Homework Answers
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Solution:
1)Let psun,pmon,ptue,pwed,pthu,pfri and psat be the proportion of traffic accident on Monday,Tuesday,Wednesday,Thursday,Friday and Saturday respectively.
Null hypothesis:Ho:psun=pmon=ptue=pwed=pthu=pfri =psat=1/7
Alterative hypothesis:Ha:Not all the proportions are equal.
Let us find chi square test statistics.
Formula is X2=
where
fo is the observed frequency and fe is the
expected frequency.
| Week | Observed frequency(fo) | Expected frequency(fe) | (fo-fe) | (fo-fe)2 | (fo-fe)2/fe |
| Sunday | 65 | 420/7=60 | (65-60)=5 | 52=25 | 25/60=0.417 |
| Monday | 52 | 420/7=60 | (52-60)=-8 | (-8)2=64 | 64/60=1.067 |
| Tuesday | 54 | 420/7=60 | (54-60)=-6 | (-6)2=36 | 36/60=0.600 |
| Wednesday | 45 | 420/7=60 | (45-60)=-15 | (-15)2=225 | 225/60=3.750 |
| Thursday | 52 | 420/7=60 | (52-60)=-8 | (-8)2=64 | 64/60=1.067 |
| Saturday | 72 | 420/7=60 | (72-60)=12 | (12)2=144 | 144/60=2.400 |
| 420 | 420 | 0 |
=9.300 |
From the above table, the value of X2=9.3
Degree of freedom,df=n-1=7-1=6
For 
=0.05 and df=6, we have P-value is 0.157.
Since p-value=0.157 is greater than
=0.05, we fail to reject the null hypothesis.
Do not reject Ho.
b)Percentage of traffic accidents occurring on each day of the week.
| Week | Percentage |
| Sunday | (65/420)*100=15.48% |
| Monday | (52/420)*100=12.38% |
| Tuesday | (54/420)*100=12.86% |
| Wednesday | (45/420)*100=10.71% |
| Thursday | (52/420)*100=12.38% |
| Saturday | (72/420)*100=17.14% |
Saturday has the highest percentage of traffic accidents.
Yes,this seems to be reasonable.
Saturday has the highest percentage of traffic accidents, followed by Sunday.This makes sense because these are times when people go out and may be consuming alcohol, having distractions such as more people in their car,and speeding.
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