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Construct a 95% confidence interval for n = 100, p = 0.8? In a survey of a random sample of 1000 teenag..?
1.) Construct a 95% confidence interval for n = 100, p = 0.8
-85.4 to 120.6
-67.13 to 99.8
-70.42 to 94.85
-72.16 to 87.84
2.) In a survey of a random sample of 1000 teenagers in Nanaimo, B.C., it was found that 125 of these teenagers had never travelled outside of their own province. Construct a 95% confidence interval for this data.
-111.2 to 137.2
-104.502 to 145.498
-87.502 to 145.534
-98.487 to 151.673
2 Answers
- paramvenuLv 79 years agoFavorite Answer
1) 95% confidence interval for the population proportion is
Sample proportion +/- z-score for 95% confidence*Standard error of p
z-score = 1.96
Standard error of p = sqrt (p*(1-p)/n) = sqrt 0.8*0.2/100) = 0.04
Therefore the confidence interval is
0.8 +/- 1.96*0.04
0.8 +/- 0.0784
lower limit is 0.8-0.0784 = 0.7216 and in terms of percentage 72.16%
upper limit is 0.8+0.0784 = 0.8784 and in terms of percentage 87.84%
The confidence interval is 72.16 to 87.84
2) Sample proportion = 125/1000 = 0.125
Standard error of p = sqrt (0.125*0.875/1000) = 0.010458
95% confidence interval for population proportion is
0.125 +/- 1.96*0.010458
0.125 +/- 0.020498
lower limit is 0.125 - 0.020498 = 0.104502
upper limit is 0.125 + 0.020498 = 0.145498
The confidence interval is 0.104502 to 0.145498
and in terms of percentages 10.4502 to 14.5498
BEWARE OF USING THE DECIMAL POINT
Show the interval either in decimal values or in percentages
The interval is normally shown in decimal values only.
