Question

For the population of Marionasty agents, let X = {Gun Type} and Y = {Knife Type}.“You found that bullet,” marveled Dr. Witless, “hidden in the victim’s auditory canal?” “Alimentary, my dear Witless,” the mirthful Hopeless did grin, “alimentary!”

                                      Knives

	Dagger	Stiletto	Switchblade	totals
Guns:
None	38	47	15	100
Revolver	17	17	16	50
automatic	25	16	9	50
totals	80	80	40	200

a. Do these data provide strong evidence that the proportion of agents who carry

switchblades is less than one quarter?

i. Test the hypothesis that less than one quarter of agents carry switchblades;

ii. Find a comparable confidence interval for the proportion of agents who carry

switchblades;

iii. Compare your interval and test results.

b. Use these data to test the hypothesis that there is an association between

the types of knives and guns that Marionasty agents carry.

Answer 1

a)

i)

Ho :   p =    0.25                  
H1 :   p <   0.25       (Left tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   40                  
Sample Size,   n =    200                  
                          
Sample Proportion ,    p̂ = x/n =    0.2000                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.031                  
Z Test Statistic = ( p̂-p)/SE = (   0.2000   -   0.25   ) /   0.0306   =   -1.6330
                          

                          
p-Value   =   0.051   [excel function =NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       
There is not enough evidence that the proportion of agents who carry switchblades is less than one quarter

ii)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   40          
Sample Size,   n =    200          
                  
Sample Proportion ,    p̂ = x/n =    0.2000          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.028284          
margin of error , E = Z*SE =    1.960   *   0.02828   =   0.0554
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.20000   -   0.05544   =   0.14456
Interval Upper Limit = p̂ + E =   0.20000   +   0.05544   =   0.25544
                  
95%   confidence interval is (   0.145   < p <    0.255   )

iii)

since, CI contains 0.25, so, null hypothesis is not rejected

both results agree with each other

b)

Chi-Square Test of independence
	Observed Frequencies
	0
0	Dagger	Stiletto	Switchblade				Total
None	38	47	15				100
Revolver	17	17	16				50
automatic	25	16	9				50
Total	80	80	40				200
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	Dagger	Stiletto	Switchblade				Total
None	80*100/200=40	80*100/200=40	40*100/200=20				100
Revolver	80*50/200=20	80*50/200=20	40*50/200=10				50
automatic	80*50/200=20	80*50/200=20	40*50/200=10				50
Total	80	80	40				200
	(fo-fe)^2/fe
None	0.1000	1.2250	1.250
Revolver	0.4500	0.4500	3.600
automatic	1.2500	0.8000	0.1000

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   9.225  
      
Level of Significance =   0.05  
Number of Rows =   3  
Number of Columns =   3  
Degrees of Freedom=(#row - 1)(#column -1) = (3- 1 ) * ( 3- 1 ) =   4  
      
p-Value =   0.0557151   [Excel function: =CHISQ.DIST.RT(χ²,df) ]
Decision:    p value > α , do not reject Ho  

there is no enough evidence to conclude that

there is an association between

the types of knives and guns that Marionasty agents carry