonal. Yes. Every species of bacteria is tested with each bactericide. 4. Have non-parametric data that can be ranked. Yes. The scale of measurement is mm which is an interval scale which is ordered in numerical order. Incorrect. Criterion Is it met? 1. Wish to test for differences in population medians. Yes. 2. Have two treatment variables each with at least two categories. Yes. The two variables are 'species of bacteria' and 'possible bactericides' 3. The design is orthogonal. Yes. Every species of bacteria is tested with each bactericide. 4. Have non-parametric data that can be ranked. Yes. The scale of measurement is mm which is an interval scale which is ordered in numerical order.

4

Q W2.4 Carry out the Scheirer- Ray- Hare test and complete this ANOVA table. Have you completed the table? a) Yes b) No Table W8.8: ANOVA table for analysis of data from Example W8.3. The inhibitory effect of oregano oil and selected antibiotics on certain bacteria Source of variation SS s2 Hcalculated Hcritical p Between species (columns) 2032.03 1   23.95 10.83 0.001 Between bactericides (rows) 7.03 1   0.083 3.84 0.05 Interaction Species × bactericides. 413.31 1   4.87 3.84 0.05 Total 2629.97 31 84.84       Complete the table before checking our answer!

Full calculation for Q W2.4

1. General hypotheses to be tested

(Columns)

H₀: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides.

H₁: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) between Escherichia coli and Pseudomonas aeruginosa when treated with possible bactericides.

(Rows)

H₀: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria.

H₁: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) resulting from treatment by oregano oil compared to streptomycin on plates inoculated with bacteria.

(Interaction)

H₀: There is no difference between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide.

H₁: There is a difference between the medians of the 'mean diameter' of the zone of inhibition (mm) due to an interaction between bacteria species and putative bactericide.

2. How to work out F calculated

1. Calculate general terms

In this test we use only the ranks that have been assigned to each observation and not the actual observations (Table W8.7).

Table W8.7: Calculation table for a Sheirer-Ray-Hare test using the ranks for observations from Example W8.3. The inhibitory effect of oregano oil and selected antibiotics on certain bacteria

	E. coli	P. aeruginosa
	Rank	Rank
Streptomycin	23 20 20 17 20 20 20 23	16 13 13 6 6 13 13 13
Summary terms for each sample	n = 8 = 163 = 3347 s = 1.92261 mean = 20.375	n = 8 = 93 = 1173 s = 3.6228 mean = 11.625
Oregano oil	27 30 27 27 32 30 25 30	6 1 6 6 6 6 6 6
Summary terms for each sample	n = 8 = 228 = 6536 s = 2.32993 mean = 28.5	n = 8 = 43 = 253 s = 1.76777 mean = 5.375

1. First add together every observation in the complete data set (grand total) ∑x_T.
   
   23 + 20 + 20 + ........... 6 + 6 + 6 = 527

2. Square each and every observation and add all these squared values together ∑x²_T
   
   23² + 20²+ 20² + ........... 6² + 6² + 6² = 11309

3. Add all the observations in a sample (∑x_s). This is the sample total. Do this for each sample. Square each sample total (∑x_s)². Add these together. Divide this by the number of observations in a sample (n_s).
   
   
   
   
4. For each column add all the observations together (∑x_c) and square the total (∑x_c)² . Add these squared column values together and divide this total by the number of observations in the column (n_c).
   
   

5. For each row add all the observations (∑x_R) and square the total (∑x_R)². Add these squared values together and divide this value by the number of observations in the row (n_R).
   
   

6. Square the result from step 1 and divide this by N. N is the total number of observations in all the samples.
   
   527²/32= 8679.03125

2. Calculate the Sums of Squares (SS)

7. SS_Total = result from step 2 - result from step 6

= 11309 - 8679.03125 = 2629.96875

8. SS_samples = result from step 3 - result from step 6

= 11131.375 - 8679.03125 = 2452.34375

9. SS_{treatment 1 columns} = result from step 4 - result from step 6

= 10711.0625 - 8679.03125 = 2032.03125

10. SS_{treatment 2 rows} = result from step 5 - result from step 6

= 8686.0625 - 8679.03125 = 7.03125

11. SS_Interaction = result from step 8 - result from step 9 - result from step 10

= 2452.34375 - 2032.03125 - 7.03125 = 413.31125

12. SS_within = result from 7 - result from 8

= 2629.96875 - 2452.34375 = 177.625

3. Construct and complete an ANOVA calculation table

1. Draw an ANOVA table
2. Transfer the results from the calculations for SS_{treatment 1 columns}, SS_{treatment 2 rows}, SS_within, and SS_total into the ANOVA Table in column 1 (see Table W8.8).

3. Calculate the degrees of freedom ()

_{treatment 1 columns} = number of columns - 1 = 2 - 1 = 1 _{treatment 2 rows} = number of rows - 1 = 2 - 1 = 1 _interaction = (number of columns - 1)(number of rows - 1)

= (2 - 1) × (2 - 1) = 1

_within = N - (number of rows × number of columns)

= 32 - (2 × 2) = 32 - 4 = 28

_total = N - 1 = 32 - 1 = 31

Enter these results in column 2. (Table W8.8)

Table W8.8: ANOVA table for analysis of data from Example W8.3. The inhibitory effect of oregano oil and selected antibiotics on certain bacteria., showing how some of the terms are calculated

Source of variation	SS		s2	Hcalculated	Hcritical	p
Between species (columns)	2032.03125	1		2032.03125 ÷ 84.83770 = 23.95198	10.83	0.001
Between bactericides (rows)	7.03125	1		7.03125 ÷ 84.83770 = 0.08288	3.84	0.05
Interaction Species × bactericides.	413.31125	1		413.31125 ÷ 84.83770 = 4.87178	3.84	0.05
Total	2629.96875	31	84.83770

At this point the analysis diverges from that described for a two-way parametric ANOVA in Chapter 7 and you should refer to 8.7.2. Firstly an s²_Total is calculated by adding all the SS values together and dividing the total by the Total degrees of freedom (Table W8.8).

s²_Total = 2629.96875 / 31 = 84.83770

3. To work out the values for H_calculated

H values are calculated instead of an F value as SS/s²_Total (Table W8.8).

4. To find F_critical

The critical values for H are found in a chi-squared table for the degrees of freedom associated with the SS (Table W8.8).

5. The rule

The Rule is the same as that given for the parametric two-way ANOVA (7.7, BOX 7.6) in that if H_calculated is greater than H_critical then you may reject the null hypothesis.

In our example it can be seen that there is a highly significant difference (H = 23.95, p = 0.01) between the median diameter (mm) of the zone of inhibition between species of bacteria when treated with bactericides; there is no significant difference (H = 0.08, p = 0.05) in the effectiveness of the bactericides; but there is a significant interaction (H = 4.87, 0.05 > p > 0.01) between the bacteria species and their response to the bactericides.

6. What does this mean in real terms?

This investigation has demonstrated that oregano oil can act as an effective bactericide under laboratory conditions but the effectiveness of the bactericides is dependent on the species of bacteria.

How to calculate Q W2.4 in Excel

There is no direct way of completing this calculation this using Excel. However the ANOVA two way with replication data analysis tool can be used to calculate the SS values and thus shortcut the calculation.

Step 1: Put the data into the spreadsheet using appropriate row and column headings

Step 2: From the top tool bar select, Tools then Data Analysis from the drop down menu.

Step 3: In the Data Analysis box, scroll down and select the Analysis Tool: Anova: Two-Factor With Replication. Click OK.

Step 4: Input values as indicated in the box.

Ensure that the cursor is flashing in the Input Range box. Input the cell references of the data by clicking on cell A2 (this includes the data or sample label) and dragging down and across the entire data set. The area on the spreadsheet will now be highlighted and the cell references shown in the input box.

Step 5: Click in the box marked Rows per sample: and type in 8 for the number of rows of values for each sample

Step 6: The default value for alpha (p) is 0.05 but this can be changed if required, but you will not normally require to do so.

Step 7: Next, select the output options. To return the output data below the input data, select 'OutputRange:' and then click in the box (the cursor will now flash in the box). Scroll over an area where you want the results to be displayed. Note that you could just select a couple of cells - Excel will determine the actual size that it requires for the results table. Note too, that it is essential to click in the box as well as selecting the 'Output Range' button. If this is not done the location is entered into the 'Input Range' box and the analysis cannot be completed.

Step 8: You can choose to have the results entered on to a 'New Worksheet Ply', in which case the results will be given on a fresh sheet, accessed by the tabs at the bottom of the current sheet. Alternatively a 'New Workbook' can be selected. We have opted to put the results table beneath the original data on the spreadsheet.

Step 9: Click OK and the Results table will be returned.

Step 10: Select the required values from the results table. Here we only require the SS values for Sample, Columns and Interaction from the ANOVA table.

Step 11: Then continue as in Table W8.8. ANOVA table for analysis of data from Example W8.3.

[Note that there isn't an easy way to do this calculation in SPSS or Minitab.]

5

Q W2.5 How do the outcomes from the non-parametric ANOVA and the Scheirer-Ray-Hare test compare? [If you would like to save a record of your answer, please type it into this Word document instead of the text box below] The outcomes from the hypotheses testing is the same in all cases and the calculated values (K and H) are very similar.