Answer:
Since the computed value of t= -4.654 does not fall in the critical region we therefore accept the null hypothesis . We conclude that there is sufficient evidence to indicate a difference in the means time served for fraud is less than that for firearms offenses.
Step-by-step explanation:
Fraud Firearm
15.2 9.2 20.1 15.7
11.2 15.8 20.4 9.8
7.2 5.2 13.1 13.5
7.7 4.9 20.7 23.1
<u>7.4 9.8 10.4 22.2 </u>
<u>∑ 93.6 169 </u>
<u />
Fraud (X1i)² Firearm(X2j)²
231.04 84.64 404.01 246.49
125.44 249.64 416.16 96.04
51.84 27.04 171.61 182.25
59.29 24.01 428.49 533.61
<u>54.76 96.04 108.16 492.84</u>
<u>∑1003.79 3079.66 </u>
<u />
We formulate the null and alternative hypothesis as
H0: μ1 - μ2 ≤ 0 i.e. the fraud meantime is less than that for firearms offenses.
Ha: μ1- μ2 > 0 i.e. the fraud meantime is greater than that for firearms offenses.
We set the significance level at ∝= 0.01
The test statistic, if H0 is true , is
t= x1`1- x`2/ Sp√1/n1+ 1/n2
which has a student t- distribution with ν= n1+ n2 -2 = 18 degrees of freedom
The critical region consists of all t- values which are greater than or equal to t> t(0.01)(18)= 2.522
Computations :
X1`= ∑X1i/n1 = 93.6/10= 9.36
X2`=∑X2j/n2=169/10 = 16.9
∑( X1i- x`1)² = ∑X1i²- (∑X1i)²/n1
= 1003.79 - 876.096/10
= 100.379- 87.6096
=12.7694
∑( X2j- x`2)² = ∑X2j²- (∑X2j)²/n2
= 3079.66 - 28561/10
= 3079.66- 2856.1
=223.56
Sp²= 12.7694+223.56/18= 13.129
Sp = √13.129 =3.623
t= 9.36-16.9/ 3.623√0.1+0.1
t= -7.54/ 1.62
t= -4.654
Conclusion: Since the computed value of t= -4.654 does not fall in the critical region we therefore accept the null hypothesis . We conclude that there is sufficient evidence to indicate a difference in the means time served for fraud is less than that for firearms offenses.
The meantime served for fraud is less than that for firearms offenses.
