**Place this file in the folder with your data. Write the names of your own data file below instead of the one shown.Then compile the script using R studio. This will build a data report designed to guide you through the analysis.**

```
d <- read.csv("filename.csv") ## Put your own file name here
xtext <- "Mussel length"
ytext <- "Mussel body volume"
source("StatsExpert.R")
```

StatsExpert is a script written in R that has been designed to help you to choose the right analysis for your own data by writing out suggestions and running some of the most appropriate analyses.

**It is not guaranteed to be foolproof**. In most cases the suggestions will be reasonable, but if the data are very unusual some of the statements might appear constradictory. Statistical analysis is not an exact science that can be turned into simple rules. However the script has been designed to help you avoid the worst pitfalls.

The StatExpert script will first run through the flow diagram shown below.

You will need to provide descriptive statistics for each variable regardless of the outcome of the rest of the analysis.

```
summary(d)
```

```
## Lshell BTVolume
## Min. : 61.9 Min. : 5.0
## 1st Qu.: 97.0 1st Qu.:21.0
## Median :106.9 Median :28.0
## Mean :106.8 Mean :27.8
## 3rd Qu.:118.7 3rd Qu.:35.0
## Max. :132.6 Max. :59.0
```

This table shows you the minimum, maximum and interquartile range. You should also look at the mean and median for each variable. If there is a big difference between them it could be an indication of skew.

The reliability of your analysis depends critically on sample size. If the sample size is low, and the intrinsic variability is high then you cannot expect to find a relationship, even if one exists.

Yor sample size was 113 This is a large sample, so providing there is not too much variability you may be able to show an effect, if one exists.

It is very important to produce boxplots and histograms for each variable. You should look very carefully at these figures in order to spot potential problems with the analysis.

For correlation and regression analysis the raw variables do **not** have to be normally distributed. However they should usually be more or less **symmetrically** distributed. So we need to check for skew, and we also need to look carefully at the most extreme values (outliers)

The StatsExpert script will try to help to find problems, but always be aware that it cannot be expected to solve all the issues nor can it know **why** some of the values may be outside the expected range.

```
par(mfrow = c(2, 2))
hist(x, col = "grey", main = xtext)
boxplot(x, col = "grey", main = xtext)
hist(y, col = "grey", main = ytext)
boxplot(y, col = "grey", main = ytext)
```

```
```

Things to watch for are outliers and skew. If a boxplot has many points beyond the whiskers then you may have problems with outliers. If the histogram does not look symmetrical then you have a potential problem with skew. These two issues go together. They are very difficult to separate cleanly, so only use StatsExperts advice as a rough guide. Careful inspection of the figures is the best way of understanding your data.

To help you to interpret your figures and get an idea of what you should be looking for examine the examples below. Hopefully your figures are closer to the blue histograms and boxplots than the red ones.**If the data are seriously skewed you have not necessarily made a mistake!** Data like these are very commonly found in ecology.

StatsExpert will have a look at your data for you and do its best to make some helpful suggestions.

**Minor violations of the assumptions used by a statistical test are often unimportant. StatsExpert will do all that it can to guide you and help you to avoid you making major mistakes.**

Nearly all data sets have some outliers. They are not necessarily mistakes. If a variable is strongly skewed then it will innevitably also appear to have many outliers.

The variable has no sign of skew.

There are only a small number of outliers. These can probably be safely ignored. If you are sure that they are mistakes you could remove them.

The variable has no sign of skew.

There are only a small number of outliers. These can probably be safely ignored. If you are sure that they are mistakes you could remove them.

**Correlation analysis always starts with a scatterplot.** You should look at this plot very carefully yourself and try to understand the relationship between the two variables by eye. Things that you should look for are

- Evidence that the points fall along a strait line. This will result in a clear correlation.
- The direction of the slope. If the slope rises from the origin you have a positive correlation. If the slope falls then the correlation will be negative.
- Effects of skew and outliers. If the scatterplot is “fan shaped” this is evidence that at least one of the variables is skewed. You may need to use data transformation. StatsExpert.R will advise on this later on in the analysis.
- Evidence of a curvilear relationship. Correlation and classical regression assume strait line relationships between your variables. If the pattern of the scatter forms a curve these methods may fail to detect a relationship even when it is very clearly there. StatsExpert.R will check this later on and advise.

```
plot(y ~ x, pch = 21, bg = 2, xlab = xtext, ylab = ytext)
```

There is a suggestion of a fan shaped pattern.

If StatsExpert has already warned you about some problems with skew or outliers you can still trust this test. The procedure is very robust as it is based on ranks. It will answer the fundamental question. **Is there a relationship of some form between the two variables? Does an increase in x lead to an increase in y?**

We want to know if there is a significant correlation between the ranks of the two variables. We also want some measure of the strength of this correlation. This test provides this important information **even if there are problems with skew**. If you have included many mistaken measurements then I am afraid that all bets are off. StatsExpert cannot know that the numbers are completely wrong. But this procedure will ignore the influence of a few erroneous outliers.

```
cor.test(x, y, method = "spearman")
```

```
##
## Spearman's rank correlation rho
##
## data: x and y
## S = 26143, p-value < 2.2e-16
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
## rho
## 0.8913
##
```

OK, so how do we interpret this output?

**The p value is <0.001**.

This is an extremely significant result.

**We can reject the null hypothesis.**

The test has told us that, if all the assumptions have been met, there would be a less than one in a thousand chance of obtaining these data if the null were true. So the results are very unlikely to be attributable to chance.

As you used a non parametric test this result is quite robust.

The coefficient of determination is 0.794

This is a strong positive correlation.

The logic behind this test is shown by the scatterplot below. The test uses ranks.

```
xlb <- paste("Ranks of (", xtext, ")")
ylb <- paste("Ranks of (", ytext, ")")
plot(rank(y) ~ rank(x), pch = 21, bg = 2, xlab = xlb, ylab = ylb)
title(main = "Input to Spearman's test")
```

There are no units attached to ranks, so all we can say is that an increase in x tends to leads to an increase in y. However if the aim of your study was only to establish whether a correlation was detectable or not, this test is very useful.

If the scatterplot of ranks looks very different to the scatterplot of your raw data then the results of this test will be different to those obtained from the parametric test, which we will now run.

This procedure is less robust. If StatsExpert has warned you already that there are many outliers or serious skew in either of the variables then you probably should not fully trust this test. You may also have seen a difference between the scatterplots.

However if the test is more or less valid we can obtain a more detailed interpretation of the relationship using regression. So, it is very important to run both of these correlation tests and compare the results.

StatsExpert will provide some automatated text output that will help you understand the result.

```
cor.test(x, y, method = "pearson")
```

```
##
## Pearson's product-moment correlation
##
## data: x and y
## t = 19.3, df = 111, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.8271 0.9142
## sample estimates:
## cor
## 0.8777
##
```

We interpret the p-value using the standard rules.

**The p value is <0.001**

This is a highly significant result.

**We can reject the null hypothesis.**

The test has told us that, if all the assumptions have been met, there would be a less than one in a thousand chance of obtaining these data if the null were true. So the results are unlikely to be attributable to chance.

Because you have used a parametric test you must look carefully at the data and justify your choice of test.

You should look at the influence of any outliers and the effects of skewed data.

It is always important to look at the strength of the correlation.

The coefficient of determination is 0.77

This is a strong positive correlation.

The two tests have produced very similar results.

Although there may have been some minor violations of the assumptions used in the parametric test they were not

severe enough to cause any problems. You may use regression analysis.

The purpose of regression analysis is to quantify the relationship between your variables.

We are trying to fit a “statistical model” to the data of the form

y=a+bx

The intercept is a and b is the slope of the line. Although a regression line can be found for any data, the analalysis only really makes sense if there is a significant correlation between the variables. We have found a p-value of ** <0.001** using the parametric test for correlation. We will go ahead and run the analysis. If the p-value was higher than 0.05 we will not gain anything from doing so.

We will plot the data again and add the line of best fit.

```
plot(y ~ x, pch = 21, bg = 2, xlab = xtext, ylab = ytext)
mod <- lm(y ~ x)
abline(mod)
title(main = "Line of best fit")
```

The most important information is that for every increase of one unit in `Mussel length`

we will expect a change of `0.598`

units of `Mussel body volume`

. Only you can tell if this result makes sense within the context of your study.

The full results of the analysis are shown below.

```
summary(mod)
```

```
##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.828 -2.672 0.147 2.235 17.404
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -36.024 3.339 -10.8 <2e-16 ***
## x 0.598 0.031 19.3 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.86 on 111 degrees of freedom
## Multiple R-squared: 0.77, Adjusted R-squared: 0.768
## F-statistic: 372 on 1 and 111 DF, p-value: <2e-16
##
```

If you look carefully you will see that you are provided with an R squared value (coefficient of determination), a p-value and some additional information.

How do you interpret the intercept, that is given here as `-36.024`

? Well, in some cases it represents the value of y when x=0. However if all your x values values fall a long way from the origin you cannot extrapolate the line back in this way. In this case you are probably safe to simply ignore the intercept.

You should always show confidence bands when plotting a regression line. So we will plot the data again with these added.

```
plot(y ~ x, pch = 21, bg = 2, xlab = xtext, ylab = ytext)
mod <- lm(y ~ x)
ablines(mod, lwd = 2)
addRsq(mod)
a <- round(coef(mod)[1], 3)
b <- round(coef(mod)[2], 3)
tx <- paste(ytext, "=", a, "+", b, xtext)
title(main = tx)
```

The scatter has a fan shape (heterscedasticity).

This is a potential problem, but a much more advanced analysis would be needed to correct for this.