Confidence interval?
Q. a sample of 10 observations is taken from a normal population for which the population standard deviation is known to be 5. the sample mean is 20. find the 95% confidence interval for the population mean
Asked by George - Tue May 27 17:07:00 2008 - - 1 Answers - 0 Comments
A. SMALL-Sample confidence interval for a poplation mean 95% Confidence Interval = x-bar +/- 1.96 * sigma x-bar = sample mean [20] sigma = population standard deviation [5] n = number of samples [10] 95% Confidence Interval: 20+/-1.96 *5 = [10.25 , 29.75] That is with a confidence interval of approximately 95% the "true mean" is within the interval of [10.25 , 29.75] and that the sample mean (which is an estimate of the "true mean") is 20.
Answered by M - Fri May 30 19:14:58 2008
Q. a sample of 10 observations is taken from a normal population for which the population standard deviation is known to be 5. the sample mean is 20. find the 95% confidence interval for the population mean
Asked by George - Tue May 27 17:07:00 2008 - - 1 Answers - 0 Comments
A. SMALL-Sample confidence interval for a poplation mean 95% Confidence Interval = x-bar +/- 1.96 * sigma x-bar = sample mean [20] sigma = population standard deviation [5] n = number of samples [10] 95% Confidence Interval: 20+/-1.96 *5 = [10.25 , 29.75] That is with a confidence interval of approximately 95% the "true mean" is within the interval of [10.25 , 29.75] and that the sample mean (which is an estimate of the "true mean") is 20.
Answered by M - Fri May 30 19:14:58 2008
confidence interval?
Q. When trying to estimate a population mean, when will we use a z-value based confidence interval? When do we use a t-value based confidence interval?
Asked by slim - Mon Jun 2 18:32:03 2008 - - 1 Answers - 0 Comments
A. "Rule of Thumb" (when specified) small sample size (n < 30): utilize "t - value" large sample size (n > 50): utilize "z - value" Consequences of misuse of "z - value" for small sample size include: - overstated variation ("sigma") - inflated size of Confidence Interval
Answered by M - Tue Jun 3 09:56:13 2008
Q. When trying to estimate a population mean, when will we use a z-value based confidence interval? When do we use a t-value based confidence interval?
Asked by slim - Mon Jun 2 18:32:03 2008 - - 1 Answers - 0 Comments
A. "Rule of Thumb" (when specified) small sample size (n < 30): utilize "t - value" large sample size (n > 50): utilize "z - value" Consequences of misuse of "z - value" for small sample size include: - overstated variation ("sigma") - inflated size of Confidence Interval
Answered by M - Tue Jun 3 09:56:13 2008
What happens to a confidence interval when you change the sample size?
Q. Statistics people, I need help! The question asks what would happen if you had a random sample of 50, then changed it to a sample of 200, what would be the effect on the confidence interval? I have the answer already because we are correcting answers we got wrong on this test, but I have to explain WHY it's true. The answer is: The width of the new interval would be about one-half the width of the original interval. Can anyone tell me why it is one-half, and not one-fourth like I thought?
Asked by lovetolaughh - Thu Apr 29 00:02:55 2010 - - 1 Answers - 0 Comments
A. The equation for the length of a confidence interval is equal to (2 * t(n-1, 1-a/2) * s)/(sqrt(n)) where t(n-1,1-a/2) is looked up in a t-distribution table, s is equal to the standard deviation, n is equal to the sample size, and a is the significance level. Since t(n-1, 1-a/2) and s are the same for both n=50 and n=200 because they use the same underlying data, those two parts of the equation can be assumed to be 1. Therefore, you have the following when the two n values are plugged in: 2/sqrt(n) 2/sqrt(50) = 0.283 2/sqrt(200) = 0.141 So the confidence interval of n=200 is about 1/2 the C.I. for n=50.
Answered by Todd - Thu Apr 29 00:18:33 2010
Q. Statistics people, I need help! The question asks what would happen if you had a random sample of 50, then changed it to a sample of 200, what would be the effect on the confidence interval? I have the answer already because we are correcting answers we got wrong on this test, but I have to explain WHY it's true. The answer is: The width of the new interval would be about one-half the width of the original interval. Can anyone tell me why it is one-half, and not one-fourth like I thought?
Asked by lovetolaughh - Thu Apr 29 00:02:55 2010 - - 1 Answers - 0 Comments
A. The equation for the length of a confidence interval is equal to (2 * t(n-1, 1-a/2) * s)/(sqrt(n)) where t(n-1,1-a/2) is looked up in a t-distribution table, s is equal to the standard deviation, n is equal to the sample size, and a is the significance level. Since t(n-1, 1-a/2) and s are the same for both n=50 and n=200 because they use the same underlying data, those two parts of the equation can be assumed to be 1. Therefore, you have the following when the two n values are plugged in: 2/sqrt(n) 2/sqrt(50) = 0.283 2/sqrt(200) = 0.141 So the confidence interval of n=200 is about 1/2 the C.I. for n=50.
Answered by Todd - Thu Apr 29 00:18:33 2010
How does this change the confidence level affect the length of the confidence interval?
Q. Suppose we consider a confidence intercal estimate for a mean, for some sample data. Suppose next we change ONLY the confidence interval. How does the length of the confidence intercal change, if at all? How does this change the confidence level affect the length of the confidence interval? Explain the connection.
Asked by Gandul-NCC1701 - Sat Mar 28 01:53:30 2009 - - 2 Answers - 0 Comments
A. So you're asking... 1.if we change only the confidence interval, how does the length of the confidence interval change??? What?? 2.How does the change in confidence level affect the length (or range) of a confidence interval? If we increase the confidence level (suppose 90% to 95%), the range of a confidence interval is increased. BECAUSE, in order for us to be MORE confident, we must make the range BIGGER so that we are less likely to be wrong.
Answered by alex austral - Sat Mar 28 09:57:58 2009
Q. Suppose we consider a confidence intercal estimate for a mean, for some sample data. Suppose next we change ONLY the confidence interval. How does the length of the confidence intercal change, if at all? How does this change the confidence level affect the length of the confidence interval? Explain the connection.
Asked by Gandul-NCC1701 - Sat Mar 28 01:53:30 2009 - - 2 Answers - 0 Comments
A. So you're asking... 1.if we change only the confidence interval, how does the length of the confidence interval change??? What?? 2.How does the change in confidence level affect the length (or range) of a confidence interval? If we increase the confidence level (suppose 90% to 95%), the range of a confidence interval is increased. BECAUSE, in order for us to be MORE confident, we must make the range BIGGER so that we are less likely to be wrong.
Answered by alex austral - Sat Mar 28 09:57:58 2009
How to calculate a confidence interval given only sample data?
Q. I'm given 20 data points from a sample in a population that is normally distributed and asked to calculate a 94% confidence interval. My question is what to do with standard deviation since it's not given. Can I still use a Z score in calculating margin of error or do I need to calculate a t-score, and if so, how do I do that and what is the formula?
Asked by Drew - Wed Dec 16 22:02:19 2009 - - 1 Answers - 0 Comments
A. I am sorry, i know what you are talking about but I do not know how to explain. have u tried some math sites? like regentsprep.org? or something? look in the internet for lessons, I think that would help
Answered by ZzzZzz - Wed Dec 16 22:07:41 2009
Q. I'm given 20 data points from a sample in a population that is normally distributed and asked to calculate a 94% confidence interval. My question is what to do with standard deviation since it's not given. Can I still use a Z score in calculating margin of error or do I need to calculate a t-score, and if so, how do I do that and what is the formula?
Asked by Drew - Wed Dec 16 22:02:19 2009 - - 1 Answers - 0 Comments
A. I am sorry, i know what you are talking about but I do not know how to explain. have u tried some math sites? like regentsprep.org? or something? look in the internet for lessons, I think that would help
Answered by ZzzZzz - Wed Dec 16 22:07:41 2009
How is a 95% confidence interval any different from the area under a curve plus/minus 1.96 std deviatiations?
Q. I have been using JMP in my biological statistics class and have been a little confused. I thought that since the area under a normal curve is 95% within 1.96 positive and negative standard deviations, that same area would be the 95% confidence interval. Is that the case? If not, how do I determine a confidence interval?
Asked by Jagamus Prime - Mon Feb 25 12:17:53 2008 - - 3 Answers - 0 Comments
A. When you are looking at the normal curve, you are finding an interval centered at the population mean and which contains 95% of the area under the curve. If the curve is the distribution for x-bar (with standard deviation sigma/sqrt(n) ), then the probability of a randomly chosen sample of size n having a sample mean in this interval is 0.95. But a confidence interval is centered at the SAMPLE mean. If you take the reasoning from the preceding paragraph and turn it inside out, so to speak, we can say that if we consider the probability experiment of drawing a sample of size n and then constructing a confidence interval in this manner, the probability of the interval enclosing the population mean is 0.95. In the first case, we are playing… [cont.]
Answered by Michael M - Mon Feb 25 13:26:42 2008
Q. I have been using JMP in my biological statistics class and have been a little confused. I thought that since the area under a normal curve is 95% within 1.96 positive and negative standard deviations, that same area would be the 95% confidence interval. Is that the case? If not, how do I determine a confidence interval?
Asked by Jagamus Prime - Mon Feb 25 12:17:53 2008 - - 3 Answers - 0 Comments
A. When you are looking at the normal curve, you are finding an interval centered at the population mean and which contains 95% of the area under the curve. If the curve is the distribution for x-bar (with standard deviation sigma/sqrt(n) ), then the probability of a randomly chosen sample of size n having a sample mean in this interval is 0.95. But a confidence interval is centered at the SAMPLE mean. If you take the reasoning from the preceding paragraph and turn it inside out, so to speak, we can say that if we consider the probability experiment of drawing a sample of size n and then constructing a confidence interval in this manner, the probability of the interval enclosing the population mean is 0.95. In the first case, we are playing… [cont.]
Answered by Michael M - Mon Feb 25 13:26:42 2008
How do you calculate a confidence interval and confidence levels?
Q. A market researcher wants to estimate the mean number of years of school completed by the residents of a particular neighborhood. A simple random sample of 90 residents is taken, the mean years of school completed being 8.4 and the standard deviation being 1.. The neighborhood contains about 2,500 residence. If the interval is found to be 8.057...8.743 what is the confidence level
Asked by christnogol dilynwr - Mon Apr 26 10:51:28 2010 - - 1 Answers - 0 Comments
A. The confidence interval is equal to (Test Statistic - critical value*standard error, Test Statistic + critical value*standard error). The Test statistic is the sample mean = 8.4 The standard deviation is the sample standard deviation divided by the square root of the number of people in the sample = 1 / sqrt(90) = .1054 Let X = critical value, Then our confidence interval of (8.057, 8.743) = (8.4 - .1054*X, 8.4 + .1.54x) Solving 8.4 - .1054*x = 8.057; X = 3.25427 We can find the confidence level by looking at a t distribution chart (or possibly a Z, usually you use a Z if sample is large, but "large" is not precisely defined). We use a t-chart with 89 degrees of freedom (sample size - 1). There is an area of .9992 to the left of 3. [cont.]
Answered by mikethechampion - Tue Apr 27 12:55:03 2010
Q. A market researcher wants to estimate the mean number of years of school completed by the residents of a particular neighborhood. A simple random sample of 90 residents is taken, the mean years of school completed being 8.4 and the standard deviation being 1.. The neighborhood contains about 2,500 residence. If the interval is found to be 8.057...8.743 what is the confidence level
Asked by christnogol dilynwr - Mon Apr 26 10:51:28 2010 - - 1 Answers - 0 Comments
A. The confidence interval is equal to (Test Statistic - critical value*standard error, Test Statistic + critical value*standard error). The Test statistic is the sample mean = 8.4 The standard deviation is the sample standard deviation divided by the square root of the number of people in the sample = 1 / sqrt(90) = .1054 Let X = critical value, Then our confidence interval of (8.057, 8.743) = (8.4 - .1054*X, 8.4 + .1.54x) Solving 8.4 - .1054*x = 8.057; X = 3.25427 We can find the confidence level by looking at a t distribution chart (or possibly a Z, usually you use a Z if sample is large, but "large" is not precisely defined). We use a t-chart with 89 degrees of freedom (sample size - 1). There is an area of .9992 to the left of 3. [cont.]
Answered by mikethechampion - Tue Apr 27 12:55:03 2010
How do I calculate the confidence interval?
Q. I have a data set of 380 customers, 28 of which did not pay. Using a 95% confidence level, how do I determine if the sample is statistically significant and/or what the confidence interval would be for a population of 10,000 customers?
Asked by Matt - Tue Aug 11 11:35:12 2009 - - 1 Answers - 0 Comments
A. ANSWER: 95% confidence interval [0.052, 0.096] (5.2%, 9.6%) Why??? population proportion, confidence interval, normal distribution 95% confidence interval p +/- (z critical value) * SQRT[p * (1 - p)/n] p = population proportion [0.074] (7.4%) 28/380 = 0.074 z critical value [1.645] from Table Look-up of Normal Distribution n = sample size [380] 95% confidence interval = 0.074+/- 1.645 * SQRT [ 0.074 * (1 - 0.074)/380] = 0.074 +/- 0.022 = [0.052, 0.096] (5.2%, 9.6%)
Answered by M - Tue Aug 11 16:15:49 2009
Q. I have a data set of 380 customers, 28 of which did not pay. Using a 95% confidence level, how do I determine if the sample is statistically significant and/or what the confidence interval would be for a population of 10,000 customers?
Asked by Matt - Tue Aug 11 11:35:12 2009 - - 1 Answers - 0 Comments
A. ANSWER: 95% confidence interval [0.052, 0.096] (5.2%, 9.6%) Why??? population proportion, confidence interval, normal distribution 95% confidence interval p +/- (z critical value) * SQRT[p * (1 - p)/n] p = population proportion [0.074] (7.4%) 28/380 = 0.074 z critical value [1.645] from Table Look-up of Normal Distribution n = sample size [380] 95% confidence interval = 0.074+/- 1.645 * SQRT [ 0.074 * (1 - 0.074)/380] = 0.074 +/- 0.022 = [0.052, 0.096] (5.2%, 9.6%)
Answered by M - Tue Aug 11 16:15:49 2009
What level of confidence did the statistician use in constructing this interval?
Q. On the basis of a survey of 545 television viewers, a statistician has constructed a confidence interval and estimated that the proportion of people who watched the season finale of Lost is between .16 and .24. What level of confidence did the statistician use in constructing this interval?
Asked by Det - Tue Jun 1 23:49:29 2010 - - 1 Answers - 0 Comments
A. ANSWER: 98% Confidence Interval Why??? population proportion, confidence interval, normal distribution, LOOK-Up proportion from normal distribution table 98% Resulting Confidence Interval for 'true mean': p +/- (z critical value) * SQRT[p * (1 - p)/n] p = 0.2 [mid-point of interval 0.16 and 0.24] (z critical value) = 0.4/SQRT[p * (1 - p)/n] = (SQRT[0.2 * (1 - 0.2)/545] = 2.33 LOOK-UP VALUE z critical = 2.33 [98%]
Answered by M - Wed Jun 2 21:03:05 2010
Q. On the basis of a survey of 545 television viewers, a statistician has constructed a confidence interval and estimated that the proportion of people who watched the season finale of Lost is between .16 and .24. What level of confidence did the statistician use in constructing this interval?
Asked by Det - Tue Jun 1 23:49:29 2010 - - 1 Answers - 0 Comments
A. ANSWER: 98% Confidence Interval Why??? population proportion, confidence interval, normal distribution, LOOK-Up proportion from normal distribution table 98% Resulting Confidence Interval for 'true mean': p +/- (z critical value) * SQRT[p * (1 - p)/n] p = 0.2 [mid-point of interval 0.16 and 0.24] (z critical value) = 0.4/SQRT[p * (1 - p)/n] = (SQRT[0.2 * (1 - 0.2)/545] = 2.33 LOOK-UP VALUE z critical = 2.33 [98%]
Answered by M - Wed Jun 2 21:03:05 2010
How do you find a 95% confidence interval for the mean interval?
Q. Quality control studies for speedy jet computer printers show the lifetime of the printer follows a normal distribution with average lifetime of 2.75 years and standard deviation s=0.61, for 14 printers randomly selected. how can i find the 95% confidence interval for the mean lifetime?? Please explain! Thank you so much!
Asked by FeFe - Thu Nov 27 18:22:42 2008 - - 1 Answers - 0 Comments
A. Use the formula mean z*s/sqrt(n) for 95% confidence interval, z = 1.96 2.75 1.96*(0.61)/sqrt(14) 2.75 0.3195
Answered by Josh S - Thu Nov 27 18:33:51 2008
Q. Quality control studies for speedy jet computer printers show the lifetime of the printer follows a normal distribution with average lifetime of 2.75 years and standard deviation s=0.61, for 14 printers randomly selected. how can i find the 95% confidence interval for the mean lifetime?? Please explain! Thank you so much!
Asked by FeFe - Thu Nov 27 18:22:42 2008 - - 1 Answers - 0 Comments
A. Use the formula mean z*s/sqrt(n) for 95% confidence interval, z = 1.96 2.75 1.96*(0.61)/sqrt(14) 2.75 0.3195
Answered by Josh S - Thu Nov 27 18:33:51 2008
How do i calculate the confidence interval?
Q. I have 100 patients in sample men score of 80 and standard deviation of 20 what is the 95% confidence interval?
Asked by Christian S - Tue Feb 13 05:23:32 2007 - - 6 Answers - 0 Comments
A. do your own homework?
Answered by Cheese A - Tue Feb 13 05:26:12 2007
Q. I have 100 patients in sample men score of 80 and standard deviation of 20 what is the 95% confidence interval?
Asked by Christian S - Tue Feb 13 05:23:32 2007 - - 6 Answers - 0 Comments
A. do your own homework?
Answered by Cheese A - Tue Feb 13 05:26:12 2007
How do you calculate a confidence interval?
Q. Could you explain it to me how to calculate the confidence interval, and give me an example? Thanks.
Asked by SadWoman - Sat Sep 1 13:17:19 2007 - - 2 Answers - 0 Comments
A. Suppose you wanted to estimate the mean, ue, of a population of college placement exams. You take a random sample of n = 36 student's scores and find the mean of these scores, Xbar = 75 with a standard deviation, s = 10. Find the - 1 = 95% confidence interval around ue. = 0.05 and /2 = 0.025 The general formula: Xbar Z /2[s/ n], Where Z /2 =1.96 is found in the standard normal table for a probability of 0.025 Evaluating: 75 1.96[10/6] and the interval is: 71.73 < ue < 78.27 If your sample size is less than 30, then, instead of Z /2, use a t /2 found in a t-table for /2 and n - 1 degrees of freedom.
Answered by cvandy2 - Sat Sep 1 13:57:16 2007
Q. Could you explain it to me how to calculate the confidence interval, and give me an example? Thanks.
Asked by SadWoman - Sat Sep 1 13:17:19 2007 - - 2 Answers - 0 Comments
A. Suppose you wanted to estimate the mean, ue, of a population of college placement exams. You take a random sample of n = 36 student's scores and find the mean of these scores, Xbar = 75 with a standard deviation, s = 10. Find the - 1 = 95% confidence interval around ue. = 0.05 and /2 = 0.025 The general formula: Xbar Z /2[s/ n], Where Z /2 =1.96 is found in the standard normal table for a probability of 0.025 Evaluating: 75 1.96[10/6] and the interval is: 71.73 < ue < 78.27 If your sample size is less than 30, then, instead of Z /2, use a t /2 found in a t-table for /2 and n - 1 degrees of freedom.
Answered by cvandy2 - Sat Sep 1 13:57:16 2007
What impact do a large quantity of low values do to a confidence interval?
Q. For example, say we have a confidence interval for the sales prices of houses of $90,000 to $160,000. What impact would a lot of $0 sales do to the confidence interval?
Asked by byap85 - Sun Mar 16 20:16:22 2008 - - 1 Answers - 0 Comments
A. It would mess it up pretty bad. No more offering to buy lunches, no more taking the kids out to the movies EVERY weekend, no more just joyriding, buying generic instead of name-brand. Whoa! Don't get me started on what the love-life would be like at home. Sheesh! I hope this helps you.
Answered by Generically_Speaking - Sun Mar 16 20:28:52 2008
Q. For example, say we have a confidence interval for the sales prices of houses of $90,000 to $160,000. What impact would a lot of $0 sales do to the confidence interval?
Asked by byap85 - Sun Mar 16 20:16:22 2008 - - 1 Answers - 0 Comments
A. It would mess it up pretty bad. No more offering to buy lunches, no more taking the kids out to the movies EVERY weekend, no more just joyriding, buying generic instead of name-brand. Whoa! Don't get me started on what the love-life would be like at home. Sheesh! I hope this helps you.
Answered by Generically_Speaking - Sun Mar 16 20:28:52 2008
Statistics: What is degrees of freedom & when is it used? How do you calculate the confidence interval?
Q. Statistics: What is degrees of freedom & when is it used? How do you calculate the confidence interval? I do not know what df means, what its purpose is, and how to recognize when to use it. I also don't know how to calculate confidence intervals when given sample data values or a sample summary (like how many successes, how many sampled, etc). And from what I've read, degrees of freedom does or at least can relate to the confidence interval--I would like to know how.
Asked by Steven C. - Wed May 5 01:30:52 2010 - - 1 Answers - 0 Comments
A. The definition of degrees of freedom can be found on For myself, it is the difference between the number of observations or sample size n and the number of parameters estimated. For example when you're doing single parameter estimation, (test mean height of koalas = 10 inches using the average of 50 observations as the mean estimate) therefore your degrees of freedom is 50-1 = 49. There is only one parameter that is the mean. From the degrees of freedom, you could use it to find the critical t value for the confidence interval using the formula average plus minus (critical t times standard error). What is confidence interval. It is an interval within which you are sure containing the parameter. For example if you have a 95%… [cont.]
Answered by Lost in Translation - Thu May 6 09:03:44 2010
Q. Statistics: What is degrees of freedom & when is it used? How do you calculate the confidence interval? I do not know what df means, what its purpose is, and how to recognize when to use it. I also don't know how to calculate confidence intervals when given sample data values or a sample summary (like how many successes, how many sampled, etc). And from what I've read, degrees of freedom does or at least can relate to the confidence interval--I would like to know how.
Asked by Steven C. - Wed May 5 01:30:52 2010 - - 1 Answers - 0 Comments
A. The definition of degrees of freedom can be found on For myself, it is the difference between the number of observations or sample size n and the number of parameters estimated. For example when you're doing single parameter estimation, (test mean height of koalas = 10 inches using the average of 50 observations as the mean estimate) therefore your degrees of freedom is 50-1 = 49. There is only one parameter that is the mean. From the degrees of freedom, you could use it to find the critical t value for the confidence interval using the formula average plus minus (critical t times standard error). What is confidence interval. It is an interval within which you are sure containing the parameter. For example if you have a 95%… [cont.]
Answered by Lost in Translation - Thu May 6 09:03:44 2010
What is the confidence interval for the mean of the meass measurement would be zero?
Q. THe glass of ten glass beads was measured five times instead of using five sets of ten randomly selcted glass beads. Would you expect that the confidence interval for the mean of the mass emasurement would be zero? how come.
Asked by nedjine05 - Tue Sep 19 20:51:48 2006 - - 2 Answers - 0 Comments
A. Even if the same individual measured the same ten glass beads all five times, chances are that there would be some variance in the measurement. You didn't specify the type of measurement, but here are some factors to consider. An instrument will fluctuaute in its readings due to changes in humidity, temperature, back ground electric radiation (wires, radio, computer, etc.), sample preparation, and other input variances. If using a simple ruler to measure diameter, the observer will not see and measure each glass bead exactly the same way every time. The use of a micrometer will improve accuracy, but deciding when to take the measurement is subjective.
Answered by L96vette - Wed Sep 20 04:43:13 2006
Q. THe glass of ten glass beads was measured five times instead of using five sets of ten randomly selcted glass beads. Would you expect that the confidence interval for the mean of the mass emasurement would be zero? how come.
Asked by nedjine05 - Tue Sep 19 20:51:48 2006 - - 2 Answers - 0 Comments
A. Even if the same individual measured the same ten glass beads all five times, chances are that there would be some variance in the measurement. You didn't specify the type of measurement, but here are some factors to consider. An instrument will fluctuaute in its readings due to changes in humidity, temperature, back ground electric radiation (wires, radio, computer, etc.), sample preparation, and other input variances. If using a simple ruler to measure diameter, the observer will not see and measure each glass bead exactly the same way every time. The use of a micrometer will improve accuracy, but deciding when to take the measurement is subjective.
Answered by L96vette - Wed Sep 20 04:43:13 2006
How do you construct a confidence interval?
Q. Construct a 99% confidence interval for the population mean of 910.8 SD=303.4 and please explain how you did it so I can understand. Thank you!
Asked by VAMP - Tue Sep 8 13:21:57 2009 - - 1 Answers - 0 Comments
A. ANSWER: 99% Confidence Interval = [128, 1694] normal distribution, standardized variable z, probability "LOOK-UP" P = 0.99 P = normal distribution probability [0.99] "LOOK-UP" Table shows "z" = (approx.) 2.58 z = (x - )/ z = standardized variable for normal distribution [2.58] = population mean [910.8] = population standard deviation [303.4] solve for x = independent variable upper limit: x = * z + [1694] (303.4 * 2.58 + 910.8) lower limit: x = * -z + [128] (303.4 *(-2.58) + 910.8)
Answered by M - Wed Sep 9 21:49:46 2009
Q. Construct a 99% confidence interval for the population mean of 910.8 SD=303.4 and please explain how you did it so I can understand. Thank you!
Asked by VAMP - Tue Sep 8 13:21:57 2009 - - 1 Answers - 0 Comments
A. ANSWER: 99% Confidence Interval = [128, 1694] normal distribution, standardized variable z, probability "LOOK-UP" P = 0.99 P = normal distribution probability [0.99] "LOOK-UP" Table shows "z" = (approx.) 2.58 z = (x - )/ z = standardized variable for normal distribution [2.58] = population mean [910.8] = population standard deviation [303.4] solve for x = independent variable upper limit: x = * z + [1694] (303.4 * 2.58 + 910.8) lower limit: x = * -z + [128] (303.4 *(-2.58) + 910.8)
Answered by M - Wed Sep 9 21:49:46 2009
How does the percentage of Confidence Interval effect the equation?
Q. Was just doing Statistics, got help with a question but didnt see how the 98% confidence interval came into play. The question was Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p n=97 x=46 98% confidence. was wondering how my answer would vary had it been 95% confidence, or 99% or anything else for that matter. The person who answered it did a great job showing me how to get the formula, but Im still unsure about where the 98% came in.
Asked by BeAuTiFuL dIsAsTeR - Mon Mar 8 01:40:13 2010 - - 1 Answers - 0 Comments
A. As the percentage becomes smaller, we are less and less concerned that our statistic is within the interval, so the interval becomes smaller. Conversely, as the percentage becomes larger, we want to be more sure, and we are forced to include more and more really unlikely values, that is, our interval becomes larger. Think about some trivial cases, just to help you remember. How big will the interval have to be to be 100% sure, say, for a normal distribution? Well, you can't really 100% rule out any value, so you have to include them all, i.e, you need (-infinity, infinity). So, the largest possible percentage results in the largest possible interval. Consider also, if we take the smallest possible "interval" (it depends on your… [cont.]
Answered by alwbsok - Mon Mar 8 01:57:15 2010
Q. Was just doing Statistics, got help with a question but didnt see how the 98% confidence interval came into play. The question was Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p n=97 x=46 98% confidence. was wondering how my answer would vary had it been 95% confidence, or 99% or anything else for that matter. The person who answered it did a great job showing me how to get the formula, but Im still unsure about where the 98% came in.
Asked by BeAuTiFuL dIsAsTeR - Mon Mar 8 01:40:13 2010 - - 1 Answers - 0 Comments
A. As the percentage becomes smaller, we are less and less concerned that our statistic is within the interval, so the interval becomes smaller. Conversely, as the percentage becomes larger, we want to be more sure, and we are forced to include more and more really unlikely values, that is, our interval becomes larger. Think about some trivial cases, just to help you remember. How big will the interval have to be to be 100% sure, say, for a normal distribution? Well, you can't really 100% rule out any value, so you have to include them all, i.e, you need (-infinity, infinity). So, the largest possible percentage results in the largest possible interval. Consider also, if we take the smallest possible "interval" (it depends on your… [cont.]
Answered by alwbsok - Mon Mar 8 01:57:15 2010
What is the upper end point in a 99% confidence interval for the average income?
Q. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the upper end point in a 99% confidence interval for the average income?
Asked by Kitty - Wed Dec 31 05:49:55 2008 - - 1 Answers - 0 Comments
A. 15000+(2.5758*1000) = 17600 (3 significant figures)
Answered by Chuck Norris - Wed Dec 31 07:45:42 2008
Q. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the upper end point in a 99% confidence interval for the average income?
Asked by Kitty - Wed Dec 31 05:49:55 2008 - - 1 Answers - 0 Comments
A. 15000+(2.5758*1000) = 17600 (3 significant figures)
Answered by Chuck Norris - Wed Dec 31 07:45:42 2008
How to calculate the confidence interval?
Q. We are measuring the IQ of students at a college. We choose a sample of 30 students, and find for the same a mean IQ of 118.3, with a standard deviation of 11.4. Find the 95% confidence interval for the population mean. I have the answer [114.221 to 122.379] but I want to know the step by step formula to use. I don't know the statistical symbols so if someone can write it out for me I would greatly appreciate it.
Asked by Will B - Thu Aug 23 19:24:27 2007 - - 2 Answers - 0 Comments
A. Hi, Sounds as if you want a blow-by-blow account, so I'll try to give it to you. First of all, technically, to use the normal distribution, we need n>30 and the population standard deviation. So, even though it going to make very little difference, we'll use the Student t distribution. Secondly, because of the limitations of this word processor, I need to define a couple terms: a) Let a (should be alpha) be 1-degree of confidence, 1-.95 = .05. b) Let t(a/2) be the t-value for a two-tailed distribution. c) x-bar is the sample mean, an x with a bar over it. Now, let's do the work. 1) The maximum error is: E = t(a/2) * s/sqrt(n) We know that s = 11.4 and n = 30, but we need t(a/2). 2) To find t(a/2), we look in a Student t… [cont.]
Answered by formeng - Thu Aug 23 21:38:43 2007
Q. We are measuring the IQ of students at a college. We choose a sample of 30 students, and find for the same a mean IQ of 118.3, with a standard deviation of 11.4. Find the 95% confidence interval for the population mean. I have the answer [114.221 to 122.379] but I want to know the step by step formula to use. I don't know the statistical symbols so if someone can write it out for me I would greatly appreciate it.
Asked by Will B - Thu Aug 23 19:24:27 2007 - - 2 Answers - 0 Comments
A. Hi, Sounds as if you want a blow-by-blow account, so I'll try to give it to you. First of all, technically, to use the normal distribution, we need n>30 and the population standard deviation. So, even though it going to make very little difference, we'll use the Student t distribution. Secondly, because of the limitations of this word processor, I need to define a couple terms: a) Let a (should be alpha) be 1-degree of confidence, 1-.95 = .05. b) Let t(a/2) be the t-value for a two-tailed distribution. c) x-bar is the sample mean, an x with a bar over it. Now, let's do the work. 1) The maximum error is: E = t(a/2) * s/sqrt(n) We know that s = 11.4 and n = 30, but we need t(a/2). 2) To find t(a/2), we look in a Student t… [cont.]
Answered by formeng - Thu Aug 23 21:38:43 2007
For the same sample size and population standard deviation, the confidence interval becomes narrower when the?
Q. For the same sample size and population standard deviation, the confidence interval becomes narrower when the probability of Type I error increases. True or False? Please let me know what website you found this information in. Thank you so much in advance!
Asked by pattyboop - Sun Oct 18 23:49:54 2009 - - 1 Answers - 0 Comments
A. True.
Answered by Hamad - Mon Oct 19 19:17:38 2009
Q. For the same sample size and population standard deviation, the confidence interval becomes narrower when the probability of Type I error increases. True or False? Please let me know what website you found this information in. Thank you so much in advance!
Asked by pattyboop - Sun Oct 18 23:49:54 2009 - - 1 Answers - 0 Comments
A. True.
Answered by Hamad - Mon Oct 19 19:17:38 2009
From Yahoo Answer Search: 'confidence interval'
Tue Jul 20 08:18:16 2010 [ refresh local cache ]
[Hide]▼
nature.com
Thu, 22 Jul 2010 15:52:04 GMT+00:00
Nature.com Shaded areas indicate the 95% confidence intervals , and rugs below and above the graph represent the distribution of the body mass data for <2000 and 2000 ...
cs.princeton.edu
44px x 223px | 2.80kB
[source page]
measurements are statistically independent 2 has a Chi square distribution with 2n degrees of freedom This enables us to compute a 95 confidence interval for 2 as follows where 22n is the quantile of a Chi square distribution with 2n degrees of freedom
ukbankswebsites.com
admin
Mon, 31 May 2010 18:25:32 GM
The time that customers have to queue for service at a bank is measured. From a sample of 100 customers, it is found that the mean waiting time is 4 minutes with a standard deviation of 2 minutes. What is the 95% . confidence interval. for ...
[Hide]▲
