CONFIDENCE INTERVALS?
Q. In a survey of 200 household, 72 had central air conditioning. Find a 90% confidence interval for the proportion of homes WITH central air conditioning. How would you go about solving this question? Thank you in advance
Asked by Katarina D - Tue Jun 15 11:54:19 2010 - - 1 Answers - 0 Comments

A. Let p the sample proportion of households with air condition. p = 72 / 200 = 0.36 Test Statistic: p z [p(1 - p) / n] 0.36 1.645 [0.36(1 - 0.36) / 200] =(0.304 , 0.4158) We are 90% confident that the true proportion of households with central air conditioning is between 0.304 and 0.4158
Answered by The Calculus Alchemist - Tue Jun 15 12:02:48 2010

How do I compute confidence intervals?
Q. I need to develop 95% confidence intervals for the mean age and household income of a survey I am analyzing. I'm massively confused? Where do I start and what are the appropriate formulas to use? Thanks.
Asked by Jason C - Mon Jun 4 13:38:56 2007 - - 1 Answers - 0 Comments

A. See this link:
Answered by Jason H - Mon Jun 4 15:45:01 2007

Can a quantile be outside confidence intervals?
Q. Hello, I have found 95% confidence intervals for a quantile in exponential distribuition. But the intervals I have arrived at suggests that the quantile is outside the 95% confidence intervals. Can anyone throw some light on whether this is true? i.e quantile can be outside a particular confidence interval? Advance thanks please
Asked by harsha j - Thu Nov 20 06:53:06 2008 - - 2 Answers - 0 Comments

A. Well a 95% confidence interval operates on the basis that 95% of sampled data are going to fall into this range so whilst it is unlikely as you increase your confidence level it is still going to happen occasionally
Answered by AeternusVeritas - Thu Nov 20 08:56:59 2008

In statistics why are estimations and confidence intervals important?
Q. Why are estimations and confidence intervals important?
Asked by GinnyGirl - Wed Feb 25 14:20:01 2009 - - 1 Answers - 0 Comments

A. Estimations are an inferential statistic that are used when you know there is an effect (or you found an effect) and you want to figure out the size of the effect in an unknown population. The importance is that you do not have all the information to actually quantify so you must use an estimate to infer information about the population. Confidence intervals are important to qualify your estimation. Without a confidence interval an estimation is meaningless.
Answered by jackpock - Thu Feb 26 02:24:38 2009

How do I calculate standard error when I only have OR (risk estimate) and confidence intervals?
Q. How do I calculate st error when I only have OR (risk estimate) and confidence intervals? Is it true you do something like this se=max-min/1.96 OR se=max-min/1.96 x2 If neither is right how do I do it???
Asked by lynnio - Thu May 22 21:16:13 2008 - - 1 Answers - 0 Comments

A. I assume you are using Normal Distribution with a 2 sided 95% confidence interval to get 1.96 . So you have max = mean + 1.96 * se min = mean - 1.96 * se subtracting these 2 equations gives max - min = mean+1.96 se - (mean -1.96 * se) = 2 *1.96 * se so se = (max - min) / (1.96 * 2), so it looks like your second answer is right
Answered by manic - Fri May 23 21:02:42 2008

How do i get the low end and high end confidence intervals for this problem?
Q. A quality Assurance technician tested 10 samples of a plastic to determine the crystallinity index. She calculated that the mean of the sample was 6.1 with a sample standard deviation of 0.5. Find a 95% confidence interval for the mean crystallinity of all of the plastic.
Asked by Dal - Thu May 6 23:12:41 2010 - - 1 Answers - 0 Comments

A. ANSWER: 95% Resulting Confidence Interval for 'true mean': = [5.74, 6.46] Why??? small sample, confidence interval, normal population distribution x-bar = Sample mean 6.1 s = Sample standard deviation0.5 n = Number of samples 10 df = degrees of freedom 9 significant digits2 Confidence Level95 "Look-up" Table 't-critical value'2.26 Look-up Table of t critical values for confidence and prediction intervals. Central two-side area = 95% with df = 9. Another Look-up method is to utilize Microsoft Excel function: TINV(probability,degrees_ freedom) Returns the inverse of the Student's t-distribution 95% Resulting Confidence Interval for 'true mean': x-bar +/- ('t critical value') * s/SQRT(n)= 6.1 +/- 2.26 * 0.5/SQRT(10) = [5.74, 6.4 [cont.]
Answered by M - Sat May 8 20:27:11 2010

Statistics and Confidence Intervals: Can you help me with this question?
Q. A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 200 individuals who had been taking the drug is questioned; 156 of them found relief from depression. Based upon this, compute a 95% confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. Thanks for any help guys!
Asked by Jones G - Tue May 26 22:16:17 2009 - - 1 Answers - 0 Comments

A. ANSWER: 95% confidence interval for individuals taking drug feeling relief = [0.732, 0.828] (73.2%, 82.8%) population proportion, confidence interval, normal distribution 95% confidence interval p +/- (z critical value) * SQRT[p * (1 - p)/n] p = population proportion [0.780] (156/200 = 78%) z critical value [1.645] n = sample size [200] 95% confidence interval = 0.780 +/- 1.645 * SQRT [ 0.780 * (1 - 0.780) / 200] = [0.732, 0.828] (73.2%, 82.8%) conclusion: 95% confidence interval of true population proportion = [0.732, 0.828] (73.2%, 82.8%). Drug manufacturer can claim to 95% confidence, sample statistic of individuals taking the drug find relief is within this confidence interval.
Answered by M - Sat May 30 05:21:22 2009

Hi evry1, im really stuck with this question for confidence intervals as i dont have a sample stand. deviation?
Q. A study showed that out of 500 people that took a weight loss drug, 70 developed side effects.Compute a 95% confidence interval for the proportion of people that developed side effects after taking the drug. How do i do this without a standard deviation of any kind? Thanks so much!
Asked by Carla - Tue Aug 11 08:49:05 2009 - - 1 Answers - 0 Comments

A. ANSWER: Manufacturer can claim to a 95% confidence interval only (11.4% , 16.6%) of people will develop side effects after taking the drug. Why??? population proportion, confidence interval, normal distribution 95% confidence interval p +/- (z critical value) * SQRT[p * (1 - p)/n] p = population proportion [0.14] (14%) 70/500 = 0.14 z critical value [1.645] from Table Look-up Normal Distribution n = sample size [500] 95% confidence interval = 0.14+/- 1.645 * SQRT [ 0.14 * (1 - 0.14)/500] = [0.114,0.166] (11.4% , 16.6%) conclusion: 95% confidence interval of true population proportion = [0.114, 0.166] (11.4% , 16.6%) Manufacturer can claim to a 95% confidence interval only (11.4% , 16.6%) of people will develop side effects after… [cont.]
Answered by M - Tue Aug 11 12:55:46 2009

How to find confidence intervals?
Q. So I am trying to solve the following problem, but I'm having a hard time because I don't really understand how to find confidence intervals. Could someone help enlighten me a bit? =) Problem: "A survey of 49 people revealed that the mean number of minutes a person talks on his or her phone in one day in a particular locality is 26. Assume the population standard deviation is 2.3. Give a 99% confidence interval for the average daily usage."
Asked by Marlo - Thu Jun 11 00:04:12 2009 - - 1 Answers - 0 Comments

A. The formula for confidence intervals is: sample mean +/- z*standard deviation / sqrt(sample size) For a 99% confidence interval, z = 2.565 Therefore, your confidence interval is: 26 +/- 2.565*2.3/sqrt(49) = [25.16, 26.84] You can be 99% confident that the true mean number of minutes a person talks on his or her phone per day is between 25.16 minutes and 26.84 minutes.
Answered by Guy - Thu Jun 11 01:48:03 2009

please help me on this statistical analysis question - confidence intervals?
Q. A random sample of size n = 144 yielded p = 0.76 (i) Find a 95% confidence interval for p. (ii) Interpret the 95% confidence interval. I would appreciate it if someone could give me a thorough walkthrough of the method needed to answer this question including how to use the appropriate tables. Thank you.
Asked by Arthur - Thu Mar 4 18:10:23 2010 - - 1 Answers - 0 Comments

I need help on a stats problem that involves confidence intervals?
Q. ok heres the problem...You have a 6 sided numbered die(numbers 1-6) you want to roll a six a total of three times. How many times do you need to roll the die so that you have a confidence interval of at least 95% that you will roll a six, three times. Thanks for your help!
Asked by katya K - Tue Apr 8 16:30:02 2008 - - 1 Answers - 0 Comments

A. As I understand the question you have a fair die and roll it three times. This is one experiment. How many experiments do you have to conduct so that you have a 95% probability of seeing 3 6's? Well... the probability of rolling 3 6's is 1/216 P( at least one 666 ) = 1 - P( no 666) = 1 - ( 1 - 1/216 ) ^ n > 0.95 0.05 > (1 - 1/216)^n log(0.05) > n * log(215/216) not that log(215/216) < 0 so reverse the inequality when dividing: n > log(0.05) / log(215/216) n > 645.5791 n must be integer valued. always take the ceiling so that the probability is correct. n 646 If you check this buy letting X be the number of experiments until the first success of rolling 666, then you have X ~ Geometric( 1/216 ) Looking at the cumulative… [cont.]
Answered by Merlyn - Wed Apr 9 20:31:14 2008

1. If you were to construct 500 95% confidence intervals for a population mean, how many of them would we ex?
Q. 1 If you were to construct 500 95% confidence intervals for a population mean, how many of them would we expect to contain ue? Give a reason for your answer. How many would contain _ X ?
Asked by kris h - Wed Nov 25 23:57:33 2009 - - 1 Answers - 0 Comments

A. ask in stat or maths section.
Answered by v_anandpatin - Sat Nov 28 23:54:10 2009

how do i calculate confidence intervals in MS excel ?
Q. I have 4 groups of numbers for a quality score - calculating the mean score is easy enough but I need 95% confidence intervals, I know there is a NORMINV function that can be used for this purpose ... how do I use this function to calculate what I need ? Is there any other way of doing it ?
Asked by shehan s - Tue Oct 3 16:06:49 2006 - - 2 Answers - 0 Comments

A. This link should take you through what you need to do. Good luck!
Answered by Jemima - Tue Oct 3 16:10:52 2006

what is the difference in calculating confidence intervals for different percents?
Q. for example i have to construct a 90% confidence interval for the population mean and i have to construct a 95% and 99% one as well. Whats the difference in calculating them???
Asked by craterface - Tue Jul 29 15:59:41 2008 - - 2 Answers - 0 Comments

A. Suppose I know that something is normally distributed about a mean m with a standard deviation of s. I know that there's a 68.26% chance that it lies within m s. I know that there's a 95.44% chance that it lies within m 2s. I know that there's a 99.74% chance that it lies within m 3s. I got those numbers by looking at a Standard-Normal table. Okay? so if m were, say, 4.5 and s were 0.5, then my 68.26% confidence interval would be [4, 5], my 95.44% confidence interval would be [3.5, 5.5], and my 99.74% confidence interval would be [3, 6]. I'm almost completely sure that it will be somewhere between 3 and 6, and it will probably be somewhere between 4 and 5 -- but only probably. Now, there are all sorts of different random… [cont.]
Answered by Drostie - Tue Jul 29 16:15:24 2008

how do political polls calculate confidence intervals?
Q. When statistical techniques are used to calculate confidence intervals for political polls?
Asked by GES - Thu Feb 7 23:59:39 2008 - - 1 Answers - 0 Comments

A. Say that p is the true proportion, expressed as a fraction, of those who support political party A. This is what you would like to know but can't know with certainty until there is a real election. You ask a number of people whether they support A and suppose m say yes out of n definite replies. (Ignore don't knows and people who don't want to tell you.) If you did the sampling process many times you would get different values of m for the same n value. This variation would be expected to relate to the Binomial distribution which can be approximated by the Normal distribution with large numbers such as >1000 in political sampling. The mean of the Binomial is np and the standard deviation sqrt[np(1 - p)]. Of course we don't know p but m/ [cont.]
Answered by mathsmanretired - Fri Feb 8 11:32:06 2008

Can you combine 2 different confidence intervals?
Q. E.G. For catalog sales, $ per book depends on two things. The proportion of people who order from the mailed catalogs and the average $ per order. I know how to find the confidence interval for proportions and how to find the confidence interval for averages. Is there a formula for the confidence interval of a mixed calculation?
Asked by MusicMan10 - Tue Jun 20 15:37:04 2006 - - 3 Answers - 0 Comments

A. If you want to compute confidence intervals in this case, you will want to do a multiple regression. You cannot combine confidence intervals in general anyway, but espcially since both variables have different units, it doesn't make sense to do, say, a two sample confidence interval.
Answered by blahb31 - Tue Jun 20 15:41:07 2006

If the statistical mean falls between two negitive 95% confidence intervals, can it be considered usual?
Q. ie. if the statistical mean is -3 and the upper interval is -2 and the lower interval is -6 , can this be considered usual?
Asked by tigeress - Fri Mar 14 00:13:21 2008 - - 1 Answers - 0 Comments

A. If ue is inside the 95% confidence limits, then the interval contains ue with probability 95% (so it can be trusted for 95%)
Answered by edward - Sat Mar 15 11:19:56 2008

What effect does standard deviation have on confidence intervals?
Q. What effect does standard deviation have on confidence intervals?
Asked by St. Nick - Sun Aug 3 00:14:52 2008 - - 1 Answers - 0 Comments

A. A confidence interval for a mean is given by u z where u denotes the estimated mean (should be mu-bar but I can't find that symbol), z is the z-score corresponding to the confidence level, and is the standard deviation. If increases, then the length of the confidence interval increases and if decreases, then the length of the confidence interval decreases. Maybe this is easier to understand - in effect, if the data has a small variance (standard deviation squared) you would expect to be more confident given the same range for data with a larger variance. Hope this helps you!
Answered by Amy J - Sun Aug 3 00:57:02 2008

Help me with confidence intervals and statistics?
Q. I need help solving this problem-- you don't have to completely answer it for me, but a walk-through would be nice :) "Obesity is defined as a body mass index (BMI) of 3 kg/m^2 or more. A 95% confidence interval for the percentage of U.S. adults aged 20 years and over who were obese was found to be 22.4 % to 23.5 %. What was the sample size?" Thanks!
Asked by jules - Sun Jul 12 19:01:13 2009 - - 1 Answers - 0 Comments

A. The confidence interval formula is m ts/ n where m is the sample mean, s is the sample standard deviation, n is the sample size, and t is the table value for 95% confidence and n - 1 degrees of freedom. We can find the sample mean, since it's halfway between 22.4% and 23.5%, which is 22.95%. Hence, the interval is 22.95% 0.55% Now we know that ts/ n = 0.55%, but without knowing what s is, we're stuck. We need one more piece of information to finish solving the problem.
Answered by statman - Sun Jul 12 20:14:25 2009

Confidence intervals and marginal error?
Q. According to Business Week magazine, the typical chief financial officer (CFO) earned RM 961 000 in 2001. Assume that this mean is based on a random sample of 60 CFO's and that the sample standard deviation is RM 180 000. a) What is the point estimate of the corresponding population mean? What is the margin error for this estimate? b) Make a 90% confidence interval for the corresponding population mean. Could someone help me solve this problem. Thanks.
Asked by celine - Wed Jul 30 08:20:39 2008 - - 1 Answers - 0 Comments

A. = 961 000 = 18000 z = (x - 961,000)/180,000 A 90% confidence interval is the mean 45%, from P(z1) = 5% toP(z2) = 95% Looking those up in a table we see that z = 1.65 standard deviations from the mean. -1.65 = (x - 961,000)/18000 -297000 = x - 961000 x = 664,000 If z = + 1.65 we get x = 1,258,000 Some where in that range is certainly a good place to be.
Answered by Peter m - Wed Jul 30 08:40:55 2008