When to use Students’ t

In the cases we see here, we assume that the underlying distribution is Normal

(that is usually the case for averages)

We need to know the population mean and standard deviation

Students’ t when we don’t know population standard deviation

When we have a sample of size 𝑛, we calculate

• sample mean
• sample standard deviation

and we use them to approximate the population values

Then we use Student’s t distribution

Example: gene expression

Visually

Using a Linear Model

We want to find the change in gene expression due to genotype and stress condition

A common approach is to model gene expression as \[\text{Gene expression} = β_0 + β_1⋅\text{Genotype} + β_2⋅\text{Stress}\]

Here Genotype and Stress are either 0 or 1, depending on the case, and we want to find \(β_0, β_1,\) and \(β_2\) (the coefficients)

Model Matrix

Fitting the model in R

Most people use the R package limma for gene expression

Here we show a simple version, with one single gene

model <- lm(logExpr ~ Genotype + Stress)
model

Call:
lm(formula = logExpr ~ Genotype + Stress)

Coefficients:
(Intercept)   GenotypeKO   StressHigh  
     5.1325       0.4941       2.6293  

Estimated coefficients are the averages from the sample

Looking at the fitted model

summary(model)

Call:
lm(formula = logExpr ~ Genotype + Stress)

Residuals:
    Min      1Q  Median      3Q     Max 
-0.8916 -0.6917 -0.3380  0.8641  1.4262 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   5.1325     0.4553  11.273 1.31e-06 ***
GenotypeKO    0.4941     0.5257   0.940 0.371872    
StressHigh    2.6293     0.5257   5.001 0.000738 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.9106 on 9 degrees of freedom
Multiple R-squared:  0.7421,    Adjusted R-squared:  0.6848 
F-statistic: 12.95 on 2 and 9 DF,  p-value: 0.002247

Estimated coefficients follow a t distribution

Confidence interval for the coefficients

Estimated coefficients follow a t distribution

We have 12 values, and we evaluate 3 coefficients

Thus, we have 12-3=9 degrees of freedom

Then we can use the 9 DF table of Student’s t to get a confidence interval

confint(model)

                 2.5 %   97.5 %
(Intercept)  4.1025560 6.162410
GenotypeKO  -0.6952039 1.683310
StressHigh   1.4400824 3.818597

Interpretation

                 2.5 %   97.5 %
(Intercept)  4.1025560 6.162410
GenotypeKO  -0.6952039 1.683310
StressHigh   1.4400824 3.818597

(Intercept) is the base expression level for Wild Type and Low Stress

GenotypeKO is the differential expression due to gene KO. It can be 0, so we do not have enough evidence to say “there is an effect”

StressHigh is the differential expression due to stress. It seems greater than 0 at 95% confidence. Will still be >0 at 99% confidence?

𝑝-values

Call:
lm(formula = logExpr ~ Genotype + Stress)

Residuals:
    Min      1Q  Median      3Q     Max 
-0.8916 -0.6917 -0.3380  0.8641  1.4262 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   5.1325     0.4553  11.273 1.31e-06 ***
GenotypeKO    0.4941     0.5257   0.940 0.371872    
StressHigh    2.6293     0.5257   5.001 0.000738 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.9106 on 9 degrees of freedom
Multiple R-squared:  0.7421,    Adjusted R-squared:  0.6848 
F-statistic: 12.95 on 2 and 9 DF,  p-value: 0.002247