Question

The mean annual consumption of beer per person in the US is 22.0 gallons. A random sample of 300 Washington D.C. residents yielded a mean annual beer consumption of 27.8 gallons per person. Calculate the Z-value at the 10% significance level. Assume that the standard deviation of annual beer consumption for Washington D.C. residents is 55 gallons. Round your answer to 2 decimal places.

Answer 1

Answer:

$z=\frac{27.8-22}{\frac{55}{\sqrt{300}}}=1.83$    

Step-by-step explanation:

Data given and notation  

$\bar X=27.8$ represent the sample mean

$\sigma=55$ represent the population standard deviation

$n=300$ sample size  

$\mu_o =22$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.  

t 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 conduct a hypothesis in order to check if the true mean is equal to 22, the system of hypothesis would be:  

Null hypothesis:$\mu = 22$  

Alternative hypothesis:$\mu \neq 22$  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic 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:  

$z=\frac{27.8-22}{\frac{55}{\sqrt{300}}}=1.83$