# Students's t-Test for x̄ = 11, μ = 15, s = 5.6, n = 28 & α = 50%

*mean*of small samples from population with unknown

*variance*by using t-statistic (t

_{0}) for sample mean x̄ = 11, population mean μ = 15, sample

*standard deviation*s = 5.6 &

*sample size*n = 28 and critical value (t

_{e}) from t-distribution table at 50% significance level for the degrees of freedom ν = 27.

Calculation Summary | |
---|---|

sample mean (x̄) | 11 |

mean (μ) | 15 |

standard deviation (s) | 5.6 |

sample size (n) | 28 |

Significance Level (α) | 0.5 |

t_{0} | 3.7115 |

t_{e} | 1.18 |

## Work with Steps for n = 28, x̄ = 11, s = 5.6, μ = 15 & α = 50%

The below is the example work with steps for test of significance (H_{0}) for mean of small samples from population with unknown variance is generated for x̄ = 11, μ = 15, s = 5.6 & n = 28 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

sample mean (x̄) = 11

mean (μ) = 15

standard deviation (s) = 5.6

sample size (n) = 28

Significance level (α) = 0.5

step 2 Substitute sample mean, mean, standard deviation, sample size value in the below formula for test of significance for mean.

=11 - 15(5.6 / √28- 1)

step 3 To find t_{0}, simplify the above expression

=4(5.6 / √28- 1)

=4(5.6 / √27)

=4 x 5.19625.6

=20.78465.6

t_{0}= 3.7115

step 4 Find the expected or critical value (t_{e}) from one tailed t-distribution table for degrees of freedom ν = n-1 = 27

The critical value t_{e} for d.f 27 at 0.5 significance level is 1.18

t _{e} = 1.18

__Inference__

t_{0} > t_{e}

The null hypothesis H_{0} is rejected since t_{0} = 3.7115 is greater than the critical value for degrees of freedom t_{e}(26) = 1.18. Therefore, there is significant difference between the sample & population *means*.