Demographics and Population Statistics. In Illinois, a typical DUI (Driving Unde

Question

Demographics and Population Statistics. In Illinois, a typical DUI (Driving Under the Influence) offender is a 34-year-old male, arrested between 11 p.m.and 4 a.m.on a weekend, and has a BAC (blood alcohol content) of 0.16. Sixteen percent of all drivers arrested in Illinois are first-time offenders. Suppose 41 people arrested for DUI in Illinois are selected at random.

(a) What is the probability ( ±0.0001) that at least 14 are first-time offenders?

P(X14) =

(b) What is the probability ( ±0.0001) that between 7 and 10 (inclusive) are first-time offenders?

P(7X10) =

(c) Suppose 4 of those arrested are first-time offenders. Is there any evidence to suggest that the proportion of first-time offenders arrested for DUI in Illinois has changed??

P(X4) ( ±0.0001) =

Conclusion:

There is evidence to suggest the claim is false

There is no evidence to suggest the claim is false

There is evidence to suggest the claim is false

There is no evidence to suggest the claim is false

Explanation / Answer

Given 16% are the first time offenders ; P=0.16

total people arrested 'n' =41; Let 'X' be the number of first time offenders follows binomomial distribution with P=0.16, n=20, Q=1-0.16=0.84

The probability ( ±0.0001) that at least 14 are first-time offenders is =P(X>=14) = 1-P(X<=13)=1-0.9966=0.0034

The probability ( ±0.0001) that between 7 and 10 (inclusive) are first-time offenders P(7<=x<=10)=P(X<=10)-P(X<=6)

=0.9467-0.5049=0.4418

c) If 4 are selected are first time offenders then p =4/41 = 0.0976

Null hypothesis : There is no change in proportion; Alternative hypothesis: There is change in proportion.

To test the claim, the test statistic z = (p-P)/SE; Where SE = sqrt(P*Q/n) = sqrt(0.16*0.84/41)=0.0573

so z =(0.0976-0.16)/0.0573 = -1.0906

P-value is 0.2755; We accept the null hypothesis.

There is evidence to suggest the claim is false

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