Hi, I just need help finding only Ho and Ha of this problem.. I already did the
ID: 3229413 • Letter: H
Question
Hi, I just need help finding only Ho and Ha of this problem.. I already did the calculations and everything... feel free to do the rest of the problem so I can check my work but this is just me asking for help on Ho and Ha!
Problem: Suppose a safety officer proposes that bicycle accidents at a certain location over a given year is expected to occur with the following distributions:
Midnight - 3AM = .05
3AM - 6AM = .05
6AM - 9AM = .10
9AM - Noon = .10
Noon - 3PM = .15
3PM - 6PM = .15
6PM - 9PM = .20
9PM - Midnight = .20
Let's also suppose that actual observed data of 715 bicycle accidents were as follows:
from Midnight - 3AM there were 38 accidents
from 3AM - 6AM there were 29 accidents
from 6AM - 9AM there were 66 accidents
from 9AM - Noon there were 77 accidents
from Noon - 3PM there were 99 accidents
from 3PM - 6PM there were 127 accidents
from 6PM - 9PM there were 166 accidents
from 9PM - Midnight there were 113 accidents
Use Chi- Square Analysis to test that the observed values equal the expected distribution @ alpha = .05 level
Explanation / Answer
Ho: obsereved values in time interval follows the given distribution as proposed by safety officers,
Ha: at least one of the time interval does not follow the given distribution as proposed by safety officers,
for above chi stat 16.5268 and 7 degree of freedom ; p value =0.0207 which is less then 0.05 level . Hence we reject null hypothesis
observed Expected Chi square Probability O E=total*p =(O-E)^2/E midnight-3AM 0.050 38.000 35.75 0.14 3AM-6AM 0.050 29.000 35.75 1.27 6AM-9AM 0.100 66.000 71.50 0.42 9AM-Noon 0.100 77.000 71.50 0.42 Noon -3PM 0.150 99.000 107.25 0.63 3PM-6PM 0.150 127.000 107.25 3.64 6PM-9PM 0.200 166.000 143.00 3.70 9PM-Midnight 0.200 113.000 143.00 6.29 1 715 715 16.5268Related Questions
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