Question

In: Statistics and Probability

In a survey, 668 out of 1112 people said they support Medicare for all. Find the...

In a survey, 668 out of 1112 people said they support Medicare for all. Find the rule of thumb 95% confidence interval estimate of the true proportion of supporters.

Expert Solution

Solution :

Given that,

n = 1112

x = 668

Point estimate = sample proportion = = x / n = 668/1112=0.601

1 - = 1- 0.601 =0.399

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * $\sqrt$ ((( * (1 - )) / n)

= 1.96 ( $\sqrt$ ((0.601*0.399) /1112 )

E = 0.029

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.601-0.029 < p < 0.601+0.029

(0.572, 0.63)

orchestra answered 2 years ago

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