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1 answer:
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Answer:
a.d= 0.0204
b.[0.4728;0.5072]
c.A.Yes, the confidence interval contains .50.
Step-by-step explanation:
Hello!
you have a sample of 2298 voters and the sample proportion of voters that agree with the president's performance is 49%, which means that the sample proportion of voters that oppose the president will be 51%.
a) The margin of error for a 95% CI to estimate the proportion of voters that approve the president.
The formula for the CI for the proportion is:
^ρ±*√(^ρ(1-^ρ))/n
The margin of error "d" is half the amplitude of the interval (or, to put it simply, is what you subtract and add to the sample proportion)
d= *√(^ρ(1-^ρ))/n
d= 1.96*√(0.49(1-0.49))/2298
d= 0.0204
b) Construct a 90% CI for the true proportion
[^ρ±*√(^ρ(1-^ρ))/n]
[0.49±1.648*√(0.49(1-0.49))/2298]
[0.4728;0.5072]
c) The CI interval I've made in b. is for the voters that agree with that president, but since you want to see if the proportion of people on favor/opposing is 0.50 whether the interval is made with the information of the people who agree or oppose doesn't change the answer.
The interval contains 0.50 if you were to make the hypothesis that the proportion of people who agree with the president H₀:ρ=0.50, at complementary confidence level α: 0.10, you can say that the proportion of people who agree with the president is 50%. Then you can also say that the proportion of opposition voters is 50%.
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