Z-test of one-sided hypothesis, given sample mean.

Question

Q. A manufacturer of flashlight batteries claims that its batteries will last an average of $\mu=34$ hours of continuous use. Of course, there is some variability in life expectancy with $\sigma=3$ hours. During consumer testing, a sample of $30$ batteries lasted an average of only $32.5$ hours. How likely is it to obtain a sample that performs this badly if the manufacturer's claim is true?

How to see this question asking for the probability of the normal score, $X \leq 32.5$. That is, we have to find $P(Z \leq \frac{32.5-34}{3 / \sqrt{30}})$, where $Z$ is the standard score. How to say we have considered the situation that the manufacturer's claim is true?

Answer 1

Thank you for showing what you have tried. I guess you are to assume batteries have normally distributed life times with standard deviation $\sigma = 3.$

You want to test $H_0: \mu = 34$ against the alternative $H_a: \mu < 34.$ If the null hypothesis is true, then individual batteries have lifetimes $X_i \sim\mathsf{Norm}(\mu = 34, \sigma=3).$ Consequently, the average lifetime of $n=30$ batteries is distributed $\bar X \sim\mathsf{Norm}(\mu = 34, \sigma = 3/\sqrt{30}= 0.5477).$

Finally, you seek the P-value of the one-sided test of $H_0$ against $H_a,$ which is $P(\bar X \le 32.5) = 0.003 < 0.01 = 1\%.$ Thus you would reject $H_0$ at the 1% level of significance in favor of $H_a.$

This probability can be found directly in R as follows:

pnorm(32.5, 34, 0.5477)  
[1] 0.003083892

Using printed tables of the standard normal CDF, you would get about the same answer (perhaps with a little rounding error from using the table). In R:

pnorm((32.5-34)/(3/sqrt(30)))
[1] 0.00308495