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Tutors Answer Your Questions about Confidence-intervals (FREE)
Question 1209947: You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 4.
While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 80.5%.
(Report answer accurate to three decimal places with appropriate rounding.)
ta/2 +/- =
Click here to see answer by CPhill(1959)   |
Question 1177421: A population consist of five persons those age are 10,20,30,40, and 50.
A random sample of size two is to be selected from the population with replacement.
1: list all the possible sample of two persons age computing the mean of each sample.
2: Obtain the sampling distribution of the mean.
3:Draw a line graph representing distribution and discribe it.
Click here to see answer by CPhill(1959) |
Question 1178025: Given the following observation in a simple random sample from a population that is approximately normally distributed. Construct and interpret a 90% confidence interval for the mean 67,79,71,98,74,70,59,102, 92,96
Click here to see answer by CPhill(1959) |
Question 1178026: An airline has surveyed a simple random sample of air travelers to find out whether they would be interested in paying a higher fare in order to have access to email during their flight:Of the 400 travelers surveyed, 80 said email access would be worth a slight extra cost. Construct a 95% confidence interval for the population proportion of air travelers to who are in favour of the airline's email idea.
Click here to see answer by CPhill(1959) |
Question 1179922: Doctors nationally believe that 70% of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At is this hospital's success rate different from the national average?
Click here to see answer by CPhill(1959) |
Question 1184225: Using the 955 confidence level scores 97,93, 95, 97, 97,94, 92, 92,98,92, 93,94, 95, 96, 96, 96 you would expect the lower end of the confidence interval to be approximately what value?
Click here to see answer by CPhill(1959) |
Question 1184231: Using the 95% confidence level and the writing scores 84, 80, 90, 90, 98, 94, 91, 86, 90, 91, 90 you would expect the lower end of the confidence interval to be approximately what value?
Click here to see answer by CPhill(1959) |
Question 1186540: The economic dynamism, which is the index of productive growth in dollars for countries that are designated by the World Bank as middle-income are in table #8.3.9 ("SOCR data 2008," 2013). Compute a 95% confidence interval for the mean economic dynamism of middle-income countries.
Table #8.3.9: Economic Dynamism ($) of Middle Income Countries
25.8057 37.4511 51.915 43.6952 47.8506 43.7178 58.0767
41.1648 38.0793 37.7251 39.6553 42.0265 48.6159 43.8555
49.1361 61.9281 41.9543 44.9346 46.0521 48.3652 43.6252
50.9866 59.1724 39.6282 33.6074 21.6643
Click here to see answer by CPhill(1959) |
Question 1186541: In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%.
Click here to see answer by CPhill(1959) |
Question 1191336: he personnel director of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers reveals the following:
• Absenteeism: = 9.7 days, S = 4.0 days.
• 12 clerical workers were absent more than 10 days.
a. Construct a 95% confidence interval estimate for the mean number of absences for clerical workers during the year.
b. Construct a 95% confidence interval estimate for the population proportion of clerical workers absent more than 10 days during the year. Suppose that the personnel director also wishes to take a survey in a branch office. Answer these questions:
c. What sample size is needed to have 95% confidence in estimating the population mean absenteeism to within ±1.5 days if the population standard deviation is estimated to be 4.5 days?
d. How many clerical workers need to be selected to have 90% confidence in estimating the population proportion to within ± 0.075 if no previous estimate is available?
e. Based on (c) and (d), what sample size is needed if a single survey is being conducted?
Click here to see answer by CPhill(1959) |
Question 1193448: Randomly selected 130 student cars have ages with a mean of 7.9 years and a standard deviation of 3.4 years, while randomly selected 95 faculty cars have ages with a mean of 5.4 years and a standard deviation of 3.3 years.
1. Use a 0.05 significance level to test the claim that student cars are older than faculty cars.
- The test statistic is ______. (5 sig. figs)
- The critical value is ______. (5 sig. figs)
2. Construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean age of student cars and μ2 is the mean age of faculty cars.
- ________ <(μ1−μ2)< _________ (5 sig. figs)
Click here to see answer by ElectricPavlov(122) |
Question 1207432: Kara surveyed 200 people, chosen at random on thompson drive, to try to determine what % of thompsonites are skiers. 86 people in her sample said they were skiers.
A.) What is the 95% confidence interval for the percent of Thompsonites who are skiers ? (Formula: 95% confidence interval = mean ± 1.96(standard deviation)
B.) Calculate the margin of error.
Click here to see answer by Theo(13342)  |
Question 1206865: Assuming the population has an approximate normal distribution, if a sample size
n=28 has a sample mean x¯=39 with a sample standard deviation s=2, find the margin of error at a 80% confidence level. Round the answer to two decimal places.
Click here to see answer by MathLover1(20849)  |
Question 1205082: Anthrax is a disease that broke out in Sinazongwe distric recently. Assume that the bacteria that causes anthrax attacks one in three of the people exposed to it. Assume further that the entire area of Sinazongwe district is exposed to the bacteria. If samples of 70 people of Sinazongwe are taken repeatedly, what would be:
A. Mean of the sampling distribution of sample proportions?
B. Standard deviation of the mean of the sampling distribution of sample proportions?
Click here to see answer by Boreal(15235)  |
Question 1205292: a shoe manufacturing company is producing 50,000 pairs of shoes daily. from a sample of 500 pairs, 2% are found to be of substandard quality. at 95% level of confidence, estimate the number of pairs of shoes that are reasonably expected to be spolied in the daily production.
Click here to see answer by Theo(13342) |
Question 1205116: Anthrax is a disease that broke out in Sinazongwe distric recently. Assume that the bacteria that causes anthrax attacks one in three of the people exposed to it. Assume further that the entire area of Sinazongwe district is exposed to the bacteria. If samples of 70 people of Sinazongwe are taken repeatedly, what would be:
Mean of the sampling distribution of sample proportions?
Standard deviation of the mean of the sampling distribution of sample proportions?
Click here to see answer by Theo(13342) |
Question 1205076: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions.
Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions .Interpret your answer in part a above?
Click here to see answer by Theo(13342) |
Question 1203076: The average math score for the class 6 is 23.6. A researcher is interested to check if the score is higher in class 6 that in the USA. How many students are needed to ensure that a two-sided test of hypothesis has 80% power to detect a difference in score of 2 marks? If the population standard deviation is 5.7.
Click here to see answer by Jason57t(3) |
Question 1201497: You wish to test the following claim (Ha) at a significance level of
α=0.02.
Ho:μ=73.2
Ha:μ>73.2
You believe the population is normally distributed and you know the standard deviation is σ=11.5. You obtain a sample mean of M=78.8for a sample of size
n=17.
What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =
What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =
The p-value is...
less than (or equal to) α
OR
greater than α
This test statistic leads to a decision to...
reject the null
OR
accept the null
OR
fail to reject the null
Click here to see answer by Theo(13342) |
Question 1201399: Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of the lifetime of the AAA batteries it produces. A random sample of 15 AAA batteries produced by this manufacturer lasted a mean of 9.5 hours with a standard deviation of 1.7 hours. Find a 95% confidence interval for the population standard deviation of the lifetimes of AAA batteries produced by the manufacturer. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
Click here to see answer by math_tutor2020(3816) |
Question 1201399: Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of the lifetime of the AAA batteries it produces. A random sample of 15 AAA batteries produced by this manufacturer lasted a mean of 9.5 hours with a standard deviation of 1.7 hours. Find a 95% confidence interval for the population standard deviation of the lifetimes of AAA batteries produced by the manufacturer. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
Click here to see answer by Theo(13342) |
Question 1201047: A survey showed that 73% of the random sample of 1506 people interviewed
favored drug tests for professional athletes. 68% said that professional athletics
using drug for the first time should be banned or suspended from the professional
sports. Find 95% confidence interval for the proportion of the public who favored
drug tests for professional athletes.
Click here to see answer by math_tutor2020(3816) |
Question 1200442: Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 203 with 39% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
Click here to see answer by Theo(13342) |
Question 1199321: In a city, out of 400 people sampled, 152 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.
Round your answers to three decimal places.
Click here to see answer by Theo(13342) |
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)  |
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 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(3816) |
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 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(20054)  |
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(13198)  |
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(52776)  |
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 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 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 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 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 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 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 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) |
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