Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 229 with 44 successes. Enter your answer as an open- interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
99% C.I. =
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

B. Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 298 with 256 successes. Enter your answer as a tri- linear inequality using decimals (not percents) accurate to three decimal places.
_____< p <______

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

C. We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 405 had kids. Based on this, construct a 95% confidence interval for the proportion pp of adult residents who are parents in this county.
Give your answers as decimals, to three places.
_____ < pp <______

D. If n=12, ¯xx¯(x-bar)=32, and s=13, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population.
Give your answers to one decimal place.

Answer 1

Answer:

A. 99% CI for p: 0.125 < p < 0.259

B.  99% CI for p: 0.807 < p < 0.911

C. 95% CI for p: 0.776 < p < 0.844

D. 95% CI using t-distribution: 23.7 < µ < 40.3

Step-by-step explanation:

A.

$p=\frac{x}{n} =\frac{44}{229} = 0.192139738$

Confidence = 99% = 0.99

$\alpha$=1 - confidence = 1 - 0.99 = 0.01

Critical z-value =$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.192139738(1-0.192139738)}{229} }$

$SE=\sqrt{\frac{0.155222059}{229} }$

SE = 0.026035

E = Margin of Error

E = z-value * SE = 2.58 * 0.026035

E = 0.067170516

Lower Limit = p - E = 0.192139738 - 0.067170516 = 0.125

Upper Limit = p + E = 0.192139738 + 0.067170516 = 0.259

99% CI for p: 0.125 < p < 0.259

B.

$p=\frac{x}{n} =\frac{256}{298} = 0.8590604$

Confidence = 99% = 0.99

$\alpha$=1 - confidence = 1 - 0.99 = 0.01

Critical z-value =$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.8590604(1-0.8590604)}{298} }$

$SE=\sqrt{\frac{0.121075629}{298} }$

SE = 0.020157

E = Margin of Error

E = z-value * SE = 2.58 * 0.020157

E = 0.052004

Lower Limit = p - E = 0.8590604 - 0.052004 = 0.807

Upper Limit = p + E = 0.8590604 + 0.052004 = 0.911

99% CI for p: 0.125 < p < 0.259

C.  

$p=\frac{x}{n} =\frac{405}{500} = 0.81$

Confidence = 95% = 0.95

$\alpha$=1 - confidence = 1 - 0.95 = 0.05

Critical z-value =$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.81(1-0.81)}{500} }$

$SE=\sqrt{\frac{0.1539}{500} }$

SE = 0.017544

E = Margin of Error

E = z-value * SE = 1.96 * 0.017544

E = 0.034387

Lower Limit = p - E = 0.81 - 0.034387  = 0.776

Upper Limit = p + E = 0.81 + 0.034387  = 0.844

95% CI for p: 0.776 < p < 0.844

D.

x = sample mean = 32

s = sample standard deviation = 13

n = sample size = 12

Confidence = 95% = 0.95

$\alpha$=1 - confidence = 1 - 0.95 = 0.05

Being the sample size less than 30, we use the statistical t.

df = degree of freedom = n - 1 = 12 - 1 = 11

Critical t-value =$t_{\alpha/2,df}=t_{0.05/2,11}=t_{0.025,11}=2.201$ (from t-table, two tails, df=11)

SE = standard error

$SE=\frac{s_{x} }{\sqrt{n} } =\frac{13 }{\sqrt{12} }=3.752777$

E = margin of error

E = t-value * SE = 2.201 * 3.752777 = 8.259862

Lower Limit = x - E = 32 - 8.259862  = 23.7

Upper Limit = x + E = 32 + 8.259862  = 40.3

95% CI using t-distribution: 23.7 < µ < 40.3

Hope this helps!