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 0.8 ounces.

Use a 0.01 significance level (α = 0.01) to test the agency's claim that the brewery is cheating its customers (that the average volume of their beers is less than 24 ounces).H0:Ha:test statistic, z =critical value =p-value =Conclusion:

Answer 1

Answer:

Null hypothesis:  $\mu \geq 24$    

Alternative hypothesis :$\mu < 24$  

$z=\frac{23.6-24}{\frac{0.8}{\sqrt{50}}}=-3.54$    

On this case since we have a right tailed test we need to look into the normal standard distribution and find a value that accumulates 0.01 of the area on the right of th distribution or 0.99 of the area on the left and for this case we see that $z_{critical}=2.33$  

If we compare the p value and a significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the content of the bottles is significantly lower than 24 at 1% of significance.    

Step-by-step explanation:

Data given and notation    

$\bar X=23.6$ represent the average score for the sample    

$s=0.8$ represent the sample standard deviation    

$n=50$ sample size    

$\mu_o =24$ represent the value that we want to test    

$\alpha=0.01$ represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a one lower tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  $\mu \geq 24$    

Alternative hypothesis :$\mu < 24$    

Compute the test statistic  

The statistic for this case is given by:    

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$z=\frac{23.6-24}{\frac{0.8}{\sqrt{50}}}=-3.54$  

Critical value

On this case since we have a right tailed test we need to look into the normal standard distribution and find a value that accumulates 0.01 of the area on the left of the distribution or 0.99 of the area on the right and for this case we see that $z_{critical}=-2.33$  

Give the appropriate conclusion for the test  

Since is a one side left tailed test the p value would be:    

$p_v =P(Z$    

Conclusion    

If we compare the p value and a significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the content of the bottles is significantly lower than 24 at 1% of significance.