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

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

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

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

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

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

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