Answer:
Step-by-step explanation:
From the given information:
The null and the alternative hypothesis can be well written as:
[tex]H_o:P=0.6[/tex]
[tex]H_1:P>0.6[/tex]
Given that:
n = 200
x = 135
Alpha ∝ = 0.05 level of significance
Then;
⇒ [tex]n \times p\times (1-P)[/tex]
= 200 × 0.6 × (1 -0.6)
= 200 × 0.6 × 0.4
= 48 ≥ 10
The sample proportion [tex]\hat P = \dfrac{x}{n}[/tex]
[tex]= \dfrac{135}{200}[/tex]
= 0.675
The test statistics [tex]Z = \dfrac{\hat P - P}{\sqrt{ \dfrac{P(1-P)}{n} }}[/tex]
[tex]Z = \dfrac{0.675 - 0.6}{\sqrt{ \dfrac{0.6 \times 0.4}{200} }}[/tex]
[tex]Z = \dfrac{0.075}{\sqrt{ \dfrac{0.24}{200} }}[/tex]
Z = 2.165
The P-value = P(Z > 2.165)
= 1 - P(Z < 2.165)
From the z tables
= 1 - 0.9848
= 0.0152
Reject the null hypothesis since P-Value is lesser than alpha. ( i.e. 0.0152 < 0.05).
Thus, there is enough evidence to conclude that the value of the population proportion is greater than 0.6
