|
Tutors Answer Your Questions about Confidence-intervals (FREE)
Question 1114168: A fence must be built to enclose a rectangular area of 45,000 ft squared. Fencing material costs $3 per foot for the two sides facing north and south and $6 per foot for the other two sides. Find the cost of the least expensive fence.
Click here to see answer by ikleyn(52898)   |
Question 1117349: Hello! I have a problem that I'm a little confused on and I calculated the first part. However, I'm getting confused on the second because I am not sure what it is asking me or if I am doing this problem correctly. I will be so thankful if I could have some feedback on my work and any help is welcomed and appreciated. Thank you so much! See, I already did the problem but there is one thing that is kind of confusing me. Here's the problem:
A scientific experiment found that the average weight of adult male emperor penguins is 74 pounds with a margin of error of 25 pounds.
a) What is a confidence interval for the average weight of adult male emperor penguins?
Here is my answer:
I know confidence interval is (sample statistic-margin of error) and (sample statistic+margin of error)
So I did this: (74-25)=49
(74+25)=99
Then it asks me this: If we found a flock of male emperor penguins that had an average weight of 51 pounds would this be unusual? Why or why not?
I checked the confidence interval and it looks like it wouldn't be unusual. That's the part that is really tricking me. I'm not sure if that's correct.
Any help is greatly appreciated! Thank you so much! This really helps me a lot!
Click here to see answer by stanbon(75887) |
Question 1121781: A random sample of 49 shoppers showed that they spend an average of $23.45 per visit at the Saturday Mornings Bookstore. The standard deviation of the sample was $2.80. What is the 90% confidence interval of the true mean?
Click here to see answer by Boreal(15235)  |
Question 1127983: I am in serious need of help. How do i build a 95% confidence interval with only just orange M&M's? Number of M&Ms
---------------------------
ORANGE TOTAL
6 20
5 17
3 17
1 17
5 19
6 18
2 20
4 19
6 18
3 20
6 17
3 16
4 18
3 17
5 17
5 18
4 19
3 19
4 20
2 17
4 18
7 19
4 18
6 19
5 18
---------------------------
106 455
======== ========
p-hat = 0.23
n = 455
Construct the 95% confidence interval around the true population porportion of orange colored M&Ms in a fun-sized bag of plain M&Ms.
Click here to see answer by MathLover1(20850)  |
Question 1140866: You want to estimate the proportion of restaurants in New York City who serve breakfast in the mornings. How many restaurants must be included in the sample if you want to be 80% confident that the sample proportion is in error by no more than 0.05?
Click here to see answer by Boreal(15235) |
Question 1140865: Jennie surveyed a random sample of Californians and asked them if they believed the minimum wage should be increased. She used a calculator to find a 90% confidence interval estimating the proportion of Californians who believe minimum wage should be raised.
a.) Use the results shown to report the confidence interval in proper form.
b.) Can Jennie conclude that more than 50% of Californians believe that the minimum wage should be increased? Justify your answer.
Click here to see answer by Boreal(15235) |
Question 1140864: In a recent study of 86 random teenagers in Utah, the mean number of hours per week that they played video games was 21.6 with σ=5.7 hours.
a.) State and verify the requirements for a valid confidence interval. Then find and report the 95% confidence interval estimating the mean number of hours per week that teenagers in Utah play video games. (Please state the calculator option you used. Round to one decimal place).
b.) Larry, a statistics student, uses the above CI to estimate the mean number of hours per week that American teenagers played video games. Would the estimate be valid? Justify.
Click here to see answer by Boreal(15235) |
Question 1142670: You are testing the claim that the proportion of men who own cats is larger than the proportion of women who own cats.
You sample 90 men, and 85% own cats.
You sample 50 women, and 40% own cats.
Find the pooled value of p, as a decimal, rounded to two decimal places.
Click here to see answer by Theo(13342)  |
Question 1146801: please help me solve this question
CNNBC recently reported that the mean annual cost of auto insurance is 976 dollars. Assume the standard deviation is 298 dollars. You take a simple random sample of 77 auto insurance policies.
Find the probability that a single randomly selected value is less than 986 dollars.
P(X < 986) =
Find the probability that a sample of size
n
=
77
n=77
is randomly selected with a mean less than 986 dollars.
P(
¯¯¯
X
X¯
< 986) =
Click here to see answer by Boreal(15235) |
Question 1155965: Given the line segment of length a and b as shown. construct the Geometric
mean between a and b.
._________.______________.
(letter "a" is between the shorter length and "b" is between the longer length. I'm sorry I couldn't put the letters between the lines because whenever I sent my question for help it keeps misplacing them.)
Click here to see answer by MathLover1(20850) |
Question 1156423: SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 88% confidence interval to 20 points, how many students should the administrator sample? Make sure to give a whole number answer.
Click here to see answer by Boreal(15235) |
Question 1156422: SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 88% confidence interval to 20 points, how many students should the administrator sample? Make sure to give a whole number answer.
Click here to see answer by Boreal(15235) |
Question 1162081: A pilot study shows that 64% of people living in the downtown core are single. A market research company wants to verify this claim. The company requires a 95% confidence interval. How many residents should be interviewed to keep the margin of error within 0.02 of the population proportion?
Click here to see answer by Boreal(15235) |
Question 1162079: A random sample of 300 drivers revealed that 96 of them had received a speeding ticket in the last 3 months. Construct a 95% confidence interval for the number of drivers who receive speeding tickets over a three-month period.
Click here to see answer by Boreal(15235) |
Question 1162994: The ABC battery company claims that their batteries last at least 100 hours, on average. Your experience with their batteries has been somewhat different, so you decide to conduct a test to see if the company's claim is true. You believe that the mean life is actually less than the 100 hours the company claims. You decide to collect data on the average battery life (in hours) of a random sample of n = 20 batteries.
Click here to see answer by Boreal(15235) |
Question 1163623: 1.The Java computer language, developed by Sun Microsystems, has the advantage that its programs can run on types of hardware ranging from mainframe computers all the way down to handheld computing devices or even smart phones. A test of 100 randomly selected programmers revealed that 71 preferred Java to their other most used computer languages. Construct a 95% confidence interval for the proportion of all programmers in the population from which the sample was selected who prefer Java.
Click here to see answer by Boreal(15235) |
Question 1170272: In a study of brain waves during sleep, a sample of 29 college students were randomly
separated into two groups. The first group had 15 people and each was given ½ liter of
red wine before sleeping. The second group had 14 people and were given no alcohol
before sleeping. All participants when to sleep at 11 PM and their brainwave activity was
measured from 4-6 AM. The group drinking alcohol had a mean brainwave activity of
19.65 hertz and a standard deviation of 1.86 hertz. The group not drinking alcohol had a
mean of 6.59 hertz and standard deviation of 1.91 hertz. Compute a 90% confidence
interval for the difference in population means of groups drinking alcohol before
sleeping and those not drinking alcohol before sleeping. Explain the meaning of the
confidence intervaL
Click here to see answer by Boreal(15235) |
Question 1170271: The shoulder height for a random sample of six (6) fawns (deer less than 5 months old) in
a national park was , 𝑥 = 79.25 cm with population standard deviation 𝞂= 5.33 cm.
Compute an 80% confidence interval for the mean shoulder height of the population of
all fawns (deer less than 5 months old) in this national park. Analyze the result to
interpret its meaning
Click here to see answer by Boreal(15235) |
Question 1171311: Create a 99% confidence interval of the number of calories in 3 ounces of french fries
based on a sample of 8 batches of french fries, weighing 3-ounce each, taken from
different restaurants. The sample has a mean of 244.5 and a sample standard deviation,
s, of 21.7. Interpret the results.
Click here to see answer by Boreal(15235) |
Question 1174171: 5. Suppose that a market research firm is hired to estimate the percent of adults living in
a large city who have cell phones. Five hundred randomly selected adult residents in
this city are surveyed to determine whether they have cell phones. Of the 500 people
sampled, 415 responded yes - they own cell phones. Using a 95% confidence level,
compute a confidence interval estimate for the true proportion of adult residents of this
city who have cell phones.
Click here to see answer by ewatrrr(24785)  |
Question 1176585: Surveyors asked a random sample of women in a major city what factor was the most important in deciding where to shop. The results appear in the following table. If the sample size was 1200, estimate with 95% confidence the proportion of women who identified price and value as the most important factor.
Factor Percentage (%)
Price and Value 40
Quality and selection of merchandise 30
Service 15
Shopping environment 15
Click here to see answer by ewatrrr(24785) |
Question 1176585: Surveyors asked a random sample of women in a major city what factor was the most important in deciding where to shop. The results appear in the following table. If the sample size was 1200, estimate with 95% confidence the proportion of women who identified price and value as the most important factor.
Factor Percentage (%)
Price and Value 40
Quality and selection of merchandise 30
Service 15
Shopping environment 15
Click here to see answer by Boreal(15235) |
Question 1177422: A survey reports it results by starting that standard error of the mean to be is 20.
The population standard deviation is 500.
1:How large is the sample used in this surveying?
2: What is the probability that the sample mean will be within 25 of the population mean?
Click here to see answer by Theo(13342) |
Question 1177860: We know that proportion of diamond cards in a regular deck of cards is 13/52=0.25. Assume for a moment that we do not know it and we wish to use statistical methods to find a confidence interval for proportion of red cards in a deck. We will use 10 regular decks of card and select random sample of 100 cards.
Please use the above information to answer questions:
Question 1
The population will consider of 10 regular decks of cards well shuffled..
As described, we randomly draw 100 cards from the population and record a sample (D - diamond card, ND - not diamond card}. In the sample we have 24 diamond cards and 76 of the other suite.
Answer the following questions:
Number of elements in the population is
Number of elements in the sample is
Proportion of diamond cards in the sample is (express it as a decimal)
Click here to see answer by ewatrrr(24785) |
Question 1178027: A machine that stuffs a cheese filled snack product can be adjusted for the amount of cheese injected into each unit. A simple random sample of 30 units is selected, and the average amount of cheese injected is found to be X= 3.5 grams. If the process standard deviation is known to be o =0.25grams, construct a 99% confidence interval for the average amount of cheese being injected by machine.
Click here to see answer by Theo(13342) |
Question 1179919: According to Beautiful Bride magazine, the average age of a groom is now 26.2 years. A sample of 16 prospective grooms in Chicago revealed that their average age was 26.6 years with a standard deviation of 5.3 years. What is the test value for a t test of the claim?
Click here to see answer by Boreal(15235) |
Question 1181286: In order to estimate the average age of onset of a certain type of disease a researcher collects a sample of 15 people with the disease and produces the following 95% confidence interval (83.9219, 90.6782). Find a 99% confidence interval for the average age based on the same data set
Click here to see answer by robertb(5830)  |
Question 1183970: In a study of brain waves during sleep, a sample of 29 college students were randomly separated into two groups. The first group had 15 people and each was given ½ litre of red wine before sleeping. The second group had 14 people and were given no alcohol before sleeping. All participants when to sleep at 11 PM and their brainwave activity was measured from 4-6 AM. The group drinking alcohol had a mean brainwave activity of 19.65 hertz and a standard deviation of 1.86 hertz. The group not drinking alcohol had a mean of 6.59 hertz and standard deviation of 1.91 hertz. Compute a 90% confidence interval for the difference in population means of groups drinking alcohol before sleeping and those not drinking alcohol before sleeping. Explain the meaning of the confidence interval. (10 points)
Click here to see answer by Boreal(15235) |
Question 1183968: A random sample of 732 judges found that 405 were introverts. Construct a 95% confidence interval for the proportion. Interpret the meaning of the confidence interval. Justify your use of a confidence interval based on a normal distribution for data regarding proportions that are normally following a binomial distribution. (10 points)
Click here to see answer by Boreal(15235) |
Question 1183978: Susan invested a total of $10000 in two accounts. One account pays 4 interest and the other account pays 6% Interest. The total retum from both accounts was $560.
Answer the following questions.
How much was invested at 4%?
How much was invested at 6%?
Click here to see answer by ikleyn(52898) |
Question 1183978: Susan invested a total of $10000 in two accounts. One account pays 4 interest and the other account pays 6% Interest. The total retum from both accounts was $560.
Answer the following questions.
How much was invested at 4%?
How much was invested at 6%?
Click here to see answer by greenestamps(13214)  |
Question 1184232: Using the 95% confidence level and the writing scores 91,91, 94, 91, 90, 81, 84, 91, 92, 91, 91, 98, 94, 91, 85, 91 you would expect the upper end of the confidence interval to be approximately what value?
Click here to see answer by Edwin McCravy(20064)  |
Question 1187550: Out of 300 people sampled, 177 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A.
Use a 90% confidence level, and give your answers as decimals, to three places.
_______ < P > ________
Click here to see answer by Theo(13342) |
Question 1190707: A medical researcher conducted a study to understand the efficacy of a new medicine. for the study he took a sample of 55 patients and found the margin of error for the estimate to be 12.As he aware that the margin of error is related to the size of sample being considered, if the researcher wants to have more precision in his measurement of efficacy by reducing the margin of error of estimate to 4,what is the sample size he needs to consider?
Click here to see answer by math_tutor2020(3817) |
Question 1190678: Parle Agro India Pvt. Ltd. wants to conduct a survey to find whether more people prefer Frooti to Slice. It was estimated before the survey that half of the population prefers Frooti and half prefers Slice. How large a sample would Parle Agro India Pvt. Ltd. have to take to estimate the proportion of people who prefer Frooti within +/- 0.03 of the actual value, with 98% confidence? considering Z value is 2
Click here to see answer by Boreal(15235) |
Question 1197051: a. A statistics practitioner calculated the mean and standard deviation from a sample of 51. They are x bar = 120 and s = 15. Estimate the population mean with 95% confidence.
b. Repeat part (a) with a 90% confidence level.
c. Repeat part (a) with an 80% confidence level.
d. What is the effect on the confidence interval estimate of decreasing the confidence level?
Click here to see answer by ewatrrr(24785) |
|
|
| |
