Answer:
The calculated value Z=1.2413 < 1.96 at 0.05 level of significance
Null hypothesis is accepted
An instructor in the English department at CBC believes rates at CBC may be equal to the state average
Step-by-step explanation:
Step(i):-
Given Population proportion
P=68%= 0.68
Sample proportion
[tex]p^{-} =\frac{x}{n} = \frac{179}{250} = 0.716[/tex]
Null hypothesis : H₀ : p = 0.68
Alternative Hypothesis :H₁ : p > 0.68
Step(ii):-
Test statistic
[tex]Z = \frac{p^{-}-P }{\sqrt{\frac{P Q}{n} } }[/tex]
[tex]Z = \frac{0.716-0.68 }{\sqrt{\frac{0.68 X 0.32}{250} } }[/tex]
Z = 1.2413
Level of significance
[tex]Z_{0.05} = 1.96[/tex]
The calculated value Z=1.2413 < 1.96 at 0.05 level of significance
Null hypothesis is accepted
An instructor in the English department at CBC believes rates at CBC may be equal to the state average