Chapter 12, Problem 1CQQ

Measures of Voltage The author measured voltage amounts from three different sources: (1) electricity supplied directly to the author’s home, (2) an independent gas-powered Generac generator (model PP 5000), and (3) an uninterruptible power supply (APC model CS 350) connected to the author’s home power supply. The results from analysis of variance are shown below. What is the null hypothesis for this analysis of variance test? What specific result is used in deciding to reject H₀ or fail to reject H₀? Based on the displayed results, should you reject H₀ or fail to reject H₀?

To determine

To identify: The null hypothesis.

To explain: The specific result that is used in deciding to reject H_o and fail to reject H_o.

To check: Whether the null hypothesis is rejected or fails to reject by using the results.

Answer to Problem 1CQQ

The null hypothesis is, $H_0:{\mu }_1={\mu }_2={\mu }_3$.

The specific result that is used in deciding to reject H_o and fail to reject H_o is the decision rule based on P value.

The null hypothesis is rejected.

Explanation of Solution

Given info:

The Minitab output shows the result of analysis of variance for the voltage amounts from three different sources. The sources are electricity supplied directly to the author’s home, an independent gas-powered generac generator (model PP 5000) and an uninterruptible power supply (APC model CS 350).

Calculation:

State the test hypotheses.

Let ${\mu }_1$ be the mean of the voltage amounts from electricity supplied directly to the author’s home, ${\mu }_2$ be the mean of the voltage amounts from independent gas-powered generac generator (model PP 5000) and ${\mu }_3$ the mean of the voltage amounts from the uninterruptible power supply (APC model CS 350).

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3$

Alternative hypothesis:

$H_1:$ At least one of the means is different from the other means

From the Minitab output, the P-value is 0.000.

Decision criteria for the P-value method:

If $P\text{-value}\le \alpha \text{,}$ then reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}>\alpha \text{,}$ then fail to reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, $0.000\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean voltage amounts from three sources are same.