Question

If n = 100 and X = 80, construct a 95% confidence interval for the population...

If n = 100 and X = 80, construct a 95% confidence interval for the population proportion, p. Give your answers to three decimals < p

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

We have given,          
          
x=80      
n=100      
Estimate for sample proportion=\hat{p}=   0.8
Level of significance is =1-0.95=0.05  
Z critical value(using Z table)=1.96

Confidence interval formula is          


=(0.722,0.878)          
Therefore, a 95% confidence interval for the population proportion, p is

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
If n = 370 and X = 296, construct a 95% confidence interval for the population...
If n = 370 and X = 296, construct a 95% confidence interval for the population proportion, p. Give your answers to three decimals _ < p < _
Construct a​ 95% confidence interval to estimate the population proportion using the data below.     x=23 n=80...
Construct a​ 95% confidence interval to estimate the population proportion using the data below.     x=23 n=80 N=500 The​ 95% confidence interval for the population proportion is The​ 95% confidence interval for the population proportion is :
If n=600 and X = 480, construct a 95% confidence interval. Give your answers to three...
If n=600 and X = 480, construct a 95% confidence interval. Give your answers to three decimals < p
if n=100 and X=30​, construct a 95​% confidence interval estimate for the population proportion.
if n=100 and X=30​, construct a 95​% confidence interval estimate for the population proportion.
Out of 100 people sampled, 80 had kids. Based on this, construct a 99% confidence interval...
Out of 100 people sampled, 80 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places
If n = 500 and ˆp (p-hat) = 0.75, construct a 95% confidence interval. Give your...
If n = 500 and ˆp (p-hat) = 0.75, construct a 95% confidence interval. Give your answers to three decimals < p <
1) If n=19, ¯xx¯(x-bar)=33, and s=18, construct a confidence interval at a 80% confidence level. Assume...
1) If n=19, ¯xx¯(x-bar)=33, and s=18, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. < μμ < 2) You measure 48 backpacks' weights, and find they have a mean weight of 44 ounces. Assume the population standard deviation is 2.8 ounces. Based on this, construct a 95% confidence interval for the true population mean backpack weight. Give your answers as decimals, to two...
If n=320 and p' (p-prime) = 0.23, construct a 95% confidence interval. Give your answers to...
If n=320 and p' (p-prime) = 0.23, construct a 95% confidence interval. Give your answers to three decimals. _______ < p < ________
Out of 100 people sampled, 78 had kids. Based on this, construct a 90% confidence interval...
Out of 100 people sampled, 78 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. As in the reading, in your calculations: --Use z = 1.645 for a 90% confidence interval --Use z = 2 for a 95% confidence interval --Use z = 2.576 for a 99% confidence interval. Give your answers to three decimals Give your answers as decimals, to three decimal places. [,]
In a sample of 590 adults, 425 had children. Construct a 95% confidence interval for the...
In a sample of 590 adults, 425 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT