11.10 In a test of the effectiveness of a new battery design, 16 battery-powered music boxes are randomly provided with either the old design or the new version. Hours of playing time before battery failure were as follows:

new battery type:3.3,6.4,3.9,5.4,5.1,4.6,4.9,7.2  hrs
old battery type:4.2,2.9,4.5,4.9,5.0,5.1,3.2,4.0  hrs

Assuming normal populations with equal standard deviations, use the 0.05 level to determine whether the new battery could be better than the old design. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test?

Solution

Let $X$ denote the Hours of playing time before battery failure for new battery type and let $Y$ denote the Hours of playing time before battery failure for old battery type.

Given that the sample size $n_1 = 8$, $n_2 = 8$, sample mean $\overline{x}= 5.1$, $\overline{y}= 4.2$, sample standard deviation $s_1 = 1.263$ and $s_2 = 0.824$.

Hypothesis testing problem

The hypothesis testing problem is

$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 > \mu_2$ ($\textit{right-tailed}$)

($\mu_1 > \mu_2$ means new battery perform better than the old battery)

Test statistic

The test statistic is

$$ \begin{aligned} t& =\frac{(\overline{x} -\overline{y})-(\mu_1 - \mu_2)}{sp\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \end{aligned} $$
where

$$ \begin{aligned} s_p & = \sqrt{\frac{(n_1-1)s_1^2 +(n_2-1)s_2^2}{n_1+n_2-2}}\\ & = \sqrt{\frac{(8-1)1.2627^2 +(8-1)0.8242^2}{8+8-2}}\\ & = 1.0662. \end{aligned} $$

Specify the level of significance

The significance level is $\alpha = 0.05$.

Determine the critical value

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $t$ using $\alpha = 0.05$ and degrees of freedom $n_1+n_2-2=8+8-2=14$ $\text{is}$ $\text{1.761}$.

t-test critical region for right tailed test
t-test critical region for right tailed test

The rejection region (i.e. critical region) is $\text{t > 1.761}$.

Computation

The test statistic under the null hypothesis is

$$ \begin{aligned} t&=\frac{(\overline{x} -\overline{y})-(\mu_1-\mu_2)}{sp\sqrt{\big(\frac{1}{n_1}+\frac{1}{n_2}\big)}}\\ &= \frac{(5.1-4.2)-0}{1.0662\sqrt{\big(\frac{1}{8}+\frac{1}{8}\big)}}\\ &= 1.6882 \end{aligned} $$

Decision

Traditional approach:

The rejection region (i.e. critical region) is $\text{t > 1.761}$.

The test statistic is $t =1.6882$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\textit{right-tailed}$ test, so p-value is the area to the $\textit{right}$ of the test statistic ($t=1.6882$). That is p-value = $P(t\geq 1.6882 ) = 0.0568$.

The p-value is $0.0568$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

$p$-value is $0.0568$. This $p$-value indicates that there is no significant evidence to support the alternative hypothesis. That is there is no significant evidence to support the claim that the new battery perform better than the old design.