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 = x1 - x2√(SD12/n1 + SD22/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:
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:
To perform a hypothesis test to compare two population means, $\mu_1$ and $\mu_2$, we have some assumptions:
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 |
Serbia | United States | |
Mean | 43.2 | 5.2 |
Standard Deviation | 1.2 | 8.3 |
Number of Employess | 67 | 166 |