Two populations

• Consider two populations:

  • Population 1 has mean \(\mu_1\) and standard deviation \(\sigma_1\).

  • Population 2 has mean \(\mu_2\) and standard deviation \(\sigma_2\).

• We want to compare the means by looking at the difference \(\mu_1-\mu_2.\)

Two samples

• We now take a sample from each population.

  • The sample from Population 1 has sample mean \(\bar{x}_1\), sample standard deviation \(s_1\) and sample size \(n_1\).

  • The sample from Population 2 has sample mean \(\bar{x}_2\), sample standard deviation \(s_2\) and sample size \(n_2\).

Dependent and independent samples

• We distinguish between two types of samples:

  • The two samples are independent.
  • The two samples are paired.

• Example: Suppose we consider the fuel consumption of cars.

  • If we compare two samples of cars with different engine types, then the two samples are independent, since each car can only have one of the two engine types.

  • If we compare the fuel consumption of cars at two different speed levels by testing each car at both speed levels, then then samples are paired.

Comparison of two means (Independent samples)

• We consider the situation, where we have two independent samples of a quantitative variable.

• We estimate the difference \(\mu_1-\mu_2\) by \[d=\bar{x}_1-\bar{x}_2.\]

• Assume that we can find the estimated standard error \(se_d\) of the difference.

• If the samples come from two normal distributions, or if both samples are large (\(n_1,n_2\geq 30\)), then one can show \[T_{obs}=\frac{(\bar{X}_1-\bar{X}_2)-(\mu_1-\mu_2)}{se_d}\sim \texttt{t}(df),\] where \(\texttt{t}(df)\) is a \(t\)-distribution with \(df\) degrees of freedom.

• By the usual procedure, we can use this to construct a confidence interval for the unknown population difference of means \(\mu_1-\mu_2\) by \[ (\bar{x}_1-\bar{x}_2)\pm t_{crit}se_d, \] where the critical \(t\)-score, \(t_{crit}\), is determined by the confidence level and the \(df\).

Significance test (Independent samples)

• We may be interested the testing the null-hypothesis that the population means are the same, which we can formulated as:

  • \(H_0:\mu_1-\mu_2=0.\)
    
  • \(H_a:\mu_1 - \mu_2\neq 0.\)

• If the null hypothesis is true, then the test statistic:  \[T_{obs} = \frac{(\bar{X}_1-\bar{X}_2) - 0}{se_d},\] has a \(t\)-distribution with \(df\) degrees of freedom.

• The p-value is the probability of observing something further away from 0 than \(t_{obs}\) in a \(\texttt{t}(df)\) distribution.

• It remains to find the estimated standard error \(se_d\) and the degrees of freedom \(df\). We distinguish between two cases:

  • The two populations have equal variances \(\sigma_1^2 =\sigma_2^2\).
  • The two populations have different variances \(\sigma_1^2 \neq \sigma_2^2\).

Standard error (Independent samples, equal variances)

• The standard error of \(d=\bar{x}_1-\bar{x}_2\) is given by the formula:

\[\sqrt{\tfrac{\sigma_1^2}{n_1}+\tfrac{\sigma^2_2}{n_2}}.\]

• If the variances are equal, \(\sigma_1^2=\sigma_2^2\), then we estimate the common value by the pooled variance estimate

\[s_p^2=\tfrac{(n_1-1)s^2_1+(n_2-1)s_2^2}{n_1+n_2-2}.\]

• Inserting this estimate in the formula for the standard error we obtain the estimated standard error

\[se_d=\sqrt{\tfrac{s_p^2}{n_1}+\tfrac{s^2_p}{n_2}}=s_p\sqrt{\tfrac{1}{n_1}+\tfrac{1}{n_2}}.\]

• In this situation, the degrees of freedom are \(df=n_1+n_2-2\).

Example: Comparing two means (independent samples, equal variances)

We return to the mtcars data. We study the association between the variables vs and mpg (engine type and fuel consumption). So, we will perform a significance test to test the null-hypothesis that there is no difference between the mean of fuel consumption for the two engine types.

• We will test the null-hypothesis assuming equal variances:

library(mosaic)
fv <- favstats(mpg ~ vs, data = mtcars)
fv

##   vs  min   Q1 median   Q3  max mean   sd  n missing
## 1  0 10.4 14.8   15.7 19.1 26.0 16.6 3.86 18       0
## 2  1 17.8 21.4   22.8 29.6 33.9 24.6 5.38 14       0

• Difference: \(d = 16.6167 - (24.5571) = -7.9405\).

• Sample sizes: \(n_1 = 18\) and \(n_2 = 14\).

• Estimated standard deviations: \(s_1 = 3.8607\) (not v-shaped) and \(s_2 = 5.379\) (v-shaped).

• Pooled variance: \[s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = \frac{17\cdot 3.8607^2 + 13 \cdot 5.379 ^2}{18 + 14 -2} = 20.984.\]

• Estimated standard error of difference:  \(se_d = s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = \sqrt{20.984}\sqrt{\frac{1}{18} + \frac{1}{14}} = 1.6324\).

• Observed \(t\)-score for \(H_0:\ \mu_1-\mu_2=0\) is: \(\quad t_{obs} = \frac{d-0}{se_d} = \frac{-7.9405}{1.6324} = -4.864\).

• The degrees of freedom are \(df=n_1 + n_2-2 = 30\).

• We find the \(p\)-value:

2*pdist("t", q = -4.864, df=30, xlim = c(-5, 5))

## [1] 3.419648e-05

Standard error (Independent samples, unequal variances)

• If the variances are unequal, then we simply insert the two estimates \(s_1^2\) and \(s_2^2\) for \(\sigma_1^2\) and \(\sigma_2^2\) in the formula for the standard error to obtain the estimated standard error \[ se_d=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}. \]

• The degrees of freedom \(df\) for \(se_d\) can be estimated by a complicated formula, which we will not present here (see p.365 in the book).

• Note:

  • If both \(n_1\) and \(n_2\) are above 30, then we may use the standard normal distribution to compute a \(z\)-score rather than the \(t\)-distribution to compute the \(t\)-score. This way we avoid computing \(df\).
  • If \(n_1\) or \(n_2\) are below 30, then we let R calculate the degrees of freedom and the \(p\)-value/confidence interval.

Example: Comparing two means (independent samples, unequal variances)

We return to the mtcars data. We study the association between the variables vs and mpg (engine type and fuel consumption). So, we will perform a significance test to test the null-hypothesis that there is no difference between the mean of fuel consumption for the two engine types.

• We now make the analysis without assuming equal variances:

library(mosaic)
fv <- favstats(mpg ~ vs, data = mtcars)
fv

##   vs  min   Q1 median   Q3  max mean   sd  n missing
## 1  0 10.4 14.8   15.7 19.1 26.0 16.6 3.86 18       0
## 2  1 17.8 21.4   22.8 29.6 33.9 24.6 5.38 14       0

• Difference: \(d = 16.6167 - (24.5571) = -7.9405\).

• Sample sizes: \(n_1 = 18\) and \(n_2 = 14\).

• Estimated standard deviations: \(s_1 = 3.8607\) (not v-shaped) and \(s_2 = 5.379\) (v-shaped).

• Estimated standard error of difference:  \(se_d = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3.8607^2}{18} + \frac{5.379^2}{14}} = 1.7014\).

• Observed \(t\)-score for \(H_0:\ \mu_1-\mu_2=0\) is: \(\quad t_{obs} = \frac{d-0}{se_d} = \frac{-7.9405}{1.7014} = -4.6671\).

• The degrees of freedom can be found using R (see below) to be \(df=22.716\).

• We find the \(p\)-value:

2* pdist("t", q = -4.6671, df=22.716, xlim = c(-5, 5))

## [1] 0.0001098212

• We reject the null-hypothesis and conclude that the fuel consumption is different for the two engine types.

Example: Comparing two means (independent samples)

• Now we know there is a difference between the two population means. We can also make a 95% confidence interval for how large the difference \(\mu_1 - \mu_2\) actually is by the formula

\[ d \pm t_{crit} se_d\]

qdist("t", p = 1-0.05/2, df=22.716, xlim = c(-3, 3))

## [1] 2.07009

• Inserting the values from the previous slide yields

\[[-7.94 - 2.07*1.70;-7.94 + 2.07*1.70]= [-11.5,-4.4].\]

• We are 95% confident that the difference in fuel consumption is between the two engine types is between -4.4mpg and -11.5mpg.

T-test in R (Independent samples)

• We can leave all the calculations to R by using t.test:

t.test(mpg ~ vs, data = mtcars,var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  mpg by vs
## t = -4.6671, df = 22.716, p-value = 0.0001098
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
##  -11.462508  -4.418445
## sample estimates:
## mean in group 0 mean in group 1 
##        16.61667        24.55714

• We recognize the \(t\)-score \(-4.6671\), the \(p\)-value \(0.0001\), and the confidence interval \([-11.5;-4.4]\). The estimated degrees of freedom can be found in the output to be \(df = 22.716\).

Test for equal variances (Independent samples)

• In order to decide whether to use the t-test with equal or unequal variance, we may test the hypothesis \(H_0: \sigma_1^2 = \sigma_2^2\).

• As test statistic we use \[F_{obs} = \frac{s_1^2}{s_2^2}.\]

• If the null-hypothesis is true, we expect \(F_{obs}\) to take values close to 1. Small and large values are critical for \(H_0\).

• Under \(H_0\), \(F_{obs}\) follows a so-called \(F\)-distribution with \(df_1=n_1-1\) and \(df_2=n_2-1\) degrees of freedom.

  • If \(F_{obs}<1\) we reject the null-hypothesis if two times the probability of getting something smaller than \(F_{obs}\) is less than the significance level.
  • If \(F_{obs}>1\) we reject the null-hypothesis if two times the probability of getting something larger than \(F_{obs}\) is less than the significance level.

Example: Test for equal variances (Independent samples)

• To test whether the variance is the same for the two engine types in the mtcars example, we first compute the sample variances.

var(mpg~vs,data=mtcars) 

##        0        1 
## 14.90500 28.93341

• We compute \(F_{obs} = \frac{s_1^2}{s_2^2} = \frac{14.9}{28.9} = 0.516\).

• The probability of observing something smaller than \(F_{obs}\) in an \(F\)-distribution with \(df_1=n_1-1 = 17\) and \(df_2=n_2-1 = 13\):

 pdist("f", 0.516, df1=17, df2=13)

## [1] 0.1004094

• The p-value is \(2*0.1004= 0.2008\). Here we multiply by two because the test is two-sided (large values would also have been critical).

• We find no evidence against the null-hypothesis.

Comparison of two means: paired \(t\)-test (dependent samples)

• We now consider the case where we have two samples from two different populations but the observations in the two samples are paired.

  • For each pair, we can compute the difference between the two observations.
  • We now have one sample of observed differences.
  • We apply the the one-sample t-test from Lecture 2.1 to test whether the mean difference is zero.

• Example: Suppose we make the following experiment:

  • Choose 32 students at random and measure their average reaction time in a driving simulator while they are listening to radio or audio books.
  • Later the same 32 students redo the simulated driving while talking on a cell phone.
  • We are interested in whether or not the fact that you are actively participating in a conversation changes your average reaction time compared to when you are passively listening.

• So we have 2 samples corresponding to with/without phone. In this case we have paired samples, since we have 2 measurement for each student.

• We use the following strategy for analysis:

  • For each student calculate the change in average reaction time with and without talking on the phone.
  • The changes \(d_1,d_2,\ldots,d_{32}\) are now considered as ONE sample from a population with mean \(\mu\).
  • Test the hypothesis \(H_0: \mu=0\) as usual (using a one-sample \(t\)-test).

Reaction time: data example

• Data is organized in a data frame with 3 variables:
  • student (integer – a simple id)
  • reaction_time (numeric – average reaction time in milliseconds)
  • phone (factor – yes/no indicating whether speaking on the phone)

reaction <- read.delim("https://asta.math.aau.dk/datasets?file=reaction.txt")
head(reaction, n = 3)

##   student reaction_time phone
## 1       1           604    no
## 2       2           556    no
## 3       3           540    no

• We first manually find the reaction time difference for each student and do a one sample t-test on this difference:

yes <- subset(reaction, phone == "yes")
no  <- subset(reaction, phone == "no")
reaction_diff <- data.frame(student = no$student, yes = yes$reaction_time, no = no$reaction_time)
reaction_diff$diff <- reaction_diff$yes - reaction_diff$no
head(reaction_diff)

##   student yes  no diff
## 1       1 636 604   32
## 2       2 623 556   67
## 3       3 615 540   75
## 4       4 672 522  150
## 5       5 601 459  142
## 6       6 600 544   56

t.test( ~ diff, data = reaction_diff)

## 
##  One Sample t-test
## 
## data:  diff
## t = 5.4563, df = 31, p-value = 5.803e-06
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  31.70186 69.54814
## sample estimates:
## mean of x 
##    50.625

• With a \(p\)-value of 0.0000058 we reject the null-hypothesis that speaking on the phone has no influence on the reaction time.

• We can avoid the manual calculations and let R perform the significance test by using t.test with paired = TRUE:

t.test(reaction_time ~ phone, data = reaction, paired = TRUE)

## 
##  Paired t-test
## 
## data:  reaction_time by phone
## t = -5.4563, df = 31, p-value = 5.803e-06
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -69.54814 -31.70186
## sample estimates:
## mean difference 
##         -50.625

Response variable and explanatory variable

• The situation with two populations is an example where we have: * A response variable (or outcome, dependent variable).

  • An explanatory variable (or independent variable, covariate) that divides data in 2 groups.

• We are interested in the effect of the explanatory variable on the response variable.

  • For instance in the mtcars data, mpg is the response variable and vs is the explanatory variable.

• In this lecture we consider the case with one discrete explanatory variable. Module 3 is concerned with the case of one or more continuous variables.

More than two populations

• We are now going to consider a situation where we have \(k\) populations with mean values \(\mu_1,\ldots,\mu_k\).

• We assume that each population follows a normal distribution and that the standard deviation is the same in all populations.

• We are interested in the null-hypothesis that all \(k\) populations have the same mean, i.e.
  \[H_0: \mu_1 = \dotsm=\mu_k.\] \[H_a: \text{ not all } \mu_1,\ldots \mu_k \text{ are the same}.\]

• We take out a sample from each population.

Data example

• The data set chickwts is available in R, and on the course webpage.
• 71 newly hatched chickens were randomly allocated into six groups, and each group was given a different feed supplement.
• Their weights in grams after six weeks are given along with feed types, i.e. we have a sample with corresponding measurements of 2 variables:
  • weight: a numeric variable giving the chicken weight.
  • feed: a factor giving the feed type.
• Always start with some graphics:

library(mosaic)
gf_boxplot(weight ~ feed, data = chickwts)

Estimation of mean values

• We estimate the mean in each group by the sample mean inside that group, i.e. \(\hat{\mu}_i = \bar{x}_i\), \(i=1,\ldots, k\).

• We use mean to find the mean, for each group:

mean(weight ~ feed, data = chickwts)

##    casein horsebean   linseed  meatmeal   soybean sunflower 
##  323.5833  160.2000  218.7500  276.9091  246.4286  328.9167

• We can e.g. see that the sample mean is \(323.6\), when feed=casein but \(160.2\), when feed=horsebean.
• Is this a significant difference ?

Contrasts

• If we want compare groups, it is convenient to formulate the model using contrasts.
• One group is chosen as the reference group, which all other groups are compared to.
  • Sometimes there is a group corresponding to “no treatment” and we are interested in the effect of different treatments. Other times the reference group can be arbitrary.
• If group 1 is the reference group, the mean values in the remaining groups groups can be expressed as \[\mu_i = \mu_1 + \alpha_i,\] where \(\alpha_i = (\mu_i-\mu_1)\) is the difference between group \(i\) and the reference group. The \(\alpha_i\) are called contrasts.

Example: contrast estimates

model <- lm(weight ~ feed, data = chickwts)
summary(model)

## 
## Call:
## lm(formula = weight ~ feed, data = chickwts)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -123.909  -34.413    1.571   38.170  103.091 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    323.583     15.834  20.436  < 2e-16 ***
## feedhorsebean -163.383     23.485  -6.957 2.07e-09 ***
## feedlinseed   -104.833     22.393  -4.682 1.49e-05 ***
## feedmeatmeal   -46.674     22.896  -2.039 0.045567 *  
## feedsoybean    -77.155     21.578  -3.576 0.000665 ***
## feedsunflower    5.333     22.393   0.238 0.812495    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 54.85 on 65 degrees of freedom
## Multiple R-squared:  0.5417, Adjusted R-squared:  0.5064 
## F-statistic: 15.36 on 5 and 65 DF,  p-value: 5.936e-10

• In the example the groups are different feeds. R chooses the lexicographically smallest, which is casein, to be the reference group.
• We get information about contrasts and their significance:
• Intercept is the estimated mean \(\hat{\mu}_{casein}=323.583\) in the reference group.
  • In the same line, there is also a test of the null-hypothesis \(H_0:\mu_1=0\) that the weight after 6 weeks is 0 (\(p < 2\times 10^{-16}\)) (of course, chickens grow a lot over 6 weeks).
• The line feedhorsebean estimates the contrast \(\alpha_{horsebean}\) between the casein and horsebean group to be \(\hat{\alpha}_{horsebean}=-163.383\).
  • The null-hypothesis that there is no difference between casein and horsebean (\(H_0:\alpha_{horsebean}=0\)) is rejected with p=\(2\times 10^{-9}\).

Overall test for effect

• We are now interested in testing the null-hypothesis \[H_0: \mu_1 = \mu_2 = \dots=\mu_k \quad \mbox{against} \quad H_a: \mbox{ Not all of the population means are the same}\]

• Alternatively \[H_0: \alpha_2 = \alpha_3 = \dots=\alpha_k =0, \quad H_a: \mbox{ At least one contrast is non-zero}.\]

• Idea: Compare variation within groups and variation between groups.

Test statistic

• We use the test statistic \[F_{obs} = \frac{(TSS-SSE)/(k-1)}{SSE/(n-k)}.\]

• If observations from group \(i\) are called \(x_{ij}\), \(j=1,\ldots,k\), we have:

  • \(TSS=\sum_i\sum_j(x_{ij}-\bar{x})^2\), where \(\bar{x}\) is the average of all observations from all groups.
  • \(SSE=\sum_i\sum_j(x_{ij}-\bar{x}_i)^2\).

• Interpretation:

  • TSS: error sum of squares if common mean.
  • SSE: error sum of squares if different means.
  • TSS-SSE: how much does error sum of squares increase if means are restricted to be equal.

• One can show that TSS-SSE measures the variation of group means around common mean.

• Thus, \[F_{obs} = \frac{\text{variation between groups}}{\text{variation within groups}}.\]

The \(F\)-test

• A large variation between groups compared to the variation within groups points against \(H_0\).

• Thus, large values are critical for the null-hypothesis.

• Under the null-hypothesis, \(F_{obs}\) follows an \(F\)-distribution with \(df_1=k-1\) and \(df_2=n-k\) degrees of freedom.

• A \(p\)-value for the null-hypothesis is the probability of observing something larger than \(F_{obs}\) in an \(F\)-distribution with \(df_1\) and \(df_2\) degrees of freedom.

• For instance if \(F_{obs}=15.36\) with \(df_1=5\) and \(df_2=65\) degrees of freedom:

1 - pdist("f", 15.36, df1=5, df2=65)

## [1] 5.967948e-10

Example

model <- lm(weight ~ feed, data = chickwts)
summary(model)

## 
## Call:
## lm(formula = weight ~ feed, data = chickwts)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -123.909  -34.413    1.571   38.170  103.091 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    323.583     15.834  20.436  < 2e-16 ***
## feedhorsebean -163.383     23.485  -6.957 2.07e-09 ***
## feedlinseed   -104.833     22.393  -4.682 1.49e-05 ***
## feedmeatmeal   -46.674     22.896  -2.039 0.045567 *  
## feedsoybean    -77.155     21.578  -3.576 0.000665 ***
## feedsunflower    5.333     22.393   0.238 0.812495    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 54.85 on 65 degrees of freedom
## Multiple R-squared:  0.5417, Adjusted R-squared:  0.5064 
## F-statistic: 15.36 on 5 and 65 DF,  p-value: 5.936e-10

• The last line gives us the value of \(F_{obs} = 15.36\) and the corresponding \(p\)-value (\(5.9 \times 10^{-10}\)). Clearly there is a significant difference between the types of feed.