Question

According to a polling​ organization, 24% of adults in a large region consider themselves to be liberal. A survey asked 200 respondents to disclose their political​ philosophy: Conservative,​Liberal, And Moderate. Treat the results of the survey as a random sample of adults in this region. Do the survey results suggest the proportion is higher than that reported by the polling​organization? Use an alpha equals0.05 level of significance.75- Liberal65- Moderate61- Conservative

Answer 1

`z=\frac{0.375 -0.24}{\sqrt{(0.24(1-0.24))/(200)}}=4.47`

Now we can calculate the p value based on the alternative hypothesis with this probability:

`p_v =P(z>4.47)=0.00000391`

The p value is very low compared to the significance level of
`\alpha=0.05` then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

Explanation:

Information given

n=200 represent the random sample taken

X=75 represent the number of people Liberal

`\hat p=(75)/(200)=0.375` estimated proportion of people liberal

`p_o=0.24` is the value that we want to test

z would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to verify if the true proportion of adults liberal is higher than 0.24:

Null hypothesis:
`p \leq 0.24`

Alternative hypothesis:
`p > 0.24`

The statistic is given by:

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

Replacing the info given we got:

`z=\frac{0.375 -0.24}{\sqrt{(0.24(1-0.24))/(200)}}=4.47`

Now we can calculate the p value based on the alternative hypothesis with this probability:

`p_v =P(z>4.47)=0.00000391`

The p value is very low compared to the significance level of
`\alpha=0.05` then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24