Mean 1 = 18.6

Mean 2 = 14.6

SD 1 = 12.5419

SD 2 = 9.1815

T-score = 0.5754

GENERATE WORK

GENERATE WORK

**Input Data : **

Data set x = 3, 11, 17, 28, 34

Data set y = 5, 8, 13, 19, 28

Total number of elements = 5

**Objective :**

Find the t-score by using mean and standard deviation.

**Solution :**

Mean 1 = (3 + 11 + 17 + 28 + 34)/5

= 93/5

Mean 1 = 18.6

Mean 2 = (5 + 8 + 13 + 19 + 28)/5

= 73/5

Mean 2 = 14.6

SD1 = √(1/5 - 1) x ((3 - 18.6)^{2} + ( 11 - 18.6)^{2} + ( 17 - 18.6)^{2} + ( 28 - 18.6)^{2} + ( 34 - 18.6)^{2})

= √(1/4) x ((-15.6)^{2} + (-7.6)^{2} + (-1.6)^{2} + (9.4)^{2} + (15.4)^{2})

= √(0.25) x ((243.36) + (57.76) + (2.56) + (88.36) + (237.16))

= √(0.25) x 629.2

= √157.3

SD1 = 12.5419

SD2 = √(1/5 - 1) x ((5 - 14.6)^{2} + ( 8 - 14.6)^{2} + ( 13 - 14.6)^{2} + ( 19 - 14.6)^{2} + ( 28 - 14.6)^{2})

= √(1/4) x ((-9.6)^{2} + (-6.6)^{2} + (-1.6)^{2} + (4.4)^{2} + (13.4)^{2})

= √(0.25) x ((92.16) + (43.56) + (2.56) + (19.36) + (179.56))

= √(0.25) x 337.2

= √84.3

SD2 = 9.1815

t-score = x_{1} - x_{2}√(SD1^{2}/n1 + SD2^{2}/n2)

= 18.6 - 14.6√((12.5419)^{2}/5 + (9.1815)^{2}/5)

= 4√((157.3)/5 + (84.3)/5)

= 4√(31.46 + 16.86)

= 4√(48.32)

= 46.9513

t-score = 0.5754

** t-test calculator** is an online statistics tool to estimate the significance of observed differences between the means of two samples when there is a null hypothesis that is no significant difference between the means by using standard deviation.

It is necessary to follow the next steps:

- Enter two samples (observed values) in the box. These values must be real numbers or variables and may be separated by commas. The values can be copied from a text document or a spreadsheet.
- Press the
**"GENERATE WORK"**button to make the computation. - t-Test calculator will give a test whether samples from two independent populations provide that the populations have different means.

For a hypothesis test that has null hypothesis $H_0 : \mu_1 = \mu_2$, the value of the test statistic is determined by the formula
$$t=\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$$
where $\bar X_1$ and $\bar X_2$ are sample means, $s_1$ and $s_2$ are sample standard deviations and $n_1$ and $n_2$ are sizes of samples $X_1$ and $X_2$, respectively.

A hypothesis test consists of two hypotheses, __the null hypothesis__ and __the alternative hypothesis__ or research hypothesis.

The symbol $H_0$ represents the null hypothesis. The null hypothesis reflects that there will be no observed effect on the experiment. The null hypothesis consists of an equal sign.
The alternative hypothesis reflects that there is an observed effect on the experiment. The symbol $H_a$ represents the alternative hypothesis.
The first step in testing is to determine the null hypothesis and the alternative hypothesis. Regarding the testing hypothesis, there are some important terms. __Rejection region__ is the set of values leads to rejection of the null hypothesis. __Non-rejection region__ is the set of values that leads to nonrejection of the null hypothesis. __Critical values__ are the value that separates the rejection and non-rejection regions.

The t-Test is used in comparing the means of two populations. There are two approaches:

- When the samples from the two populations are independent;
- When the samples from the two populations are depended, i.e. when they are paired.

The variable $$z=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$$ has the standard normal distribution, where $\sigma_1$ and $\sigma_2$ are the population standard deviations. We distinguish two cases:

- If populations standard deviations are equal, $\sigma_1-\sigma_2$

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

where $s_1$ and $s_2$ are sample standard deviations, respectively.If $X$ is a normally distributed variable then for independent samples of sizes $n_1$ and $n_2$ from the two populations, the variable $$t=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$ has the t-distribution with $df = n_1 + n_2 - 2$. For a hypothesis test that has null hypothesis $H_0 : \mu_1 = \mu_2$, we can use the variable $$t=\frac{\bar X_1-\bar X_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$

- If populations standard deviations are different

To perform a hypothesis test to compare two population means, $\mu_1$ and $\mu_2$, we have some assumptions:

- Simple and independent random samples;
- Normal populations or large samples.

A t-Test is one of the most frequently used tests in statistics. A t-Test is useful to conclude if the results are correct and applicable to the entire population.
If we want to analyze simple experiments or when making simple comparisons between levels of independent variable we use the t-Test. It's used in comparison between two separate groups of individuals, for example: male vs female, experimental vs
control group, etc.

**Practice Problem 1:**

There are two company A and B. We want to test average age of employees at these companies so we use a random sample of employee ages from each company.

Company A | Company B | |

Mean | 43.2 | 36.7 |

Standard Deviation | 7 | 8.3 |

Number of Employess | 50 | 66 |

We wanted to compare the average annual earnings of math professors between the two countries, Serbia and United States. We obtained earnings for a random sample of people from each country represented in thousands of US dollars

Serbia | United States | |

Mean | 43.2 | 5.2 |

Standard Deviation | 1.2 | 8.3 |

Number of Employess | 67 | 166 |

The t-test calculator, work with steps, formula and practice problems would be very useful for grade school students (K-12 education) to learn what is appropriate test statistic for t-Test, how to find it. With this statistics and probability tool, we can effortlessly make student's t-distribution calculation for given data sets. This concept can be applied in many real-life situation to test a hypothesis.