Question

To determine how well a new method of teaching vocabulary is working in a certain elementary...

To determine how well a new method of teaching vocabulary is working in a certain elementary school, education researchers plan to give a vocabulary test to a sample of 100 sixth graders. It is known that scores on this test have a standard deviation of 8. The researchers found the sample mean to be 100. Construct a 99% confidence interval for the population mean. Round your answers to whole numbers.

Homework Answers

Answer #1

99% confidence interval for is

- Z/2 * / sqrt(n) < < + Z/2 * / sqrt(n)

100 - 2.576 * 8 / sqrt(100) < < 100 + 2.576 * 8 / sqrt(100)

98 < < 102

99% CI is ( 98 , 102 )

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
There are two elementary schools in Caleb's community. Third graders in both schools recently took the...
There are two elementary schools in Caleb's community. Third graders in both schools recently took the same standardized exam. A random sample of 45 third graders from the first school has a mean score of 75. A random sample of 39 third graders from the second school has a mean score of 78. Assume the population standard deviations of the third graders' scores are known to be 10 for the first school and 5 for the second school. Construct and...
The State Department of Education is developing a new test to measure the reading level of...
The State Department of Education is developing a new test to measure the reading level of sixth graders. To pilot this test, they select a random sample of 50 sixth graders from within the state and have them take the test. The mean score for the students in this sample is 70.5 points. Suppose that it is already known that the standard deviation for all sixth graders in the state is 8.4 points. A.Construct and interpret a 98% confidence interval...
a. A random sample of elementary school children in New York state is to be selected...
a. A random sample of elementary school children in New York state is to be selected to estimate the proportion p who have received a medical examination during the past year. The survey found that x = 468 children were examined during the past year. Construct the 95 % confidence interval estimate of the population proportion p if the sample size was n=600 . Give your confidence interval bounds to at least 4 decimal places. b) Which of the following...
The quality of the elementary school teachings is very important, and the effectiveness of teaching can...
The quality of the elementary school teachings is very important, and the effectiveness of teaching can be shown by the students’ performance on the standardized test. The state requires that the average score of the 5th grade students from each school need to be at least 75. If any school’s average score is below 75, the school will be investigated further to ensure their quality of teaching. For School A, 20 5th grade students are randomly selected to take the...
An experiment is run to determine how well a new process can measure the amountof a...
An experiment is run to determine how well a new process can measure the amountof a certain pesticide in water. The experiment uses the new test and the old test to measure 10 water samples with known amounts. The data below is the number of parts per million that each test differs from for each sample. The old test produces: {30, 19, 24, 24, 24, 13, 22, 12, 19, 33 } The new test produces: { 29, 25, 29, 29,...
Try to answer what you can. Use the following information to answer questions 14 - 20...
Try to answer what you can. Use the following information to answer questions 14 - 20 The principal of a certain high school claims that the SAT Math scores for his students is 660. A random sample of 68 of these students showed a mean 650. Assume that the population standard deviation of the scores is known to be 100. Test using a level of significance of 1% whether the populations mean SAT score for the high school 660. 14.   ...
5. How heavy are the backpacks carried by college students? To estimate the weight of backpacks...
5. How heavy are the backpacks carried by college students? To estimate the weight of backpacks carried by college students, a researcher weighs the backpacks from a random sample of 58 college students. The average backpack weight ends up being 15.7 pounds, with a standard deviation of 2.4 pounds. If you use this data to construct a 90% confidence interval, what will the margin of error be for this interval? Try not to do a lot of intermediate rounding until...
Dr Susan Benner is an industrial Psychologist. She is currently studying stress among executives of internet...
Dr Susan Benner is an industrial Psychologist. She is currently studying stress among executives of internet companies. She has developed a questionnaire that she believes measures stress. A score above 80 indicates stress at a dangerous level. A random sample of 15 executives revealed the following stress level scores. 94 78 83 90 78 99 97 90 97 90 93 94 100 75 84 a. Find the mean stress level for this sample. What is the point estimate of the...
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...
The owner of a local golf course wants to determine the average age of the golfers...
The owner of a local golf course wants to determine the average age of the golfers that play on the course in relation to the average age in the area. According to the most recent census, the town has an average age of 23.44. In a random sample of 26 golfers that visited his course, the sample mean was 30.63 and the standard deviation was 8.771. Using this information, the owner calculated the confidence interval of (25.84, 35.42) with a...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT