A consumer association is testing the lifetime of a particular type of battery that is claimed to have a lifetime of $150$ hours. A random sample of $70$ batteries of this type is tested and the lifetime, $x$ hours, of each battery is measured. The results are summarized by
$\sum{x}=10317$ , $\sum{{{x}^{2}}}=1540231$.
The population mean lifetime is denoted by $\mu $ hours. The null hypothesis $\mu =150$ is to be tested against the alternative hypothesis $\mu <150$. Find the $p$-value of the test and state the meaning of this $p$-value in the context of the question.
A second random sample of $50$ batteries of this type is tested and the lifetime, $y$ hours, of each battery is measured, with results summarized by
$\sum{y}=7331$ , $\sum{{{y}^{2}}}=1100565$.
Combining the two samples into a single sample, carry out a test, at the $10\%$ significance level, of the same null and alternative hypotheses.

A consumer association is testing the lifetime of a particular type of battery that is claimed to have a lifetime of $150$ hours. A random sample of $70$ batteries of this type is tested and the lifetime, $x$ hours, of each battery is measured. The results are summarized by

$\sum{x}=10317$ , $\sum{{{x}^{2}}}=1540231$.

The population mean lifetime is denoted by $\mu $ hours. The null hypothesis $\mu =150$ is to be tested against the alternative hypothesis $\mu <150$. Find the $p$-value of the test and state the meaning of this $p$-value in the context of the question.

A second random sample of $50$ batteries of this type is tested and the lifetime, $y$ hours, of each battery is measured, with results summarized by

$\sum{y}=7331$ , $\sum{{{y}^{2}}}=1100565$.

Combining the two samples into a single sample, carry out a test, at the $10\%$ significance level, of the same null and alternative hypotheses.

Default solution

Let $X$ be the lifetime of a particular type of battery in hours.
$H_0: \mu = 150$
$H_1: \mu < 150$
$n = 70$, $\bar{x} = \frac{\sum x}{70} = \frac{10317}{70} = 147.3857$
$s^2 = \frac{1}{n-1}\left[\sum x^2 - \frac{(\sum x)^2}{n}\right] = \frac{1}{69}\left[1540231 - \frac{10317^2}{70}\right] = 284.8200$
Under $H_0$, test statistic $Z = \frac{\bar{X} - 150}{\sqrt{\frac{s^2}{70}}} \sim \text{N}(0, 1)$ approximately by Central Limit Theorem since $n = 70$ is large.
From GC, p-value $= 0.09740$
$= 0.0975$ (3 s.f.)
p-value is the smallest significance level at which the claim that the lifetime of a particular type of battery is 150 hours will be rejected.

Combining 2 samples:
$n = 70 + 50 = 120$
$\bar{x}_{\text{new}} = \frac{\sum x + \sum y}{70 + 50} = \frac{10317 + 7331}{120} = 147.0666$
$s^2_{\text{new}} = \frac{1}{n-1}\left[\sum x^2_{\text{new}} - \frac{(\sum x_{\text{new}})^2}{n}\right] = \frac{1}{119}\left[(1540231 + 1100545) - \frac{(10317 + 7331)^2}{120}\right] = 381.2056$
$H_0: \mu = 150$
$H_1: \mu < 150$
$n = 120$, $\bar{x}_{\text{new}} = 147.0666$, $s^2_{\text{new}} = 381.2056$, $\alpha = 10\%$
Under $H_0$, test statistic $Z = \frac{\bar{X}_{\text{new}} - 150}{\sqrt{\frac{s^2_{\text{new}}}{120}}} \sim \text{N}(0, 1)$ approximately by CLT since $n = 120$ is large.
From GC, p-value $= 0.04990 < 0.10$
Hence, we reject $H_0$, at 10% level of significance, we have sufficient evidence to conclude that the lifetime of the battery is less than 150 hours.

2008 H1 Q10

2008 H1 Q10

Video Solution:

Default solution