Difference of Means

Schwab

Reading

Read chapter 20 before these lectures.

Side note: prop.test()

There is also a prop.test() function that will test a proportion hypothesis from data based on mathematical formula (as opposed to bootstrapping).

New test: Difference of two means

Another hypothesis test.

  • Conditions

  • Distributions

  • New Standard Error

  • New df

Conditions

Independence Extended

Normality (no extreme outliers)

Compute the df

with no computer

t - distribution with the smaller of \(n_1-1\) and \(n_2-1\) degrees of freedom.


with R

t - distribution, let R find the degrees of freedom with t.test().

SE

\[ SE = \sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}\\ \text{or}\\ SE = \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} \] We rarely know \(\sigma\) so the second standard error is more often used.

Test Statistic

\[ \text{Test Statisitc} = \frac{\bar{x}_1 - \bar{x}_2 -0}{SE } \]

(It looks just like the difference of proportions, but with means)

Problem number 18

Chapter 20

  1. CI for the difference of means
  2. Hypothesis Test

Confidence interval 99%

Check independence extended.

No extreme outliers.

Input values & calculate SE

\(\bar{x}_1 - \bar{x}_2 \pm t^* SE\)

# Automatic car stats
x_1 = 23.7
s_1 = 3.9
n_1 = 25

# Manual car stats
x_2 = 30.9
s_2 = 5.13
n_2 = 25

# Calculate the SE
SE =  sqrt(s_1^2/n_1 +s_2^2/n_2)

# Calculate the t_star
t_star = qt(p = 0.005, df = 24, lower.tail = FALSE)

Conclusion

We are 99% sure that the true difference in mpg is between 3.6 and 10.8.

#Lower bound
x_1 - x_2 - t_star*SE
[1] -10.80477
# Upper Bound
x_1 - x_2 + t_star*SE
[1] -3.595228

Hypothesis test

Conditions are checked and values are input.

Notation:

\[ H_o: \mu_1 = \mu_2\\ H_a: \mu_1 \ne \mu_2\\ \alpha = 0.05 \]

Calculate pvalue

Test statistic:

\[ T = \frac{\bar{x}_1-\bar{x}_2 - 0 }{SE} \]

Test_stat = x_1 - x_2

2*pt(q=Test_stat, df = 24)
[1] 1.933053e-07

Conclusion:

With a pvalue close to zero we have strong evidence to reject the null hypothesis in favor of the alternative. It seems that the average fuel economy of automatic vs manual cars is different.

It is possible we have made a type 1 error.

example with t.test()

Is there a difference in the absenteeism of students from New South Wales based on reported gender?

  • absenteeism

  • check conditions, make boxplot

  • write hypothesis

  • use t.test( y\~x, alternative= "t", data=)

Solution:

We should first check out absenteeism. ?absenteeism

Independence between and within groups.

check for outliers:

library(tidyverse)
library(openintro)

ggplot(data= absenteeism, aes(x=days, color = sex))+
  geom_boxplot()

Oh no

Those outliers seem pretty extreme. We should probably not do this test with a math model, like t.test()

Let’s do it anyway for practice

\[ H_o: \mu_1 = \mu_2\\ H_a: \mu_1 \ne \mu_2\\ \alpha = 0.05 \]

t.test(absenteeism$days ~ absenteeism$sex, alternative = "t")

    Welch Two Sample t-test

data:  absenteeism$days by absenteeism$sex
t = -1.0058, df = 136.35, p-value = 0.3163
alternative hypothesis: true difference in means between group F and group M is not equal to 0
95 percent confidence interval:
 -8.096135  2.637044
sample estimates:
mean in group F mean in group M 
       15.22500        17.95455

You try one

Is there a difference in the absenteeism of students from New South Wales based on ethnicity?

Conditions

Independence

Outliers seem to still be a problem

ggplot(data= absenteeism, aes(x=days, color = eth))+
  geom_boxplot()

Doing the test anyway

\[ H_o: \mu_1 = \mu_2\\ H_a: \mu_1 \ne \mu_2\\ \alpha = 0.05 \]

t.test(absenteeism$days~absenteeism$eth, alternative = "t")

    Welch Two Sample t-test

data:  absenteeism$days by absenteeism$eth
t = 3.4358, df = 126.85, p-value = 0.0007991
alternative hypothesis: true difference in means between group A and group N is not equal to 0
95 percent confidence interval:
  3.837747 14.262384
sample estimates:
mean in group A mean in group N 
       21.23188        12.18182

Do it with a simulation

library(infer)

set.seed(9)

absenteeism_randomization <- absenteeism |>
  specify( days ~ eth )|>
  hypothesise(null= "independence") |>
  generate(reps = 5000, type = "permute")|>
  calculate(stat = "diff in means", order = c("A","N"))

Visualize and pvalue

The observed difference is \(21.2 - 12.2 \approx 9\)

absenteeism_randomization |>
  visualise()+
  shade_p_value(obs_stat = 9, direction = "right")
absenteeism_randomization|>
  get_p_value(obs_stat = 9, direction = "both")
# A tibble: 1 × 1
  p_value
    <dbl>
1  0.0004