# t-Test Example for 2, 4, 6, 8, 9, 10, 13 & 15 & x̄ = 56

Student's two tailed t-test summary & example for test of significance (hypothesis) for

*mean*of small samples from population with unknown*variance*by using t-statistic (t_{0}) for dataset {2, 4, 6, 8, 9, 10, 13 & 15}, population mean μ = 56 and critical value (t_{e}) from t-distribution table at 10% significance level for the degrees of freedom ν = 7.Calculation Summary | |
---|---|

data set value | 2, 4, 6, 8, 9, 10, 13 & 15 |

population mean | 56 |

significance level (α) | 0.1 |

t_{0} | 30.802 |

t_{e} | 1.89 |

## Solved example for dataset {2, 4, 6, 8, 9, 10, 13 & 15} & x̄ = 56

The t-test work with steps for sample data set values {2, 4, 6, 8, 9, 10, 13 & 15}, population mean = 56 to conduct the test of significance for mean may helpful for beginners or grade school students to learn how to solve t-test worksheet or homework problems efficiently by using t-statistic & critical value of t from t-distribution table.

__Workout :__

step 1 Address the formula, input parameters and values

data set value = 2, 4, 6, 8, 9, 10, 13 & 15

Significance level (α) = 0.1

population mean = 56

step 2 To find sample mean x̄ & variance s

^{2}

x | d = x-x̄ (x - 8.375) | d^{2} |

2 | -6.375 | 40.6406 |

4 | -4.375 | 19.1406 |

6 | -2.375 | 5.6406 |

8 | -0.375 | 0.1406 |

9 | 0.625 | 0.3906 |

10 | 1.625 | 2.6406 |

13 | 4.625 | 21.3906 |

15 | 6.625 | 43.8906 |

∑x = 67 | ∑d = 0 | ∑d^{2} = 133.8748 |

x̄ = ∑xn

x̄ =678

x̄ = 8.375

s

^{2}= 1/n - 1[ ∑d

^{2}- {(∑d)

^{2}/n} ]

s

^{2}= 1/8 - 1[ 133.8748 - (0

^{2}/8)]

s

^{2}= 1/7 x 133.8748

s

^{2}= 19.125

step 3 Substitute sample mean, population mean,

*standard deviation*,

*sample size*value in the below formula for test of significance for sample mean.

d̄ = x̄ - μ

t

_{0}=d̄√s

^{2}/n

=8.375 - 56√19.125/8

step 4 To find t

_{0}, simplify the above expression

=47.625√19.125/8

=47.625√2.3906

=47.6251.5462

t

_{0}= 30.802

step 5 Find expected t

_{e}with level of significance α = 0.1 , degrees of freedom ν = n - 1 = 7 from two tailed t-distribution table

t

_{e}= 1.89

__Inference__t

_{0}> t

_{e}

The null hypothesis H

_{0}is rejected since t

_{0}= 30.802 is greater than the critical value for degrees of freedom t

_{e}(6) = 1.89. Therefore, there is significant difference between the sample & population

*means*.