Answer:
Explanation:
From the given information:
The null and the alternative hypothesis can be well written as:
![H_o:P=0.6](https://img.qammunity.org/2021/formulas/mathematics/high-school/3t4tvc8l7mjf2d7ytjzhwehsour0l6srsg.png)
![H_1:P>0.6](https://img.qammunity.org/2021/formulas/mathematics/high-school/9egyrgw5tzb5j78devc2igocqgc31ezb8i.png)
Given that:
n = 200
x = 135
Alpha ∝ = 0.05 level of significance
Then;
⇒
![n * p* (1-P)](https://img.qammunity.org/2021/formulas/mathematics/high-school/re2q36zoak7z2izwqtezxxchuc6ornc9f8.png)
= 200 × 0.6 × (1 -0.6)
= 200 × 0.6 × 0.4
= 48 ≥ 10
The sample proportion
![\hat P = (x)/(n)](https://img.qammunity.org/2021/formulas/mathematics/high-school/abc9y5509xj442pq414s633nup33cwk6p6.png)
![= (135)/(200)](https://img.qammunity.org/2021/formulas/mathematics/high-school/wtfn2ge6wt02ivkb86hr5da9rq7718j00d.png)
= 0.675
The test statistics
![Z = \frac{\hat P - P}{\sqrt{ (P(1-P))/(n) }}](https://img.qammunity.org/2021/formulas/mathematics/high-school/wii4tf5qd21jtnr6fx7j5gk8469izqa9mp.png)
![Z = \frac{0.675 - 0.6}{\sqrt{ (0.6 * 0.4)/(200) }}](https://img.qammunity.org/2021/formulas/mathematics/high-school/3v717eigq0uzluwspwla5il4aids0qi6f6.png)
![Z = \frac{0.075}{\sqrt{ (0.24)/(200) }}](https://img.qammunity.org/2021/formulas/mathematics/high-school/nuwu32vtn8ydhd6zc5x4eqzqwnw6ehfy2c.png)
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