t-tests

We are trying to test the null hypothesis $$ H_0 : temperature_{2019} = temperature_{average} $$

against the null hypothesis

$$ H_0 : temperature_{2019} > temperature_{average} $$

We are going to transform this into the form

$$ H_0 : temperature_{2019} - temperature_{average} = 0 $$

which creates only ONE VECTOR which we can work with.

T_2019 <- c(-1.7, 1.7, 5.6, 9.4, 10.7, 20.7, 18.8, 18.9, 13.3, 9.5, 5.6, 1.9)
T_average <- c(-2.8, -1.1, 2.5, 7.3, 12.3, 15.5, 16.9, 16.4, 12.8, 8.0, 2.7, -1.0)

x <- T_2019 - T_average

Test normality in any way you prefer. Remember the null hypothesis is that our data is indeed normal.

if(shapiro.test(x)$p.value > 0.05){print("pass")} else{print("fail")}

[1] "pass"

Now we perform the usual type of testing, using the test statistic

$$ t = \frac{x - mean}{\frac{standard \; deviation}{\sqrt{n}}} \sim students \; distribution(df = n - 1)$$

Construct

• critical region using quantiles

• confidence interval using the formula from previous seminars

• $p$-value using the formula from previous seminars

And then check your results using the magic function t.test(). The arguments are

• paired = T indicating that our data correspond to the same months
• alternative = "greater" indicates we are testing the right-side alternative

t.test(T_2019, T_average, paired = T, alternative = "greater")

	Paired t-test

data:  T_2019 and T_average
t = 4.3394, df = 11, p-value = 0.000588
alternative hypothesis: true mean difference is greater than 0
95 percent confidence interval:
 1.216245      Inf
sample estimates:
mean difference 
          2.075 

See the structure of the result

str(t.test(T_2019, T_average, paired = T, alternative = "greater"))

List of 10
 $ statistic  : Named num 4.34
  ..- attr(*, "names")= chr "t"
 $ parameter  : Named num 11
  ..- attr(*, "names")= chr "df"
 $ p.value    : num 0.000588
 $ conf.int   : num [1:2] 1.22 Inf
  ..- attr(*, "conf.level")= num 0.95
 $ estimate   : Named num 2.07
  ..- attr(*, "names")= chr "mean difference"
 $ null.value : Named num 0
  ..- attr(*, "names")= chr "mean difference"
 $ stderr     : num 0.478
 $ alternative: chr "greater"
 $ method     : chr "Paired t-test"
 $ data.name  : chr "T_2019 and T_average"
 - attr(*, "class")= chr "htest"

Pull the value of the test statistic and compare it with the appropriate quantiles

t.test(T_2019, T_average, paired = T, alternative = "greater")$statistic

t: 4.33937779168679

Pull the confidence interval

t.test(T_2019, T_average, paired = T, alternative = "greater")$conf.int

1. 1.21624531725495
2. Inf

What number should be in the confidence interval for us to not reject the null hypothesis? Look at the formulation of the null hypothesis

$$ H_0 : temperature_{2019} - temperature_{average} = 0 $$

Pull the $p$-value

t.test(T_2019, T_average, paired = T, alternative = "greater")$p.value

0.000588009196712441

t-tests - task 2

In this task our data isn't essentially paired. Examples of paired samples would be

• patients blood pressure before and after treatement (bond = same patient)
• temperatures in two different cities on the same day (bond = same day)
• students taking a test before and after a seminar (bond = same student)

Typically the paired samples arise in these types of scenarios

• characteristic of a subject before and after an event
• repeated measures
• control group

Test the null hypothesis that the variances of the two samples are the same.

$$H_0 : \sigma^2_1 = \sigma^2_2 $$

You can use the in-built function var.test() or do in manually

mach.1 <- c(29, 27, 29, 35, 29, 32, 28, 34, 32, 33)
mach.2 <- c(31, 28, 30, 28, 37, 29, 27, 27, 39, 33,
            31, 32, 31, 29, 32, 28, 27, 28, 24, 34)

alpha <- 0.05

n1 <- length(mach.1)
sigma1 <- sd(mach.1)

n2 <- length(mach.2)
sigma2 <- sd(mach.2)

F <- sigma1^2/sigma2^2

CR_1 <- qf(alpha/2, n1-1, n2-1)
CR_2 <- qf(1 - alpha/2, n1-1, n2-1)

print(c(CR_1, F, CR_2))

[1] 0.2714929 0.5807166 2.8800520

Now use the t.test which assumes equal variances. Compute the common variance using this formula

$$ S = \sqrt{ \frac{ (n_1 - 1) \sigma_1^2 + (n_2 - 1) \sigma_2^2}{n_1 + n_2 - 2} } $$

S = sqrt(((n1-1)*sigma1^2+(n2-1)*sigma2^2)/(n1+n2-2))

The test statistic is going to be

$$ t = \frac{mean_1 - mean_2}{S \sqrt{\frac{n_1 + n_2}{n_1 n_2}}} \sim students \; distribution (n_1 + n_2 - 2)$$

Now you can compute

• critical region
• $p$-value

Use the in built function to do the same thing.

t.test(mach.1, mach.2, var.equal = T)

	Two Sample t-test

data:  mach.1 and mach.2
t = 0.4245, df = 28, p-value = 0.6744
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.103981  3.203981
sample estimates:
mean of x mean of y 
    30.80     30.25 

Poisson distribution - testing of the expected number of events

Remember that Poisson process models the number of events. Number of events in one sector is going to have Poisson distribution with parameter $\lambda$. Number of events in all sectors is going to also follow Poissons distribution with the parameter $\lambda$ multiplied by the number of sectors.

The value S plays the role of the test statistic. The parameter is going to be $\lambda$ multiplied by the number of sectors.

Exact test

S <- sum(n_bombs*n_sectors) # total number of events

# hint for computing the p-value

ppois(S, lambda*n)

# hint for visualization

qpois(ppois(S,lambda*n))

Asymptotic test

Using central limit theorem

$$ t = \frac{mean - \lambda_0}{\sqrt{\lambda_0}} \sim N(0,1) $$

Confidence interval is going to be

$$ mean \pm q_{1 - \alpha/2} \cdot \frac{mean}{n}$$

Mean here denotes the average number of events per sector!