# Solved: The heating bills for a selected sample of houses using various forms of heating are

#### ByDr. Raju Chaudhari

Oct 15, 2020

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..