Answer:
z(c) = - 1,64
We reject the null hypothesis
Step-by-step explanation:
We need to solve a proportion test ( one tail-test ) left test
Normal distribution
p₀ = 63 %
proportion size p = 51 %
sample size n = 114
At 5% level of significance α = 0,05, and with this value we find in z- table z score of z(c) = 1,64 ( critical value )
Test of proportion:
H₀ Null Hypothesis p = p₀
Hₐ Alternate Hypothesis p < p₀
We now compute z(s) as:
z(s) = ( p - p₀ ) / √ p₀q₀/n
z(s) =( 0,51 - 0,63) / √0,63*0,37/114
z(s) = - 0,12 / 0,045
z(s) = - 2,66
We compare z(s) and z(c)
z(s) < z(c) - 2,66 < -1,64
Therefore as z(s) < z(c) z(s) is in the rejection zone we reject the null hypothesis
