Respuesta :
Answer:
[tex]\bar{x}=105.7217[/tex]
Step-by-step explanation:
Fast-food franchise is considering:-
Given,
mean [tex]\mu=100[/tex]
standard deviation [tex]\sigma=22[/tex]
[tex]n=40[/tex]
Now, can be calculated that at the 5% of:-
test statistic[tex]=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{40}}}[/tex]
[tex]=\frac{\bar{x}-100}{\frac{22}{\sqrt{40}}}=\frac{\bar{x}}{\frac{22}{\sqrt{40}}}-\frac{100}{\frac{22}{\sqrt{40}}}[/tex]
[tex]=\frac{\bar{x}}{3.4785}-\frac{100}{3.4785}[/tex]
[tex]=\frac{\bar{x}}{3.4785}-28.748= f_{test}\quad \quad ...(i)[/tex]
Now we calculate [tex]f_{critical}[/tex] value is [tex]f_{\alpha}[/tex]
[tex]\alpha=0.05[/tex]
[tex]f_{\alpha}=f_{0.05}=1.6449 \quad \quad ...(ii)[/tex]
From equation [tex](i)[/tex] and [tex](ii)[/tex] we get,
[tex]\frac{\bar{x}}{3.4785}-28.748=1.6449[/tex]
[tex]\Rightarrow \frac{\bar{x}}{3.4785}=1.6449+28.748[/tex]
[tex]\Rightarrow \frac{\bar{x}}{3.4785}=30.3929[/tex]
[tex]\Rightarrow \bar{x}=30.3929\times 3.4785[/tex]
[tex]\Rightarrow \bar{x}=105.7217[/tex]
Therefore, when mean less than [tex]105.7217[/tex], we do not reject the null hypothesis at [tex]\alpha =0.05[/tex]
Hence, it is acceptable.
