Question

A local brewery distributes beer in bottles labeled 24 ounces. A government agency thinks that the brewery is cheating its customers. The agency selects 50 of these bottles, measures their contents, and obtains a sample mean of 23.6 ounces with a standard deviation of 0.70 ounce. Use a 0.01 significance level to test the agency's claim that the brewery is cheating its customers.

Answer 1

Answer with explanation:

Let $\mu$ the population mean.

The hypothesis for the given situation :-

$H_0:\mu=24\\\\H_a: \mu$, since the alternative hypothesis is left-tailed , so the test a left-tailed test.

The sample size : $n=50$, since n>30 so its a large sample . We use z-test.

Sample mean : $\overline{x}=23.6$

Standard deviation : $\sigma= 0.70$

Significance level : $\alpha=0.01$

The test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

i.e. $z=\dfrac{23.6-24}{\dfrac{0.70}{\sqrt{50}}}=-4.04$

The p-value : $P(z$

Since, the p-value is less than the significance level , so we reject the null hypothesis.

Hence, we have evidence to accept the agency's claim that the brewery is cheating its customers.