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Explain And Execute Statistical Design And Analysis Of Two Variable

Hypothesis

Dr. Nancy Agens, Head,

Technical Operations, Statswork

I. INTRODUCTION

In this blog, I will explain you how

the statistical analysis is being applied for

two independent samples. In practice, the

test statistic used for comparing the two

means from a population is by using the t-

test because t-test shrinks the data to a

single t-value and it is then compared with

the significant value for the final conclusion.

Now, Let us understand the theoretical

background in performing the t-test for two

variables.

Suppose X1 and X2 be the two

independent random variable and let ,

be the sample with size n1 and n2 from a

population with mean µ1, µ2 and variance

σ12, σ2

2 respectively. It is obvious that if the

sample size is large enough then the sample

mean will follow a normal distribution, (i.e)

and

In addition, if the means of the two

samples are said to follow normal

distribution, then the difference of mean are

also said to follow normal distribution. It is

given by

Under the null hypothesis, H0: µ1 =

µ2 which means there is no statistical

significant difference between the means.

The test statistic becomes

If suppose we come across the data

having the same variance then the test

statistics boils down to

Once the t-value is calculated, the

next step is to compare with the critical

value with alpha level of significance and if

the calculated t-value is less than the

significant value then the conclusion is to

reject the null hypothesis stating that there is

a significant difference between the means

of the population. (Cressie & Whitford,

1986)

Imagine a marketing company has

recently launched two campaigns for

advertising their product. The company’s

head wants to identify whether both the

campaign is equally effective or not. In such

case, the statistical hypothesis testing is the

essential method to give a valid inference.

Before performing any statistical hypothesis

testing, the main task is to understand the

problem statement, to frame the hypotheses

of interest, to find a suitable test statistics,

and finally to make a proper decision with

the results.

This blog will elaborate each one

with the advertising example as mentioned

above.

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II. UNDERSTANDING THE PROBLEM

STATEMENT

The primary or basic task in any

statistical data analysis is to know or find

out what the problem is and how the data is

being measured. In our example, the

manager wish to find the effectiveness of

their campaign, for this, he/she has to

consider all the information related to the

campaign and find out whether the

campaign results in a profit or loss. The only

way to test whether the two campaign is

effective is to perform a statistical test by

comparing their means.

III. CONSTRUCTION OF TEST

HYPOTHESES

Once you understand the problem at

hand, the next step is to frame an appropriate

hypothesis to test for statistical significance;

we call it as the null hypothesis and

alternative hypothesis (Flandin & Friston,

2019). The null hypothesis is something

which we claim or our belief about the

problem and is denoted by H0. That is, for

our example, the null hypothesis will be

there is no statistically significant difference

between the mean incomes from two

campaigns.

H0:μ1 = μ2

Or

H0:μ1−μ2=0

The alternative hypothesis (H1) is

simply a contrary to the null hypothesis.

That is, there is a significant difference

between the means of the two campaigns.

H1: μ1≠μ2

or

H1: μ1−μ2≠0

IV. FINDING A SUITABLE TEST

STATISTICS

For finding the suitable statistic test,

we need to find the distribution of the

data. I will illustrate with a simulated data

for two campaigns using R software.

set.seed(123)

camp1<-rt(30,29)*50+210

camp2<-rt(30,29)*48+170

Fig. 1 Histogram-Normal Distribution

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If you see the above graph, the data

is closely from a normal distribution.

However, I have taken the sample size as 30

per campaign, so we should make use of the

t-distribution for testing this problem. From

the simulated data, the mean for two

campaigns is $210.2226 with standard

deviation $60.0008 and $182.8537 with

standard deviation $47.56557 respectively.

V. CALCULATION OF TEST STATISTIC

Once you got all the necessary

values for the calculation, the next step is to

apply it into the formula of statistics test as

mentioned earlier. Here, I will illustrate

using R.

Difference<-mean(camp1)-mean(camp2)

Std.dev<-sqrt((sd(camp1)^2+sd(camp2)^2)/2)

Std.err<-

Std.dev*(1/length(camp1)+1/length(camp2))^0.5

t.value<-Difference/Std.err

The difference of mean is $27.3689 and the t-value is

1.9578.

VI. CONCLUSION OF THE PROBLEM

As a final step, we compare the

calculated t.value with the critical value. In

order to find the critical value, we need to

fix the significance level alpha. Usually, it is

considered as 5% that means we can tolerate

the probability of rejecting the null

hypothesis by 5% or 0.05 level of

significance. Next step is to check whether

the null hypothesis is one-sided or two-sided

for the concluding the problem. If you are

concerned about which campaign is higher

or smaller then the null will be one-sided.

However, in our case, it is two sided null

hypothesis stating that the means of the

campaigns are equal. An important note is

that in a two-sided test the critical region is

divided by half (5% is equally distributed in

both sides from population mean).

The rejection region can be

calculated using the confidence interval. If

the t-value lies outside the confidence limit

we will reject the null hypothesis otherwise

we accept the same. In R, there is a function

called t.test to perform the calculation and

the p-value is compared with 0.05 for the

conclusion.

Res<-

t.test(camp1,camp2,paired=FALSE,var.equal=TRU)

Res

Two Sample t-test

data: cam1 and cam2

t = 1.9578, df = 58, p-value = 0.05507

alternative hypothesis: true difference in

means is not equal to 0

95 percent confidence interval:

-0.613592 55.351410

sample estimates:

mean of x mean of y

210.2226 182.8537

From the results, the t-value (or test

statistic) is 1.9578 as we got previously and

the p-value is 0.05507, which is greater

than 0.05. Since the p-value is greater than

0.05, we accept the null hypothesis and

conclude that the difference of mean amount

from two campaign is same.To sum up, this

blog is to elaborate and explain you the

procedure used to test the two variables and

how to provide a valid inference from the

results. I hope this blog serves you better for

understanding and analyzing similar data.

REFERENCES

[1] Cressie, N. A. C., & Whitford, H. J. (1986). How to use

the two sample t‐test. Biometrical Journal, 28(2),

131–148. Retrieved from

https://onlinelibrary.wiley.com/doi/abs/10.1002/bimj

.4710280202

[2] Flandin, G., & Friston, K. J. (2019). Analysis of

family‐wise error rates in statistical parametric

mapping using random field theory. Human Brain

Mapping, 40(7), 2052–2054. Retrieved from

https://onlinelibrary.wiley.com/doi/abs/10.1002/hbm

.23839