Answer to Question #278434 in Statistics and Probability for brad

Question

The packaging on an electric light bulb states that the average length of life of bulbs is 1000 hours. A consumer association thinks that this is an overestimate and tests a random sample of 64 bulbs, recording the life x

 hours, of each bulb. The results are summarised as follows:

∑x=63910.4,∑x2=63824061

 

A significance test, at the 10% level, is done to test where the statement on the packaging in overestimating the length of life of this type of light bulb. What is the null and alternative hypothesis for this test?

Answer 1

Given $H_o$ = 650 hour.

$n=50\ bulbs \quad, \sum x=6392 \quad \sum x^{2}=62424060.$

$\begin{aligned} \\ \text { Sample mean } &=\left(\frac{\sum x}{n}\right) \\ &=\left(\frac{6392}{50}\right) \text { hours. } \\ &=127.84 \text { hours. } \end{aligned}$

Standard deviation $=\frac{\sigma}{\sqrt{n}}=\frac{1}{n}\sqrt{\left.\frac{1}{n}\left(\sum x_i-n\right)^{2}\right) }$

$=\frac{1}{\sqrt{7}} \sqrt{\frac{\sum x^{2}}{n}-x^{-2}} =\sqrt{\frac{(62424060)}{50 \times 50}-\frac{(6312)^{2}}{\left(50^{3}\right)}}$

$=156.98$

(c) $\operatorname{For} H_{0} \leq 950 ,H_{1}>950$

$t^*=\frac{950-127.84}{156.98}=5.2365$

$for\ \alpha=0.1\quad \mathrm{and} \quad d f=(n-1)=50-1=49$

t score $\approx 1.676$

Since $t^*>1.67 \rightarrow$ then we fail to reject $H_o$ and can conclude population mean or claim

by company is false.