Respuesta :
Answer:
[tex]z=\frac{50.6-50.7}{\frac{2.2}{\sqrt{140}}}=-0.54[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=50.6[/tex] represent the sample mean
[tex]\sigma=2.2[/tex] represent the population standard deviation
[tex]n=140[/tex] sample size
[tex]\mu_o =50.7[/tex] represent the value that we want to test
z would represent the statistic (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is different from 50.7, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 50.7[/tex]
Alternative hypothesis:[tex]\mu \neq 50.7[/tex]
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:
[tex]z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:
[tex]z=\frac{50.6-50.7}{\frac{2.2}{\sqrt{140}}}=-0.54[/tex]
