I hope you are all well.
I am struggling to understand a question I need doing as part of my work.
Specifically, I am asked:
A company has developed a new range of batteries. They test a sample of $50$ of these batteries and their mean lifetime to be $7.1$ hours. Their older range of batteries followed a normal distribution with mean $7$ hours and standard deviation $0.35$ hours. Test the claim made by the company that the new range of batteries are longer lasting at the $5%$ and $1%$ significance levels.
Is it correct to say the normal distribution for the old range of batteries can be written as $p~N(7, 0.35^{2})$ and a $Z$ correspondence equaling $\frac{x-7}{0.35}$?
Likewise, is it correct to say the new range of batteries follows a normal distribution $p~N(7.1, \sigma^{2})$ and a $Z$ correspondence equaling $\frac{50-7.1}{\sigma}$?
I am struggling to know how to incorporate the Significance Levels into the question as well.
Thank you for your help.
Your formula is correct for the $Z$-score for the old batteries. However, what you really want to do is a hypothesis test to see if the mean lifetime of new batteries $(\mu_2)$ is greater than the mean lifetime of old batteries.
For this test, the null hypothesis is that the new mean is not greater than the old mean, while the alternate hypothesis is that the new mean is greater than the old mean. Therefore,
$H_0:\mu_2 \le 7$
$H_a:\mu_2 \gt 7$
Generally, if the sample size is $\gt 30$ and the population standard deviation is known, we would use a $Z$-test. The $Z$-score is found using the following formula:
$$z={\bar X-\mu_0 \over \sigma \sqrt n}$$
where $\mu_0$ is the population mean, and $\bar X$ is the sample mean.
Once you have the Z-score, you would use the $Z$-table to calculate the probability that $Z \gt z$. If that probability is lower than the significance level ($.05$ or $.01$), then we will reject the null hypothesis. Otherwise, we cannot reject the null.