Answer :
Answer:
[tex]z=\frac{0.47-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-1.897[/tex]
We have aleft tailed test the p value would be:
[tex]p_v =P(z<-1.897)=0.0289[/tex]
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.
Step-by-step explanation:
Information given
n=1000 represent the random sample selected
X=470 represent the number of people who felt political news was reported fairly
[tex]\hat p=\frac{470}{1000}=0.470[/tex] estimated proportion of people who felt political news was reported fairly
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] 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:[tex]p \geq 0.5[/tex]
Alternative hypothesis:[tex]p < 0.5[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info provided we got:
[tex]z=\frac{0.47-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-1.897[/tex]
We have aleft tailed test the p value would be:
[tex]p_v =P(z<-1.897)=0.0289[/tex]
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.