One Sample T-Test - Data Transformation

Question

I have a dataset ($n=52$) of samples that I would like to test against a known value (e.g., $5$). Because I am testing my population mean against a known mean, this would be a one-sample t-test. My data is positively skewed (majority of samples are low, with some high outliers) and I understand that a normal distribution is necessary to support the use of a parametric test.

I have two questions:

1. When I log-transform my data, the histogram, Q-Q plot, kurtosis, skew, etc. all strongly suggest the data is normal enough for a t-test. From here, do I also log-transform the known mean I am testing against (e.g., $\log_{10}5$)?
2. Is something like this typically done for a one-sample t-test or should I try a non-parametric test?

Answer 1

Preliminaries. You might want to start by reviewing the standard definition of the lognormal distribution and fundamental facts about its mean and variance, either in a textbook or on the relevant Wikipedia page.

Three points are of particular importance for your question.

(1) If you take the natural log $(\log_e=\ln,$ not $\log_{10})$ of lognormal data, then you get data from the corresponding normal distribution.

(2) If the mean and SD of the normal population are $\mu$ and $\sigma,$ respectively, then the mean of the lognormal population will be $\exp(\mu+\sigma^2/2).$

(3) Because the relationship between the mean and variance of normal and corresponding lognormal distributions is not trivial, it is customary (but sometimes possibly confusing) to use the normal parameters for the corresponding lognormal distribution.

Demo with huge lognormal sample in R for $\mu=5, \sigma=1,$ for which sample mean nearly matches population mean:

set.seed(1234)
mean(rlnorm(10^7, 5, 1))
[1] 244.775
exp(5 + .5)
[1] 244.6919

Fictitious data for illustration. Consider fictitious normal data y and corresponding lognormal data x as follows:

set.seed(2021)
y = rnorm(52, 5, 1)   # normal
x = exp(y)            # lognormal

# Normal
summary(y);  length(y);  sd(y)  
    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.744   4.023   4.990   4.962   5.902   7.120 
[1] 52        # sample size
[1] 1.156031  # sample SD
shapiro.test(y)$p.val
[1] 0.177547           # not signif different from normal

# Lognormal
summary(x);  length(x);  sd(x)
  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 15.55   55.95  146.93  257.55  365.94 1236.45 
[1] 52
[1] 278.6784
shapiro.test(x)$p.val
[1] 3.559653e-07      # normality strongly rejected

The normal sample y (left panel) has a normal probability plot that is roughly linear, except possibly for a few points in the tails. The red reference line goes through the theoretical and sample upper and lower quartiles. Lognormal sample y (right panel) is clearly not normal.

R code for figure:

par(mfrow=c(1,2))
 qqnorm(y, main="Normal QQ Plot: Normal")
  qqline(y, col="red")
 qqnorm(x, main="Normal QQ Plot: Lognormal")
  qqline(x, col="red")
par(mfrow=c(1,1))

one-sample t test on normal data. You could do a one sample t test on the normal data y, as you suggest. Because these are fictitious data sampled with mean $\mu_y = 5,$ the null hypothesis $H_0: \mu_y = 5$ is not rejected in favor of the alternative $H_a: \mu_y \ne 5.$

t.test(y, mu=5)

        One Sample t-test

data:  y
t = -0.2354, df = 51, p-value = 0.8148
alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
 4.640422 5.284104
sample estimates:
mean of x              # output uses symbol 'x'
 4.962263 
sample estimates:
mean of x 
 257.5527 

However, $\mu_x = E(X) = 244.6919,$ the lognormal mean, has natural log about $5.2,$ which is not $E(Y),$ so $H_0: \mu_y = 5.2$ is strongly rejected in favor of $H_a: \mu_y \ne 5.2$ with P-value far below 5%.

t.test(y, mu=5.5)$p.val
[1] 0.001507389