Answer :
Answer:
Yes
Step-by-step explanation:
Let p1 represent proportion of women and p2 proportion of men. The null and alternative hypothesis will be as follows
Null hypothesis
[tex]H_o[/tex]=p2-p1 ≤0
Alternative hypothesis
[tex]H_a[/tex]=p2-p1>0
Sample proportion of women, p1=74/200=0.37
Sample proportion of men is p2=104/200=0.52
Level of significance is 0.01
Pooled proportion=[tex]\frac {104+74}{200+200}=0.445[/tex]
Test statistic
[tex]z=\frac {(0.52-0.37)-0}{0.445(1-0.445)*\sqrt{(1/200+1/200)}}=6.9124[/tex]
p-value=P(Z≥z)=P(Z≥6.9124)=P(Z≤-6.9124)=0
Since the value of p is less than 0.01, we reject null hypothesis. There’s sufficient evidence that a greater proportion of men is expecting to get a raise