Sunday, 1 July 2012

STA301 assign no. 4 idea solution




         Question # 1:
          a. solution:
                  n=160    n=220
12
                     x1=15.80  x2=10.25
22
                     S=64      S=47

12
                                 1-a=0.96

22
SS

12
                             (x1-x2)±Z+

a

nn

212
                Z=0.48                

a
2
a

      Now see value of  in the Z table
2

         0.48 lie in the row 2.00 and lie in the column 0.06 
Z=2.00+0.06
a

2
                           

Z=2.06

a
2
     Now put values in formula,






                                          
6447
(15.80-10.25)±2.06+

160220
135
                           (5.55)±2.06

220
(5.55)±5.1088
5.55+5.10885.55-5.1088
                  ,,,,,,,,,,         

10.65880.3912
b. Solution:

                        n=200  p=0.30 

              q=1-p=1-0.30=0.70

                  1-a=0.95
pq


                       p±Z

a
n

2
       Z

a
2
a0.95

                   ==0.475
22

a
           Now see the value of  in the Z table.

2
            0.47 lies in 1.9 rows and 0.06 columns
Z=1.9+0.06
a

2
                       

Z=1.96

a
2
          Now put values in formula,

(0.30)(0.70)
0.30±1.96

200
                         0.30±1.96(0.0324)
0.30±0.063504
0.30+0.0635040.30-0.063504
         ,,,,,,,,,,        

0.3640.236

      

            

                   







                        

Question # 2:
Solution Of part a:
               e=0.30s
1-a=0.90
                   

a=0.1
2
æZ

aö
1-

ç2÷

                       n=

ç÷
e

ç÷

èø
                 Z=Z
a0.95

1-
2

           see the value of in Z table
                    Z=1.6449=1.645a
0.95
2
æ1.645sö

n=

ç÷
                            

è0.30sø

n=30.07

Solution of part b:
                     m=30 ,  s=5  ,  n=50
s
                        s=

x
n







5
s=

x
                            50

s=0.707
x