Question

Calculating Confidence Intervals 1. The average distance driven to work for a random sample of 55...

Calculating Confidence Intervals

1. The average distance driven to work for a random sample of 55 people living in Columbus Ohio is 17 miles with a standard deviation of 2.2 miles. The average distance driven to work for a random sample of 30 people living in Cincinnati Ohio is 15.5 miles with a standard deviation of 1.7 miles. We are interested in a 95% confidence interval for the difference in average distance driven to work.

2. A study of teenage heroin use included a random sample of 112 boys and 133 girls between ages of 12 and 16 years selected randomly from public schools in Kentucky. Knowledge of heroin use by a family member was reported by 15 of the boys and 19 of the girls. We are interested in a 99% confidence interval for the difference in proportion of knowledge of heroin use by a family member between boys and girls ages 12 to 16 in Kentucky.

Homework Answers

Answer #1

Solution:-

1) 95% confidence interval for the difference in average distance driven to work is C.I = (0.646 , 2.354).

C.I = 1.50 + 1.989 × 0.42934

C.I = 1.50 + 0.85396

C.I = (0.646 , 2.354)

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
In a random sample of twelve ​people, the mean driving distance to work was 24.5 miles...
In a random sample of twelve ​people, the mean driving distance to work was 24.5 miles and the standard deviation was 5.2 miles. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 90% confidence interval for the population mean μ. Identify the margin of error. Construct a 90​% confidence interval for the population mean.
In a random sample of six ​people, the mean driving distance to work was 18.3 miles...
In a random sample of six ​people, the mean driving distance to work was 18.3 miles and the standard deviation was 4.3 miles. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 90​% confidence interval for the population mean mu. Interpret the results. Identify the margin of error.
In a random sample of five ​people, the mean driving distance to work was 22.9 miles...
In a random sample of five ​people, the mean driving distance to work was 22.9 miles and the standard deviation was 4.9 miles. Assuming the population is normally distributed and using the​ t-distribution, a 95​% confidence interval for the population mean is (16.8, 29.0) ​(and the margin of error is 6.1​). Through​ research, it has been found that the population standard deviation of driving distances to work is 6.2. Using the standard normal distribution with the appropriate calculations for a...
In a random sample of five ​people, the mean driving distance to work was 24.9 miles...
In a random sample of five ​people, the mean driving distance to work was 24.9 miles and the standard deviation was 4.3 miles. Assuming the population is normally distributed and using the​ t-distribution, a 99​%confidence interval for the population mean mu is left parenthesis 16.0 comma 33.8 right parenthesis ​(and the margin of error is 8.9​). Through​ research, it has been found that the population standard deviation of driving distances to work is 3.3 Using the standard normal distribution with...
True or false? A larger sample size produces a longer confidence interval for μ. False. As...
True or false? A larger sample size produces a longer confidence interval for μ. False. As the sample size increases, the maximal error decreases, resulting in a shorter confidence interval.True. As the sample size increases, the maximal error decreases, resulting in a longer confidence interval.    True. As the sample size increases, the maximal error increases, resulting in a longer confidence interval.False. As the sample size increases, the maximal error increases, resulting in a shorter confidence interval. True or false? If the...
1. A sample of 25 adults living in Kentucky had an average IQ of 102. The...
1. A sample of 25 adults living in Kentucky had an average IQ of 102. The margin of error for a 95% confidence interval is 18. We are interested in a 99% confidence interval for the average IQ of all adults in Kentucky. 2. We are interested in a 90% confidence interval for the proportion of EKU students who will spend summer break outside of Kentucky if the sample had 75 students of which 20 would be leaving the state....
CONFIDENCE INTERVAL PROJECT – Day 3 On this sheet, you will be calculating 95% Confidence Intervals...
CONFIDENCE INTERVAL PROJECT – Day 3 On this sheet, you will be calculating 95% Confidence Intervals for your data. Fill in all of the values for each piece, the formula, then show the full calculation for the CI. Write the CI in 2 different ways: In + notation In interval notation: ( ________ , ________ ) When you record standard deviation- round to the nearest tenth. Use the back of this sheet if extra space is needed. PROPORTION CONFIDENCE INTERVAL:...
3) A random sample of 420 American teens finds that they spend an average of 7.2...
3) A random sample of 420 American teens finds that they spend an average of 7.2 hours per day using screens, with a standard deviation of 2.4 hours. a) Use the 2SD method to find a 95% confidence interval estimate of the average (mean) hours of screen time that American teens use per day. Show all work/steps to get the 2SD estimate. Round your margin of error to three decimal places. Work and Answer: b) A cell phone company claims...
SHOW ALL WORK IN EXCEL For this assignment, submit one Excel file. Part 1 sheet is...
SHOW ALL WORK IN EXCEL For this assignment, submit one Excel file. Part 1 sheet is kept blank. Use Part 1 sheet to answer question 1. Put a box around your answer. Round your answer to 4 decimal places. The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher...
1. For a pair of sample x- and y-values, what is the difference between the observed...
1. For a pair of sample x- and y-values, what is the difference between the observed value of y and the predicted value of y? a) An outlier b) The explanatory variable c) A residual d) The response variable 2. Which of the following statements is false: a) The correlation coefficient is unitless. b) A correlation coefficient of 0.62 suggests a stronger correlation than a correlation coefficient of -0.82. c) The correlation coefficient, r, is always between -1 and 1....
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT