In: Statistics and Probability
1) Out of 200 people sampled, 14 had kids. Based on this,
construct a 95% confidence interval for the true population
proportion of people with kids.
Give your answers as decimals, to three places
< p <
2)
If n=450n=450 and ˆpp^ (p-hat) =0.91, find the margin of error at a 90% confidence level.
Round to 4 places.
z-scores may be rounded to 3 places or exact using technology.
3) In a recent poll, 110 people were asked if they liked dogs,
and 41% said they did. Find the margin of error of this poll, at
the 90% confidence level.
Give your answer to three decimals
Solution :
1 ) Given that,
n = 200
x = 14
= x / n = 14 / 200 = 0.070
1 - = 1 - 0.070 = 0.930
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.960
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.960 * (((0.070 * 0.930) / 200) = 0.035
A 95 % confidence interval for population proportion p is ,
- E < P < + E
0.070 - 0.035 < p < 0.070 + 0.035
0.035 < p < 0.105
The 95% confidence interval for the population proportion p is : ( 0.035 , 0.105)
2 ) Given that,
n = 450
= 0.91
1 - = 1 - 0.91 = 0.09
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z0.05 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 * (((0.91 * 0.09) / 450) = 0.0222
A 90 % confidence interval for population proportion p is ,
- E < P < + E
0.9100 - 0.0222 < p < 0.9100 + 0.0222
0.8878< p < 0.9322
The 90% confidence interval for the population proportion p is : ( 0.8878 , 0.9322)
3 ) Given that,
n = 110
= 41 % = 0.410
1 - = 1 - 0.410 = 0.590
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z0.05 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 * (((0.410 * 0.590) / 110) = 0.077
A 90 % confidence interval for population proportion p is ,
- E < P < + E
0.410 - 0.077 < p < 0.410 + 0.077
0.333 < p < 0.487
The 90% confidence interval for the population proportion p is : ( 0.333 , 0.487)