Question

Large Sample Proportion Problem. Surveys were conducted in multiple countries and respondents were asked if they felt political news was reported fairly. The data for the United States is that out of 1,000 sampled, 470 indicated yes, they felt political news was reported fairly. Suppose we want to determine if the proportion for the U.S. is below .50 for an alpha level of .05. What is conclusion of my test

Answer 1

`z=\frac{0.47-0.5}{\sqrt{(0.5(1-0.5))/(1000)}}=-1.897`

We have aleft tailed test the p value would be:

`p_v =P(z<-1.897)=0.0289`

The p value obtained is less compared to the significance level so then we have enough evidence to conclude that the true proportion is significantly lower than 0.5.

Explanation:

Information given

n=1000 represent the random sample selected

X=470 represent the number of people who felt political news was reported fairly

`\hat p=(470)/(1000)=0.470` estimated proportion of people who felt political news was reported fairly

`p_o=0.5` is the value that we want to test

`\alpha=0.05` represent the significance level

z would represent the statistic

`p_v` represent the p value

System of hypothesis

For this case we want to test if proportion for the U.S. is below .50 so then the system of hypothesis for this case are:

Null hypothesis:
`p \geq 0.5`

Alternative hypothesis:
`p < 0.5`

The statistic is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

Replacing the info provided we got:

`z=\frac{0.47-0.5}{\sqrt{(0.5(1-0.5))/(1000)}}=-1.897`

We have aleft tailed test the p value would be:

`p_v =P(z<-1.897)=0.0289`

The p value obtained is less compared to the significance level so then we have enough evidence to conclude that the true proportion is significantly lower than 0.5.