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