# Student t test

William Sealy Gosset in 1908 published his work under the pseudonym “Student” to solve problems associated with inference based on sample(s) drawn from normally distributed population when the population standard deviation is unknown. He developed the t-test and t-distribution, which can be used to compare two small sets of quantitative data collected independently of one another, in this case this t-test is called independent samples t-test or also called unpaired samples t-test.

Student’s t-test is the most commonly used statistical techniques in testing of hypothesis on the basis of difference between sample means. The t-test can be computed just by knowing the means, standard deviations and number of data points in both samples by using the following formula

$t=\frac{\overline{X}_1-\overline{X}_2 }{\sqrt{s_p^2 (\frac{1}{n_1}+\frac{1}{n_2})}}$

where $s_p^2$ is the pooled (combined) variance and can be computed as

$s_p^2=\frac{(n_1-1)s_1^2 + (n_2-2)s_2^2}{n_1+n_2-2}$

Using this test statistic, we test the null hypothesis $H_0:\mu_1=\mu_2$ which means that both samples came from the same population under the given level of significance or level of risk.

If the computed t-statistics from above formula is greater than the critical value (value from t-table with $n_1+n_2-2$ degrees of freedom and given level of significance, say $\alpha=0.05$), the null hypothesis will be rejected, otherwise null hypothesis will be accepted.

Note that the t-distribution is a family of curves depending of degree of freedom (the number of independent observations in the sample minus number of parameters). As the sample size increases, the t-distribution approaches to bell shape i.e. normal distribution.

Example: The production manager wants to compare the number of defective products produced on the day shift with the number on the afternoon shift. A sample of the production from 6day shifts and 8 afternoon shifts revealed the following numbers of defects. The production manager wants to check at the 0.05 significance level, is there a significant difference in the mean number of defects per shits?

 Day shift 5 8 7 6 9 7 Afternoon Shit 8 10 7 11 9 12 14 9

Some required calculations are:

Mean of samples:

$\overline{X}_1=7$, $\overline{X}_2=10$,

Standard Deviation of samples

$s_1=1.4142$, $s_2=2.2678$ and $s_p^2=\frac{(6-1) (1.4142)^2+(8-1)(2.2678)^2}{6+8-2}=3.8333$

Step 1: Null and alternative hypothesis are: $H_0:\mu_1=\mu_2$ vs $H_1:\mu_1 \ne \mu_2$

Step 2: Level of significance: $\alpha=0.05$

Step 3: Test Statistics

\begin{aligned} t&=\frac{\overline{X}_1-\overline{X}_2 }{\sqrt{s_p^2 (\frac{1}{n_1}+\frac{1}{n_2})}}\\ &=\frac{7-10}{\sqrt{3.8333(\frac{1}{6}+\frac{1}{8})}}=-2.837 \end{aligned}

Step 4: Critical value or rejection region (Reject $H_0$ if absolute value of t-calculated in step 3 is greater than absolute table value i.e. $|t_{calculated}|\ge t_{tabulated}|$). In this example t-tabulated is -2.179 with 12 degree of freedom at significance level 5%.

Step 5: Conclusion: As computed value $|2.837| > |2.179|$, which means that the number of defects is not same on the two shifts.

See some Mathematica demonstration

Student T Distribution

## Introduction

A t-test for independent groups is useful when the same variable has been measured in two independent groups and the researcher wants to know whether the difference between group means is statistically significant. “Independent groups” means that the groups have different people in them and that the people in the different groups have not been matched or paired in any way.

## Objectives

The independent t-test compares the means of two unrelated/independent groups measured on the Interval or ratio scale. The SPSS t-test procedure allows the testing of hypothesis when variances are assumed to be equal or when are not equal and also provide the t-value for both assumptions. This test also provide the relevant descriptive statistics for both of the groups.

## Assumptions

• Variable can be classified in two groups independent of each other.
• Variable is Measured on interval or ratio scale.
• Measured variable is approximately normally distributed
• Both groups have similar variances  (variances are homogeneity)

## Data

Suppose a researcher want to discover whether left and right handed telephone operators differed in the time it took them to answer calls. The data for reaction time were obtained (RT’s measured in seconds):

 Subject no. RTs (Left) Subject no. RTs (Right) 1 500 11 392 2 513 12 445 3 300 13 271 4 561 14 523 5 483 15 421 6 502 16 489 7 539 17 501 8 467 18 388 9 420 19 411 10 480 20 467 Mean 476.5 430.8 Variance Ŝ2 5341.167 5298.84

The mean reaction times suggest that the left-handers were slower but does a t-test confirm this?

## Independent Sample t Test using SPSS

Perform the Following step by running the SPSS and entering the data set in SPSS data view

1. Click Analyze > Compare Means > Independent-Samples T Test… on the top menu as shown below.

Menu option for independent sample t test

2. Select continuous variables that you want to test from the list.

Dialog box for independent sample t test

3. Click on the arrow to send the variable in the “Test Variable(s)” box. You can also double click the variable to send it in “Test Variable” Box.
4. Select the categorical/grouping variable so that group comparison can be made and send it to the “Grouping Variable” box.
5. Click on the “Define Groups” button. A small dialog box will appear asking about the name/code used in variable view for the groups. We used 1 for males and 2 for females. Click Continue button when you’re done. Then click OK when you’re ready to get the output.  See the Pictures for Visual view.

Define Group for Independent sample t test

## Output

Independent sample t test output

First Table in output is about descriptive statistics concerning your variables. Number of observations, mean, variance, and standard error is available for both of the groups (male and female)

Second Table in output is important one concerning testing of hypothesis. You will see that there are two t-tests. You have to know which one to use. When comparing groups having approximately similar variances use the first t-test. Levene’s test checks for this. If the significance for Levene’s test is 0.05 or below, then it means that the “Equal Variances Not Assumed” test should be used (second one), Otherwise use the “Equal Variances Assumed” test (first one).  Here the significance is 0.287, so we’ll be using the “Equal Variances” first row in the second table.

In output table “t” is calculated t-value from test statistics, in example t-value is 1.401

df stands for degrees of freedom, in example we have 18 degree of freedom

Sig (two tailed) means two tailed significance value (P-Value), in example sig value is greater than 0.05 (significance level).

## Decision

As the P-value 0.178 id greater than our 0.05 significance level we fail to reject the null hypothesis. (two tailed case)

As the P-value 0.089 id greater than our 0.05 significance level we fail to reject the null hypothesis. (one tail case with 0.05 significance level)

As the P-value 0.089 id smaller than our 0.10 significance level we reject the null hypothesis and accept the alternative hypothesis. (one tail case with 0.10 significance level). In this case, it means that left handler have slower reaction time as compared to right handler on average.