# Hypothesis Testing

Hypothesis testing uses concepts from statistics to determine the probability that a given assumption is true. In this chapter, you will learn about several types of statistical test, their practical applications, and how to interpret the results of hypothesis testing.

Throuh hypothesis testing one can make inference about the population parameters by analysing the difference between the observed results, the sample statistics, and the results otherwise one would expect if an underlying hypothesis is actually true.

Typically hypothesis testing starts with an assumption, or an assertion about a population parameter. For example, you may be intereted in validating the claim of philips that the average life of there bulb 10 years.

### Things You Will Master

- Applications of hypothesis testing
- Perform One-Sample t-test
- Perform two-Sample t-test
- Perform Paired t-test
- Perform ANOVA test
- Perform Chi-Square test
- Wilcoxon Signed Rank Test
- Shapiro Test
- Kolmogorov And Smirnov Test

## Applications of hypothesis testing

When you want to compare the sample mean with the population mean. For example - You would like to determine if the average life of a bulb from brand X is actually 10 years or not. In this case, when you want to check if the sample mean represents the population mean then you should run

**One Sample t-test**.When you want to compare the means of two independent variables. One of which can be categorical variable. In this case, we run

**Two sample t-test**.When you want to compare the before and after effects of an experiment or a treatment. Then in that case, we run

**Paired t-test**.When you want to compare more than two independent variables then in that case, we run

**ANOVA test**In all the above applications, we assumed that variables are numeric. However, When you want to compare two categorical variables, we run

**Chi-square test**.

## Perform One-Sample t-test

One sample ttest is a parametric test. It is used when you wish to check if the sample mean represents the population mean or not. It assumes that the data follows normal distribution.

```
set.seed(100)
# Generating random data with normal distibution
x <- round(rnorm(30, mean = 10, sd = 1),0)
# Checking if mean is really 10 years
t.test(x, mu=10)
```

```
# Output
One Sample t-test
data: x
t = 0.17668, df = 29, p-value = 0.861
alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:
9.647473 10.419193
sample estimates:
mean of x
10.03333
```

### How to interpret results?

In the above case,

**p-value = 0.861**, this values i greated than alpha value and thus we have to accept the null hypothesis. Here null hypothesis was that average value is 10 and alternative hypothesis was that it is not equal to 10.

**95 percent confidence interval**:
**9.647473 10.419193** - The 95% CI also includes the 10 and thus it is fine to state that the mean value is 10.

## Perform welch two-Sample t-test

Two sample t-tests are used to compare the means of two independent quantitative variables. By default, the `t.test()`

function runs a welch test which is a parametric test. It assumes that the two populations have normal distributions and with equal variances.

In the below example we assumed that the x and y are samples taken from populations which follow normal distribution.

```
# Generating two variable x and y
x <- c(3.80, 5.83, 11.89, 21.04, 12.45, 7.38, 6.97, 6.60)
y <- c(10.54, 8.88, 9.89, 23.74, 14.65)
# Checking if mean is really 10 years
t.test(x,y)
```

```
# Output
Welch Two Sample t-test
data: x and y
t = -1.2051, df = 7.9346, p-value = 0.2629
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-11.796332 3.706332
sample estimates:
mean of x mean of y
9.495 13.540
```

### How to interpret results?

Here, the null hypothesis is that the mean of x - mean of y = 0

and alternative hypotheis is that the mean of x - mean of y != 0

**As p-value(0.2629) is greate than alpha value(0.05)** we accept the null hypothesis and conclude that the mean of x is indeed equal to the mean of y.

**95 percent confidence interval:**
**-11.796332 3.706332** - Also, it is evident that zero did apreare in at least 95% of the experiments and thus we conclude that out decision of accepting the null hypothesis is correct.

## Perform Paired t-test

I am taking this example from datasciencebeginners.

An educational institute wants to check if their course helps in improving the scores of the students. So what they do is they give a test to a bunch of students before the class started and recorded the scores. After which all these students were trained on the subject and at the end of the course another test was given to the students, and the scores were noted. They now need to understand if the course or training has resulted in better scores.

```
# Score of students before and after the course
before <- c(73, 38, 46, 99, 5, 15, 58, 78, 99, 43)
after <- c(90, 78, 21, 70, 72, 59, 55, 39, 40, 99)
# Running the paired ttest
t.test(before, after, paired = TRUE)
```

```
# Output
Paired t-test
data: before and after
t = -0.49552, df = 9, p-value = 0.6321
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-38.39999 24.59999
sample estimates:
mean of the differences
-6.9
```

### How to interpret results?

As **p-value** came out to be 0.63 which is higher than the alpha value. We need to accept the null hypothesis and thus conclude that there is no significant change in test scores.

## Perform ANOVA test

ANOVA stands for analysis of variance and to test this we run Fishers F-test. We run this test when we want to compare the means of more than two independent variables.

For example - we may want to know if the average sepal length across three different flower species is similar or not.

Here, Null Hypothesis :: μ1 = μ2 = μ3

and, Alternative :: μ1 ≠ μ2 ≠ μ3 or μ1 = μ2 ≠ μ3 or μ1 ≠ μ2 = μ3

You need to run the post adHoc test in case you reject the null hypothesis. This is to ceratin if all groups are different or only on of them is different.

```
# Running anova
result <- aov(Sepal.Length ~ Species, data = iris)
# Checking the result
summary(result)
```

```
# Output
Df Sum Sq Mean Sq F value Pr(>F)
Species 2 63.21 31.606 119.3 <0.0000000000000002 ***
Residuals 147 38.96 0.265
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
```

### How to interpret results?

**Pr(>F) = <0.0000000000000002** which is less than alpha value we reject the null hypothesis stating that the average sepal length of three different flower species are not same.

### As we have rejected null hypothesis, we now run post AdHoc test

### Running post Adhoc test

As part of post Adhoc test We are running Tukey test.

`TukeyHSD(result) # pass the model output`

```
# Output
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = Sepal.Length ~ Species, data = iris)
$`Species`
diff lwr upr p adj
versicolor-setosa 0.930 0.6862273 1.1737727 0
virginica-setosa 1.582 1.3382273 1.8257727 0
virginica-versicolor 0.652 0.4082273 0.8957727 0
```

Each line of output in the above table can be thought of individual independent test run for each pair. The p-value for which is represented by **p adj**. Comparing the padj value against alpha value we conclude that mean of all the three flowers are different.

## Perform Chi-Square test

Chi-sqaure test uses contigency table to test if the two categorical variables are dependent on each other or not. To run the test you first need to create a contigency table between the two categorical variables. This table is then passed to the `chisq.test()`

function.

For example - Let us check if the treatment and type are dependent on each other in the CO2 dataset.

Here, null hypothesis is that they are not dependent

And, alternatuve is that they are dependent on each other

```
# Creating contigency table
tab <- table(CO2$Type, CO2$Treatment)
# Running chi-square test
chisq.test(tab)
```

```
# Output
Pearson's Chi-squared test
data: tab
X-squared = 0, df = 1, p-value = 1
```

### How to interpret results?

We again look for the p-value and compare that with the preset alpha value of 0.05. In this case, p-value is greater than alpha and thus we accept the null hypothesis. So final concusion is that the plant and treatment are not dependent on each other.

## Wilcoxon Signed Rank Test

The Wilcoxon Signed Rank test is a non parametric test. It is an alternative of one sample t-test when the data is not assumed to follow a normal distribution.

```
data <- c(30, 29, 29, 28, 26, 23, 28, 25, 24, 19)
wilcox.test(data, mu=35, conf.int = TRUE)
```

```
# Output
Wilcoxon signed rank test with continuity
correction
data: data
V = 0, p-value = 0.005857
alternative hypothesis: true location is not equal to 35
95 percent confidence interval:
23.50001 28.99999
sample estimates:
(pseudo)median
26.49994
```

## Shapiro Test

We use Shapiro test to check if the data follows normal distribution or not. In shapiro test, null hypothesis is that the data has normal distribution and alternative hypothesis is that data does not follow normal distribution.

```
set.seed(123)
data <- rnorm(50, mean = 30, sd = 2)
shapiro.test(data)
```

```
# Output
Shapiro-Wilk normality test
data: data
W = 0.98928, p-value = 0.9279
```

As p-value > 0.05, we accept the null hypothesis which states that the distribution is normally distributed. We also know that because we used `rnorm`

function which generates random numbers which follow normal distribution.

# Kolmogorov And Smirnov Test

The test is also very famous by the name k-s test. The test is done to check whether two data sets follow the same distribution or not. Here, null hypothesis is that the distribution of two samples are same and alternative hypothesis is that the distributions are different.

```
# Generating random numbers
x <- rnorm(100)
y <- runif(100)
# Performaing k-s test
ks.test(x, y)
```

```
# Output
Two-sample Kolmogorov-Smirnov test
data: x and y
D = 0.54, p-value = 0.0000000000004335
alternative hypothesis: two-sided
```

The output above suggests that the distribution of x and y are different as p-value < 0.05 and thus we reject the null hypothesis.