Independent random samples of released prisoners in the fraud and firearms offense categories yielded the given information on time served in months. At the 1% significance level, do the data provide sufficient evidence to conclude that the meantime served for fraud is less than that for firearms offenses? Assume that populations standard deviations are the same for both groups. (i. e., we assume σ1 = σ2 .)

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

7.4        9.8                            10.4            22.2            

∑ 93.6                                             169                    

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

54.76        96.04                            108.16           492.84

∑1003.79                                                  3079.66          

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.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

exa:

)

3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.