Pharmacy Calculations## Pharmaceutical Calculations: One-Way ANOVA with Microsoft Excel

**An**alysis **o**f **Va**riance (ANOVA) is a widely used statistical technique in clinical research, pharmacology, psychology, molecular medicine, and other fields of experimental science for analysing data. ANOVA can be used to determine if there is a statistically significant difference between the means of groups, due to some influence **factor**. In graphs and in figures, statistical significance is typically expressed using asterisks (*) and sometimes, authors also state the significance level (α).

In the context of ANOVA, an influence **factor** is an independent treatment variable that an experimenter can control and/or vary, which may or may not result in significant differences in means. Examples of possible factors include temperature, solvent polarity, catalyst, location, method of analysis, diet or drug treatments.

**One-way ANOVA** looks at the effects of one factor. The null and the alternative hypotheses for one-way ANOVA can be stated as follows:

**Null Hypothesis (H**: There is no difference in the means due to a factor. Using mathematical expressions, we can express H_{0})_{0}as: µ_{1}= µ_{2}= µ_{3}… = µ_{n}**Alternative Hypothesis (H**: There is a significance difference in the means due to a factor._{1})

One-Way ANOVA is also referred to as **unifactor ANOVA** or **single factor ANOVA.**

The calculation of the **F-statistic**, which is simply the ratio of the variation between sample means to the variation within the samples, is the basis of ANOVA calculations. The higher this ratio, the weaker the evidence supporting H_{0}.

In this article, we will show you how to do one-way ANOVA calculations on Microsoft Excel.

The t-test is a robust test for detecting a statistical difference in means between two populations that are normally distributed. When you are analysing more than two groups, it is strongly recommended to analyse your data using ANOVA instead. Using multiple t-tests **increases the chances of making a Type I error** (incorrect rejection of a true H_{0}!).

Research questions are often complex and typically involve more than one factor. Suppose a biochemist wants to look at the influence of stirring speed and temperature on the outcome of an assay or if a pharmacology researcher wants to determine if there is an interaction between an experimental drug treatment and gender on cytochrome P450 activity. In both scenarios, you have two factors (example 1: stirring speed and temperature & example 2: drug treatment and gender). In such scenarios, you would need to perform **multifactor ANOVA** (AKA **multiway ANOVA**) to analyse your data. You can also do Two-Way ANOVA calculations using Microsoft Excel’s “Data Analysis” but not Three-Way or greater.

In practice, doing one-way ANOVA by hand can be laborious and time-consuming compared to doing the calculations with software but yes, ANOVA by hand is possible. Aside from Microsoft Excel, **Minitab, GraphPad PRISM and SPSS**. Of course, you can do ANOVA calculations with **R**. In this tutorial, we will show you how to do ANOVA calculations with Microsoft Excel and how to interpret the results by showing you solutions to sample problems below:

You will need to enable the “Data Analysis” feature on Excel for this part. (**See our tutorial here**)

A pharmacology research laboratory is testing the effect of four drug candidates on the concentration of nitric oxide (NO) in rat plasma (n = 12). The data for the quantification of NO, in µmol/L, is displayed below. Determine if the treatments result in a significant change to the concentration of NO in rat plasma. (α = 0.05)

Although the example below uses Excel 2013, other versions of Excel work in a similar fashion.

**Step 1:**

**Enable “Data Analysis” on Excel**. If it’s already enabled, you can skip Step 1.

**Step 2:**

Tabulate your data on a completely empty spreadsheet as shown below

**Step 3:**

Select the “Data” tab > Click “Data Analysis” > Click “Anova: Single Factor” > Press OK

**Step 4:**

Click the highlighted box

**Step 5:**

Highlight your data, starting at the top left corner and ending at the bottom right corner of your data as shown below. Exclude the labels.

**Step 6:**

Set your significance level (α) under “Alpha:”

The default setting is “Alpha: 0.05”. Since α = 0.05, we will leave the “Alpha:” box as it is.

**Step 7:**

Select the circle to the left of “Output Range:”

**Step 8:**

Click the highlighted box to the right of “Output Range:”

**Step 9:**

Choose an empty cell on your spreadsheet. Bear in mind that Excel will generate the results on a 7×17 area so an area of that size to the bottom right of your chosen cell must also be empty.

**Step 10:**

Click the highlighted box

**Step 11:**

Press “OK”

**Result:**

Excel should generate the table shown below

Our p-value is **0.00281**. Excel also shows you the F-statistic.

You performed the test at a significance level (α) of 0.05. If you obtain a p-value greater than 0.05, that means there is no statistically significant difference between the means due to a factor. However, in the example shown above, we obtained a **p-value of 0.00281**, which is **lower than 0.05**, meaning **there is a statistically significant difference** (we reject H_{0}!).

Since p ≤ 0.05, we have strong statistical evidence that the factor (treatment) has an effect (concentration of NO in rat plasma) that is likely not due to chance and we may reject H_{0}. We may also state that since p ≤ 0.05, there is a statistically significant difference in the mean concentrations of NO in rat plasma due to the drug treatments (we accept H_{1}!).

Several groups of healthy male Struthio camelus (n = 15 per group, age = 25 years old) were subjected to special diets for a month to investigate the effects on body mass (kg). Determine, at a significance level of 0.05, if the special diets resulted in a significant change in body mass.

Doing Steps 1 to 11 as shown in Problem 1 should give you the following results:

Our p-value is **0.0914**

You performed the test at a significance level (α) of 0.05 and we obtained p = 0.0914. Since p is greater than 0.05, our evidence for rejecting H_{0} is weak. You can conclude that there is no statistically significant difference in mean body mass of the groups of Struthio camelus due to their diet (we accept H_{0}!).

ANOVA is an **omnibus test**, meaning ANOVA tests for an overall experimental effect. ANOVA can tell you that there is a difference between the groups due to a factor but it won’t tell you where the differences are. Tests referred to as **post hoc** tests such as **Dunnett’s test** can be used to tell you where the differences lie. Other software packages such as SPSS and GraphPad PRISM can perform post hoc tests.

Drugs, their targets and the nature and number of drug targets:

- Kim, H.-Y. Analysis of Variance (ANOVA) Comparing Means of More than Two Groups. Restor. Dent. Endod.
**2014**, 39 (1), 74. DOI: 10.5395/rde.2014.39.1.74/li> - Nayak, B.; Hazra, A. How to Choose the Right Statistical Test? Indian J. Ophthalmol.
**2011**, 59 (2), 85. DOI: 10.4103/0301-4738.77005 - Kao, L. S.; Green, C. E. Analysis of Variance: Is There a Difference in Means and What Does It Mean? J. Surg. Res.
**2008**, 144 (1), 158–170 DOI: 10.1016/j.jss.2007.02.053

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