This is R program question for my statistic class. (So, I need to use RStudio).

Question

This is R program question for my statistic class. (So, I need to use RStudio).

Binomial proportions in real life. On June 28, 2012 the U.S. Supreme Court upheld the much debated 2010 healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicated that 46% (465) of 1,012 Americans agreed with this decision. I. Use the survey results to compute a 95% confidence interval for the proportion of Americans in the population that agreed with the decision at the time. 2. What is the margin of error at the 95% confidence level? 3, What would the margin of error be at a 90% confidence level?

Explanation / Answer

Answer:

1).

R code:

# CI for proportion

p=465/1012 ;p

var =p*(1-p)/1012 ; var

se =sqrt(var);se

E=qnorm(.975)*se ; E

p+c(-E,E)

R output:

p=465/1012 ;p

[1] 0.4594862

> var =p*(1-p)/1012 ; var

[1] 0.0002454137

> se =sqrt(var);se

[1] 0.01566568

> E=qnorm(.975)*se ; E

[1] 0.03070418

> p+c(-E,E)

[1] 0.4287820 0.4901903

95 CI = (0.4288, 0.4902)

2).

E=qnorm(.975)*se ; E

[1] 0.03070418

Margin of error at 95% level =0.0307

3)

E=qnorm(.95)*se ; E

[1] 0.02576776

Margin of error at 95% level =0.0258

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