According to a polling organization, 22% 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 pollingorganization? Use an alpha =0.05 level of significance. 75- Liberal
65- Moderate
61- Conservative

Step-by-step explanation:

Using the proportion test

Null hypothesis: p <= 0.22

Alternative hypothesis: p > 0.22

Using the formula

z score = p - P /√ (P(1-P)/n)

Where p is 74/200= 0.37, P = 0.22, n = 200.

0.37-0.22 / √(0.22(1-0.22)/200)

0.15 / √(0.22(0.78)/200)

0.15 / √(0.1716/200)

0.15/ √0.000858

0.15 / 0.02929

= 5.1212

To help arrive at a conclusion, we have to find the p value, using a p value calculator at the 0.05 level of significant, the p value is less than 0.00001... Thus we would reject the null as there is sufficient statistical evidence to prove that the proportion is higher than that reported by the polling organization.

i believe that the answer is

b. y-1=-2(x-4)

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step-by-step explanation: