Question

In: Statistics and Probability

A national health organization warns that 30% of middle school students nation wide have been drunk....

A national health organization warns that 30% of middle school students nation wide have been drunk. A local health agency randomly surveys 100 students and only 25 report having been drunk. Construct an appropriate 80% confidence interval.

A. 0.25

B. 0.0555

C. 0.1946 to 0.3055

D. 0.1651 to 0.3349

Please show work and exaplain why. Thank you!

Expert Solution

Solution :

Given that,

n = 100

x = 25

= x / n = 25 / 100 = 0.250

1 - = 1 - 0.250 = 0.750

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z_0.10 = 1.282

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

= 1.282 * ( $\sqrt$ ((0.250 * 0.750) / 100)

= 0.0555

A 80 % confidence interval for population proportion p is ,

- E < P < + E

0.250 - 0.0555 < p < 0.250 + 0.0555

0.1945< p < 0.3055

Option B) is correct.

orchestra answered 1 year ago

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