The heating bills for a selected sample of houses using various forms of heating are given below. (Values are in dollars.)

Natural Gas | Central Electric | Heat Pump |
---|---|---|

84 | 95 | 85 |

64 | 60 | 93 |

93 | 89 | 90 |

88 | 96 | 92 |

71 | 90 | 80 |

a. At $\alpha = 0.05$, test to see if there is a significant difference among the average heating bills of the homes. Use the p-value approach.

b. Test the above hypotheses using the critical value approach. Let $\alpha = .05$.

#### Solution

**(a) Overall mean**

Grand mean $\overline{\overline{X}}= \frac{G}{N}=\frac{1270}{15} = 84.6666667$.

**(b) Hypothesis :**

$H_0: \mu_1 =\mu_2 =\mu_3$ (i.e., there is no significant difference among the average heating bills of the homes) against $H_1:$ there is a significant difference among the average heating bills of the homes.

**(c) Computation**

`$$SS_i = \sum_{j=1}^{n_i} (X_{ij}-\overline{X}_i)^2$$`

` $$ \begin{aligned} SS_1 &= \sum (X_{1j}-\overline{X}_1)^2\\ & = (80-80)^2 + (75-80)^2 + (76-80)^2+(89-80)^2+(80-80)^2\\ & = 586 \end{aligned} $$ `

summary | Natural Gas | Central Electric | Heat Pump |
---|---|---|---|

Total (T) | 400 | 430 | 440 |

mean | 80 | 86 | 88 |

SS | 586 | 882 | 118 |

Number of factors $= k = 3$, $n_1 = 5$, $n_2 = 5$, $n_3 = 5$ subjects. Total number of observations $N = \sum n_i = 15$

` $$ \begin{aligned} G &= \sum T \\ & = 400+430+440\\ & = 1270 \end{aligned} $$ `

$\sum X^2 = 109286$

Correction factor = CF = $\frac{G^2}{N} = \frac{1612900}{15} =107526.6667$.

Total SS is given by

` $$ \begin{aligned} TSS &= \sum X^2 - CF\\ & = 109286 - 107526.6667\\ &= 1759.3333 \end{aligned} $$ `

SS Error is given by

` $$ \begin{aligned} SS_{Error}& =\sum SS\\ &= 586+882+118\\ & =1586 \end{aligned} $$ `

SS due to factor is given by

` $$ \begin{aligned} SS_{Factor}& =\sum \frac{T_i^2}{n_i} - CF\\ &= 32000+36980+38720- 107526.6667 \\ & =173.3333 \end{aligned} $$ `

` $$ \begin{aligned} df_{\text{total}} &= N-1 = 14\\ df_{\text{Factor}} &= k-1\\ &= 2\\ df_{\text{Error}} &= df_{total} -df_{\text{Factor}}\\ &= 14 - 2\\ & =12 \end{aligned} $$ `

Mean sum of square = Sum of square divided by resective degrees of freedom.

MS due to factor is given by

` $$ \begin{aligned} MS_{Factor}& =\frac{SS_{\text{Factor}}}{\text{Factor df}}\\ &= \frac{173.3333333}{2}\\ & =86.6667 \end{aligned} $$ `

MS due to Error is given by

` $$ \begin{aligned} MS_{Error}& =\frac{SS_{\text{Error}}}{\text{error df}}\\ &= \frac{1586}{12}\\ & =132.1667 \end{aligned} $$ `

#### ANOVA table

source | SS | df | MS | F |
---|---|---|---|---|

Factor | 173.3333 | 2 | 86.6667 | 0.6557 |

Error | 1586 | 12 | 132.1667 | |

Total | 1759.3333 | 14 |

Test Statistic is $F =\frac{MST}{MSE} =\frac{86.6667}{132.1667} = 0.6557$.

**p-value approach **

The $p$-value is `$P(F_{2,12} > F_{\text{observed }}) =P(F_{2,12} > 0.6557) =0.5367$`

.

Because the $p$-value $(0.5367) > 0.05$, we do not reject he null hypothesis.

That is there is no significant difference among the average heating bills of the homes.

**Critical value approach**

The Critical value of $F$ at 0.05 level of significance and for 2 and 12 degrees of freedom is $3.8852938$.

The test statistic is less than the critical value, we do not reject the null hypothesis at 0.05 level of significance. That is there is a significant difference among the average heating bills of the homes..

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators