Question

Dr Z decided to investigate the productivity of her plutonium mines. The standard deviation for the...

Dr Z decided to investigate the productivity of her plutonium mines. The standard deviation for the output is known (fat hint: use z-tables) and is 5 tons a day. The average output in a sample of a 100 mines was 66 tons. At 95 % confidence level, what is the margin of error? (Two decimals)

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

Homework Answers

Answer #1

solutionPopulation standard deviation =    = 5

Sample size n =100

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * (5 / 100)

=0.98

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
From a normal population, Dr. Çetin takes a sample of 235. The population standard deviation is...
From a normal population, Dr. Çetin takes a sample of 235. The population standard deviation is 24, and the sample mean is 17. a.Determine the standard error of the mean. (Round your answer to 3 decimal places.) Standard error of the mean: ……? b. Determine the 95% confidence interval for the population mean. (In your calculations: round your z/t value to 2 decimals, but round all other values to 3 decimals. Round your final answer to 3 decimal places.) Confidence...
5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin...
5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin of error to 95% confidence interval to 25 points, how many students to sample? 6. Population proportion is 43%, would like to be 95% confident that your estimate is within 4.5% of the true population proportion. How large of a sample is required? 7. P(z<1.34) (four decimal places) 8. Candidate only wants a 2.5% margin error at a 97.5% confidence level, what size of...
1)Given a sample mean is82, the sample size is 100and the population standard deviation is 20....
1)Given a sample mean is82, the sample size is 100and the population standard deviation is 20. Calculate the margin of error to 2 decimalsfor a 90% confidence level. 2)Given a sample mean is 82, the sample size is 100 and the population standard deviation is 20. Calculate the confidence interval for 90% confidence level. What is the lower limit value to 2 decimals? 3)Given a sample mean is 82, the sample size is 100 and the population standard deviation is...
An IQ test scoring is designed to have mean of 100 and a standard deviation of...
An IQ test scoring is designed to have mean of 100 and a standard deviation of 15. The scores are known to be normally distributed. A principal at a private school wants to investigate whether the average IQ of the students at her school is statistically different from the designed mean. She randomly samples from the school and has 15 students take the IQ test. The average test score of the sample is 95 and the sample standard deviation is...
Intelligence of the general population has a mean of µ = 100 and a standard deviation...
Intelligence of the general population has a mean of µ = 100 and a standard deviation of σ = 15. A researcher wonders if people who exercise more than 30 min each day have an average mean IQ different from 100. He randomly selects n = 81 people and enrolls them in an exercise program for six months. After, he has each individual take an IQ test and finds the average IQ of the sample is M = 102. a)...
Q1. [5 pts] Suppose that Z follows a standard normal. Calculate the probability that: a)Z is...
Q1. [5 pts] Suppose that Z follows a standard normal. Calculate the probability that: a)Z is less than 1.25. b)Z is greater than 1.75. c)Z is between 1.25 and 1.75. d)Z is between -1.25 and 1.75. e)X is between 85 and 105, where X follows a normal with mean of 100 and SD of 10.Note: Write probabilities as decimals and to the four decimal places (e.g., 0.2538). Q2. What is the value of z such that:a)the probability of a standard...
1. Find the area under the standard normal curve (round to four decimal places) a. To...
1. Find the area under the standard normal curve (round to four decimal places) a. To the left of  z=1.65 b. To the right of z = 0.54 c. Between z = -2.05 and z = 1.05 2. Find the z-score that has an area of 0.23 to its right. 3. The average height of a certain group of children is 49 inches with a standard deviation of 3 inches. If the heights are normally distributed, find the probability that a...
What price do farmers get for their watermelon crops? In the third week of July, a...
What price do farmers get for their watermelon crops? In the third week of July, a random sample of 45 farming regions gave a sample mean of  = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.92 per 100 pounds. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each...
Do all the following problems. I. Choose the best answer for each multiple choice. Please use...
Do all the following problems. I. Choose the best answer for each multiple choice. Please use CAPITAL letters to indicate your answer and write neatly. (30 points) 1. ____ 4.____ 7. ____ 10.____ 13. ____ 2. ____ 5.____ 8. ____ 11.____ 14. ____ 3. ____ 6.____ 9. ____ 12.____ 15. ____ Whenever the population standard deviation is unknown and the population has a normal or near-normal distribution, which distribution is used in developing an interval estimation? A. standard distribution B....
Calculate the mean, median, mode, variance (population) and standard deviation (sample) for the following test grades:...
Calculate the mean, median, mode, variance (population) and standard deviation (sample) for the following test grades: 65, 60, 67, 71, 75, 80, 65 Use this information to answer questions #1-4. Question 1 (0.5 points) Saved The mode is: Question 1 options: 65 60 There is no modal category 75 71 Question 2 (0.5 points) Saved The median is: Question 2 options: 67 65 80 66 None of the above Question 3 (0.5 points) What is the approximate population variance for...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT