Respuesta :
Answer:
Yes. We have evidence to support the claim that there is a difference in the use of the four entrances.
Step-by-step explanation:
The question is incomplete:
Entrance Frequency
Main Street 81
Broad Street 129
Cherry Street 72
Walnut Street 119
Total: 401
The building maintenance supervisor wants to know if the entrances are equally utilized.
This problem can be solved using the Chi-square goodess of fit test.
The expected value for each door is
[tex]E=401/4=100.25[/tex]
The degrees of freedom are equal to the number of categories (4 doors) minus one:
[tex]df=n-1=4-1=3[/tex]
Then, the value of the chi-square statistic can be calculated as:
[tex]\chi^2=\sum \dfrac{(O_i-E)^2}{E}\\\\\\\chi^2=\dfrac{(81-100.25)^2}{100.25}+\dfrac{(129-100.25)^2}{100.25}+\dfrac{(72-100.25)^2}{100.25}+\dfrac{(119-100.25)^2}{100.25}\\\\\\\chi^2=\dfrac{370.5625+826.5625+798.0625+351.5625}{100.25}=\dfrac{2346.75}{100.25}=23.41[/tex]
The P-value for this test statistic χ^2=23.41 and df=3 is:
[tex]P-value=P(\chi^2_3>23.41)=0.00003[/tex]
This P-value is much smaller than the significance level (0.01), so the effect is significant.
We have evidence to support the claim that there is a difference in the use of the four entrances.
