Question

In: Statistics and Probability

Exercise 2. A survey of 525 homeless persons showed that 168 were not veterans. Construct a...

Exercise 2. A survey of 525 homeless persons showed that 168 were not veterans. Construct a 95% confidence interval for the proportion of homeless persons who are not veterans.

Expert Solution

Solution :

Given that,

n = 525

x = 168

Point estimate = sample proportion = = x / n = 0.32

1 - = 0.68

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * ( $\sqrt$ ((0.32*0.68) / 525)

= 0.040

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.32 - 0.040 < p < 0.32 + 0.040

0.280 < p < 0.360

The 95% confidence interval for the population proportion p is : ( 0.280 , 0.360 )

orchestra answered 1 year ago

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