Question

To estimate with a 99% confidence interval, the difference in mean earned income among men and...

To estimate with a 99% confidence interval, the difference in mean earned income among men and women based on a sample of 30 women and 20 men, would you use for the significance test or confidence interval the t-student or the Z-score values? Why?

Homework Answers

Answer #1

In the above problem we use t test for two means.

For z test we have two criterias..

1. the sample size is sufficiently large.

That is sample size greater than or equal to 30

2. The population standard deviations is known.

In the above nothing is mentioned about population standard deviations.

Therefore, we can't go with z test.

For above problem, we can make decisions usingt t test for two sample means.

We can find confidence interval using t test for two means.

Dear student,
I am waiting for your feedback. I have given my 100% to solve your queries. If you satisfied with my answer then please please like this.
Thank You

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 99% confidence interval estimate of the population mean ? can be interpreted to mean: a)...
A 99% confidence interval estimate of the population mean ? can be interpreted to mean: a) if all possible sample are taken and confidence intervals created, 99% of them would include the true population mean somewhere within their interval. b) we have 99% confidence that we have selected a sample whose interval does include the population mean. c) we estimate that the population mean falls between the lower and upper confidence limits, and this type of estimator is correct 99%...
Mary, the counselor, wants to test whether there is any meaningful difference in depression scores among...
Mary, the counselor, wants to test whether there is any meaningful difference in depression scores among women and men clients of hers. She has 13 women clients and 14 men clients. The mean depression score of men clients is 4.5, while that of women is 5.3. The pooled standard error is 0.5 of the two groups. Is there any gender difference in depression levels among her women and men clients? Indicate the null and research hypothesis. What test do you...
1a. Given: mean difference between men and women = 2.2; standard error of mean difference =...
1a. Given: mean difference between men and women = 2.2; standard error of mean difference = 0.8; n men = 61, n women = 61. Test the following null hypothesis at a=.05 Ho: u men = u women Ha: u men > u women You need to compute the t statistic, look up t critical value and make decision including likelihood of appropriate error. You don't have to compute the standard error of mean difference because it's given. 1b. Compute...
2. Using the same information as question one, compute a 99% confidence interval for the mean...
2. Using the same information as question one, compute a 99% confidence interval for the mean difference. Based on the interval, make a statistical decision about the null hypothesis and briefly state why you came to your conclusion. 1. Professional sports teams use the vertical jump to gauge the strength of athletes. This test requires a person to stand flat footed and leap vertically into the air and touch the highest reachable mark. A typical person in the population can...
A 99% confidence interval has to be created to estimate the mean of the volume of...
A 99% confidence interval has to be created to estimate the mean of the volume of contents in the bottles of an expensive medication. How large a simple random sample has to be taken so that the margin of error is no more than ±5 ml? Assume that the population standard deviation is 20 ml.
Use the sample data to find a 99% confidence interval for the average monthly student loan...
Use the sample data to find a 99% confidence interval for the average monthly student loan payment among all recent college graduates who had at least one student loan during college.Round all answers to the nearest cent (two decimal places).The margin of error for this confidence interval is: $ The 99% confidence interval is:$ <μ<<μ< $ The heights of all women in a large population follow a Normal distribution with unknown mean μμ. In a simple random sample of 44...
Previously, with a recent sample of 18 half carat diamonds, we determined a 99% confidence interval...
Previously, with a recent sample of 18 half carat diamonds, we determined a 99% confidence interval estimate of the mean price of all half carat diamonds to be ($1823.7, $2105.8). a) In 2007 the mean price of a half carat diamonds was $2150. At the 1% significance level, does our recent sample of 18 half carat diamonds with the 99% confidence interval estimate provide evidence that the mean price has changed? Show (1) hypotheses, (5) decision based on C.I., and...
Determine the margin of error for a 99​% confidence interval to estimate the population mean when...
Determine the margin of error for a 99​% confidence interval to estimate the population mean when s​ = 40 for the sample sizes below. ​a) nequals12 ​b) nequals25 ​c) nequals50 ​a) The margin of error for a 99​% confidence interval when nequals12 is _____. ​(Round to two decimal places as​ needed.) ​ b) The margin of error for a 99​% confidence interval when nequals25 is ____. ​(Round to two decimal places as​ needed.) ​ c) The margin of error for...
A 99% confidence interval estimate for a population mean  is determined to be 85.58 to 96.62. If...
A 99% confidence interval estimate for a population mean  is determined to be 85.58 to 96.62. If the confidence level is reduced to 90%, the confidence interval for  :
A manager wishes to estimate a population mean using a 99​% confidence interval estimate that has...
A manager wishes to estimate a population mean using a 99​% confidence interval estimate that has a margin of error of plus or minus±48.0 If the population standard deviation is thought to be 630​, what is the required sample​ size? The sample size must be at least?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT