Suppose a survey was done this year to find out what proportion of young Canadians would...

Suppose a survey was done this year to find out what proportion of young Canadians would like to be rich. Out of a random sample of 1100 young Canadians, 387 said yes, they would like to be rich. What is the 95% confidence interval for the true proportion of all young Canadians who would like to be rich? Show all your work.

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Expert Solution

Solution :

Given that,

n = 1100

x = 387

Point estimate = sample proportion = = x / n = 387/1100=0.352

1 - = 1- 0.352 =0.648

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 ( $\sqrt$ ((0.352*0.648) / 1100)

E = 0.028

A 95% confidence interval for proportion p is ,

- E < p < + E

0.352-0.028 < p < 0.352+0.028

0.324< p < 0.380

(0.324, 0.380)

orchestra answered 1 year ago

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