Answer:
Reject Test.
Step-by-step explanation:
Our values are:
[tex]n=200[/tex]
[tex]\sigma = 8[/tex]
[tex]\mu = 74.5[/tex]
[tex]\bar{x} = 75.9[/tex]
[tex]\alpha=0.05 \rightarrow z_{c2}=1.96, z_{c1}=1.64[/tex]
We have here a Null Hypothesis of [tex]H_0 : \mu \neq 74.5[/tex]
We calculate Z through Test Statistic,
[tex]Z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}[/tex]
[tex]Z=\frac{75.9-74.5}{8\sqrt{200}}[/tex]
[tex]Z=2.4748[/tex]
We can make now the comparation between our value with the critical value of Z_c = 1.96
a) [tex]Z_c[/tex] = 1.64 at one tailed test,
Therefore [tex]|Z|>|Z_c|[/tex][tex]\Rightarrow[/tex] we reject H_0
b) [tex]Z_c[/tex]= 1.96 at two tailed test
Therefore [tex]|Z|>|Z_c| \Rightarrow[/tex] we reject H_0
In this way the population mean [tex]\mu \neq 74.5[/tex]
We can comprobate that both these test reject the null hypothesis
