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ycow [4]
3 years ago
14

A fast-food franchise is considering building a restaurant at a certain location. Based on financial analyses, a site is accepta

ble 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
Mathematics
1 answer:
spayn [35]3 years ago
7 0

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.

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