Question

A fast-food franchise is considering building a restaurant at a certain location. Based on financial analyses, a site is acceptable only if the number of pedestrians passing the location averages more than 100 per hour. The number of pedestrians observed for 40 hours during different times of the day was recorded. Assuming that the population standard deviation is known to be 22, can we conclude at the 5% significance level that the site is acceptable

Answer 1

Answer:

$\bar{x}=105.7217$

Step-by-step explanation:

Fast-food franchise is considering:-

Given,

mean $\mu=100$

standard deviation $\sigma=22$

$n=40$

Now, can be calculated that at the 5% of:-

test statistic$=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{40}}}$

                  $=\frac{\bar{x}-100}{\frac{22}{\sqrt{40}}}=\frac{\bar{x}}{\frac{22}{\sqrt{40}}}-\frac{100}{\frac{22}{\sqrt{40}}}$

                  $=\frac{\bar{x}}{3.4785}-\frac{100}{3.4785}$

                   $=\frac{\bar{x}}{3.4785}-28.748= f_{test}\quad \quad ...(i)$

Now we calculate $f_{critical}$ value is $f_{\alpha}$

$\alpha=0.05$

$f_{\alpha}=f_{0.05}=1.6449 \quad \quad ...(ii)$

From equation $(i)$ and $(ii)$  we get,

$\frac{\bar{x}}{3.4785}-28.748=1.6449$

$\Rightarrow \frac{\bar{x}}{3.4785}=1.6449+28.748$

$\Rightarrow \frac{\bar{x}}{3.4785}=30.3929$

$\Rightarrow \bar{x}=30.3929\times 3.4785$

$\Rightarrow \bar{x}=105.7217$

Therefore, when mean less than $105.7217$, we do not reject the null hypothesis at $\alpha =0.05$

Hence, it is acceptable.