Answer:
z = 2.1784 > 1.96,
Reject the null hypothesis
Step-by-step explanation:
For the males:
n1 = 162, x1 = 63
P1 = x 1/ n1 = 0.3889
For the Females:
n2 = 333,
x2 = 97
P 2 = x2/n2
= 0.2913
P 1= P2 Null hypothesis
P 1 is not equal to P 2 alternative hypothesis
Pooled proportion:
P= (x1 + x2) /( n1+ n2)
= (63 + 97) / (162 + 333)= 0.3232
Test statistics :
Z= (p1 - p2) /√p(1-p)× (1/n1 + 1/n2)
0.3889- 0.2913 / √0.3232 × 0.6768 × (1/162 +1/333)
=2.1784
c) Critical value :
Two tailed critical value, z critical = Norm.S .INV (0.05/2) = 1.960
Reject H o if z < -1.96 or if z > 1.96
d) Decision:
z = 2.1784 > 1.96,
Reject the null hypothesis
X + 3/4 > 11/4
X > 8/4
X > 2
So make a rectangle that extends from 2 and above
In inequalities normal athematic are used the only difference is that you flip the operator when dividing over -1 or taking inverse
