Andrew Warrington

Andrew Warrington

he/him/his

Graduate of the University of Virginia | Interested in Neuroscience, Cancer Genomics, and Computational Biology

Biostatistics Final Data Analysis

Published:

Read in the data:

anole <- read.csv("data/cox_functional_ecology_data.csv")
ovx <- subset(anole, Treat == "OVX")
sham <- subset(anole, Treat == "SHAM")

Explore summaries of the two groups:

summary(ovx)
summary(sham)

The growth (in mm) five number summary for the OVX group:

  • Min: -1.0
  • 1st Quart: 2.0
  • Median: 2.0
  • 3rd Quart: 3.0
  • Max: 7.0
  • Mean: 2.309
  • SD: 1.274607
sd(ovx$Growth..mm., na.rm = TRUE)

hist(ovx$Growth..mm., main = "Change in Body Length of OVX Lizards", xlab = "Growth (mm)", col = "blue")

boxplot(ovx$Growth..mm., main = "Change in Body Length of OVX Lizards", ylab = "Growth (mm)", col = "green")
  • The histogram and box plot visualizations indicate a uni-modal, right-skewed distribution.
  • There are at least 3 high outliers and at least 2 low outliers.
  • The center appears to be around 2.0 mm of growth, with a mean value of 2.309 mm, and a median value of 2.0 mm of growth.
  • The mean being of higher value than the median is also indicative of a right-skewed distribution.

Growth (in mm) 5 number summary for the sham group:

  • Min: -1.0
  • 1st Quart: 1.0
  • Median: 1.0
  • 3rd Quart: 2.0
  • Max: 4.0
  • Mean: 1.426
  • SD: 1.042156
sd(sham$Growth..mm., na.rm = TRUE)

hist(sham$Growth..mm., main = "Change in Body Length of SHAM Lizards", xlab = "Growth (mm)", col = "cyan")

boxplot(sham$Growth..mm., main = "Change in Body Length of SHAM Lizards", ylab = "Growth (mm)", col = "red")
  • The histogram and boxplot visualizations indicate the growth for the sham group follows a bimodal distribution. - - There is at least one high outlier and one low outlier.
  • There are two centers, one around 1.0 mm and a center closer to 2.0 mm.
  • The growth for sham group has a mean growth of 1.426 mm and a median growth of 1.0 mm.

Hypothesis Test and Confidence Interval:

H_0: mu_ovx = mu_sham

H_a: mu_ovx > mu_sham

Is there sufficient evidence to conclude that the mean growth of the ovx group is significantly higher than the mean sham growth?

Choosing a two-sample t test, one-sided, right-tailed:

  • Because we have two independent random samples and Population parameters m1,s1,m2,s2 are unknown.

  • Our H_a is that the mean ovx growth in mm is significantly greater than the mean sham growth.

Assumptions:

  1. Sample Mean is Approximately Normal.
  2. Data Comes from SRS (simple random sample) or random experiment
  3. Observations are independent.
  4. This data came from a random sample of Anoles.

Even though the data itself is slightly right skewed (with outliers, both high and low), we have a large enough sample size to assume that the sample mean and standard deviation will be approximately normal. The anole observations are independent.

Calculating T test statistic:

Calling ovx group 1 and sham group 2

  • T_ts (T-test statistic)
  • ovx: n = 139, (140 observations - 1 NA)
  • sham: n = 121, (122 observations - 1 NA)
(2.309 - 1.426)

(1.274607^2) / (139) + (1.042156^2) / (121)

sqrt(0.02066388)

0.883 / 0.1437494

T_ts = 6.142634

Df = 120 (Degrees of freedom)

P-Value Calculation:

A one-sided, right-tailed t-test

pt(6.142634, df = 120, lower.tail = FALSE)

P-value = 5.445521e-09

Our cutoff: significance level of 5% or 0.05 (alpha)

5.445521e-09 < 0.05, we reject the null hypothesis, which states that the mean growth of ovx and sham group are the same.

P-value interpretation:

If the true mean growth for the ovx and sham group were equal (H_0 is true), then there is a 5.445521e-09 chance that we would have obtained the sample means we observed from both groups based on the data.

There is sufficient evidence to reject the null hypothesis (H_0), which states that the mean growth for the ovx and sham groups of anoles is the same, and conclude H_a, which states that the mean growth in mm ofr ovx group is greater than the sham group.

# Calculate t*
qt(.025, df = 120)

# t* = -1.97993

-1.97993 * 0.1437494 # equals m (margin of error) -> -0.2846137
(2.309 - 1.426) + (-0.2846137) # 0.5983863, lower bound of 95% CI
(2.309 - 1.426) - (-0.2846137) # 1.167614, upper bound of 95% CI

We are 95% confident that the difference in true mean growth for the ovx and shame group is between 0.5983863 and 1.167614 mm.

Conclusion:

There is sufficient evidence to reject the null hypothesis (H_0), which states that the mean growth for the ovx and sham groups of anoles is the same, and conclude H_a, which states that the mean growth in mm ofr ovx group is greater than the sham group.

We are 95% confident that the difference in true mean growth for the ovx and shame group is between 0.5983863 and 1.167614 mm.

These findings are relevant because they suggest that there is a reproductive cost that is associated with the growth of female anoles. When reproductive mechanisms are removed in OVX, they experience a greater growth relative to the sham group that reproduces normally. The lack of growth that occurs in reproducing anoles could contribute to their lower annual survival rate.

References

Data Reference: Cox, RM. (2010) Cox et al. Functional Ecology Data. Raw data.
Video Reference: Cox, RM. Biostats Bob Cox 100115. Retrieved from UVA Collab