Answer:
[tex]z=\frac{50.6-50.7}{\frac{2.2}{\sqrt{140}}}=-0.54[/tex] Â Â
Step-by-step explanation:
Data given and notation Â
[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 Â
[tex]\mu_o =50.7[/tex] represent the value that we want to test
z would represent the statistic (variable of interest) Â
State the null and alternative hypotheses. Â
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: Â
Null hypothesis:[tex]\mu = 50.7[/tex] Â
Alternative hypothesis:[tex]\mu \neq 50.7[/tex] Â
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: Â
[tex]z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] Â (1) Â
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". Â
Calculate the statistic
We can replace in formula (1) the info given like this: Â
[tex]z=\frac{50.6-50.7}{\frac{2.2}{\sqrt{140}}}=-0.54[/tex] Â Â