Question

The following is on fish populations. Recent data show that 0.34 of fish are female in...

The following is on fish populations. Recent data show that 0.34 of fish are female in the sample of 120 from one pond. What is the confidence interval for the population proportion based on the estimate? (We are testing at the 95% confidence level)

SD = 0.16

Homework Answers

Answer #1

Given that,

= 0.34

1- = 1-0.34 = 0.66

n = 120

At 95% confidence interval,

Z/2 = 1.960

95% confidence interval = Z/2 * √((1-)/n)

= 0.34 1.96 * √(0.34*0.66/120)

= 0.34 0.085

= (0.255, 0.425)

Therefore, the 95% confidence interval for the population proportion based on the estimate is between 0.255 and 0.425.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
a random sample of 81 executives (these 81 include both male and female) is drawn for...
a random sample of 81 executives (these 81 include both male and female) is drawn for the purpose of estimating the population proportion of females and the mean age of all female executives. The sample contains 33 female executives for those ladies, the sample mean and standard deviation are 46.5 years and 6.8 years, respectively. We first want to build a confidence interval for the proportion of female executives in the population of all executives. a.:check that the conditions to...
Salary information regarding male and female employees of a large company is shown below. Male Female...
Salary information regarding male and female employees of a large company is shown below. Male Female Sample Size: 64 36 Sample Mean Salary (in $1,000): 44 41 Population Variance:    128 72 1.) The standard error for the difference between the two means is 2.) The point estimate of the difference between the means of the two populations is 3.) At 95% confidence, the margin of error is 4.)  The 95% confidence interval for the difference between the means of the two...
Consider the following results for independent samples taken from two populations. Sample 1 Sample 2 n1...
Consider the following results for independent samples taken from two populations. Sample 1 Sample 2 n1 = 500 n2= 200 p1= 0.45 p2= 0.34 a. What is the point estimate of the difference between the two population proportions (to 2 decimals)? b. Develop a 90% confidence interval for the difference between the two population proportions (to 4 decimals). Use z-table. to c. Develop a 95% confidence interval for the difference between the two population proportions (to 4 decimals). Use z-table....
Let's say we want to estimate the population proportion of a population. A simple random sample...
Let's say we want to estimate the population proportion of a population. A simple random sample of size 400 is taken from the population. If the sample proportion is 0.32: 1) what is the point estimate of the population proportion? 2) At the 95% level of confidence, what is the margin of error? 3) Based on 2) what is a confidence interval at the 95% confidence level? 4) What is the margin of error if the level of confidence is...
Determine the margin of error for a confidence interval to estimate the population proportion for the...
Determine the margin of error for a confidence interval to estimate the population proportion for the following confidence levels with a sample proportion equal to 0.35 and n= 120 the margin of error for a confidence interval to estimate the population portion for 90% confidence level is the margin of error for a confidence interval to estimate the population portion for 95% confidence level is the margin of error for a confidence interval to estimate the population portion for 97%...
An online site presented this? question, "Would the recent norovirus outbreak deter you from taking a?...
An online site presented this? question, "Would the recent norovirus outbreak deter you from taking a? cruise?" Among the 34,577 people who? responded,64?% answered?"yes." Use the sample data to construct a 95?% confidence interval estimate for the proportion of the population of all people who would respond? "yes" to that question. Does the confidence interval provide a good estimate of the population? proportion?
The FDA’s webpage provides some data on mercury content of fish. Based on a sample of...
The FDA’s webpage provides some data on mercury content of fish. Based on a sample of 15 salmon fish in the Pacific Ocean, a sample mean was computed as 0.31 (parts per million). The sample's standard deviation was 0.07 ppm. The 15 observations ranged from 0.18 to 0.41 ppm, telling us that the sample data is symmetric and with no significant outliers; however, we don’t know if the data is Normally distributed. Construct an appropriate 95% confidence interval for the...
Consider the following results for independent samples taken from two populations. sample 1 sample 2 n1=500...
Consider the following results for independent samples taken from two populations. sample 1 sample 2 n1=500 n2=200 p1= 0.42 p2= 0.34 a. What is the point estimate of the difference between the two population proportions (to 2 decimals)? b. Develop a confidence interval for the difference between the two population proportions (to 4 decimals). (______to _______) c. Develop a confidence interval for the difference between the two population proportions (to 4 decimals). (______to________)
Independent simple random samples are taken to test the difference between the means of two populations...
Independent simple random samples are taken to test the difference between the means of two populations whose variances are not known. Given the sample sizes are n1 = 11 and n2 = 16; and the sample variances are S12 = 33 and S22 = 64, what is the correct distribution to use for performing the test? A. t distribution with 49 degrees of freedom B. t distribution with 59 degrees of freedom C. t distribution with 24 degrees of freedom...
In a recent poll of 210 south Jersey households, it was found that 152 households had...
In a recent poll of 210 south Jersey households, it was found that 152 households had at least one computer. Construct a 95% confidence interval to estimate the population proportion of south Jersey. A separate survey in Philadelphia found the sample proportion of households with more than one computer to be 0.66, what kind conclusion you can made in terms of the difference in proportion between South Jersey and Philadelphia? (Your conclusion must be based on the confidence interval you...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT