• 16. One sample t-test

Motivating Example: We’ve summarized our data (with uncertainty) and checked our assumptions. Now we can conduct a formal hypothesis test. This section provides a step-by-step.

Learning Goals: By the end of this section, you will be able to:

1. List and explain the steps of a one-sample t-test.
2. Calculate a t-statistic from summary data.
   
3. Find the p-value for a given t-statistic and degrees of freedom using the pt() function in R.
4. Interpret the p-value to make a decision about the null hypothesis.
   
5. Write a concise summary of the test results.

---

library(tidyr)
library(dplyr)
range_shift_file <- "https://whitlockschluter3e.zoology.ubc.ca/Data/chapter11/chap11q01RangeShiftsWithClimateChange.csv"
range_shift <- read_csv(range_shift_file) |>
  mutate(uphill = elevationalRangeShift > 0)|>
  separate(taxonAndLocation, into = c("taxon", "location"), sep = "_")

The one sample t-test

This feels weird. I would be shocked if you ever use the one-sample t-test in your research. That said, I feel good about teaching it in this course because understanding how the one sample t-test works will provide you the foundation needed to understand the more complex analyses that you will likely perform (or read about). So lets do it!

To conduct a one sample t-test we:

1. State the null and alternative hypotheses. This means you must say what \(\mu_0\) should be under the null.
   
2. Calculate the t-value as the difference between the estimated mean \(\bar{x}\) and its hypothesized value under the null \(\mu_0\) divided by the sample standard error, \(s_\bar{x}\). That is, \(t = \frac{\bar{x}-\mu_0}{s_\bar{x}}\).
   
3. Find the p-value as the probability, under the null t-distribution, of observing a value at least as extreme as the one from our data (i.e. the proportion of the t-distribution as or more extreme than our t).
   • You can do this with pt(): P-value = pval <- 2*pt(q = abs(t_value), df=these_df, lower.tail=FALSE).
     
   • We multiply by two because this is a two-tailed test.
     
4. Reject the null if our p-value is less than \(\alpha\). Otherwise, fail to reject the null hypothesis.
   • Note that when our t-value is greater than the critical t-value, the p-value is less than \(\alpha\) and we reject the null.
     
5. Write up your results.

Worked example

Let’s apply the one-sample t test to our elevational range shift data set.

Step 1. State statistical hypotheses

To start, let’s state the null and alternative hypotheses as we did when introducing the data:

• Null Hypothesis (\(H_0\)): The true mean elevational shift (\(\mu\)) is zero. Any observed shift from our sample is due to random chance (\(H_0: \mu=0\)).
  
• Alternative Hypothesis (\(H_A\)): The true mean elevational shift (\(\mu\)) is not zero. Species are, on average, shifting their elevation (\(H_A: \mu \neq 0\)).

Is the null really \(\mu = 0\)?

For the sake of our learning goals, we’ll assume the null expectation is \(\mu_0 = 0\) (i.e. species are equally likely to increase or decrease elevation). But that’s not necessarily a biologically realistic null without more evidence.

• If all species started at low elevations, they would have nowhere to go but up.
  
• If all species started at high elevations, it would all be downhill.

In a real analysis, I’d want to model the expected change in elevation based on the available habitat area, and use that as my null.

Step 2. Calculate the t-value

To calculate the t-value, we simply follow the equation:

\[t = \frac{\bar{x}-\mu_0}{s_\bar{x}} = \frac{39.33 - 0}{5.51} = 7.14\]

Summary for one-sample t-test

Stat	Value
mean	39.33
sd	30.66
n	31.00
se	5.51
df	30.00
t	7.14

mu_0  <- 0 # set the null

range_shift_summary <- range_shift |>
  summarise(mean     = mean(elevationalRangeShift),
            sd       = sd(elevationalRangeShift),
            n        = n(),
            se       = sd / sqrt(n),
            df       = n - 1,
            t        = (mean - mu_0) / se) 

Step 3. Find the p-value

t_30 <- tibble(t = seq(-9,9,.005))|>
  mutate(prob_density = dt(x = t, df = 30))


ggplot(t_30, aes(x = t, y = prob_density)) +
  geom_area(fill = "white", color = "black")+
  labs(title = "Comparing our t to the t-distribution", 
       y = "Probability density")+
  scale_x_continuous(breaks = seq(-8,8,4))+
  geom_vline(xintercept = c(-7.14,7.14), color = "red", linewidth = 1.2,linetype= 2)+
  theme(axis.title = element_text(size = 21),
        axis.text = element_text(size = 21), 
        title = element_text(size = 19))

Figure 1: The probability density function for a t-distribution with 30 degrees of freedom. The t-score is on the x-axis, and the probability density is on the y-axis. Dashed red lines show the value in our data.

We don’t need R to tell us this what we’ll do to the null, Figure 1 makes it clear that the p-value is very small. More generally,

• When t is less than 1.96 (the critical Z-value) we will fail to reject the null.
  
• When t is greater than 2.5 (and our sample size is greater than seven) we will reject the null at the traditional \(\alpha\) value of 0.05. Here t is much greater than two and a half (it’s seven) and our sample size is way more than seven (it’s thirty one) – we will reject the null.
  
• When t is between 1.96 and 2.5 we may or may not reject the null depending on the degrees of freedom and the value of t.

But let’s find the p-value anyways:

# Our p-value
2 * pt(7.14, df= 30, lower.tail = FALSE) 

[1] "6.08e-08"

Step 4. Make a decision

Our p-value is minuscule. We reject the null hypothesis. Only one in fifteen million samples from the null distribution would have generated a t-statistic with an absolute value of 7.14 or more.

Step 5. Write up

Let’s work on writing these results. A write up should include the biological motivation, study design, sample summaries, effect sizes, test statistics, p-values, and what we do to the null. Here is a short version:

On average, species have shifted their range to higher elevations, with a mean shift of 39.3 meters, a standard error of 2.21 meters, and a 95% confidence interval between 28.1 and 50.6 meters. Although there is some variability among species (sd = 30.7), this variability is overshadowed by the overall trend toward higher elevations (Cohen’s d = 1.28 – A very large effect size). Such a large shift is highly unlikely under the null hypothesis (t = 7.14, df = 30, p = \(6.06 \times 10^{-8}\)). This result is not driven by the UK samples, which make up half of the dataset - we observe a similar result (mean = 48.6 meters, 95% confidence interval between 27.5 and 69.7 meters) after excluding them.