It depends whether or not the observations are independent and on the distribution of the variable that is being measured or the sample size. You cannot simply assume that the observations are independent and that the distribution is Gaussian (Normal).
You can't. You need an estimate of p (p-hat) q-hat = 1 - p-hat variance = square of std dev sample size n= p-hat * q-hat/variance yes you can- it would be the confidence interval X standard deviation / margin of error then square the whole thing
A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find anoher positio. the standard deviation of the sample was 6.2 weeks. construct a 95 % confidence interval for the population. Is it reasonable that the population mean is 28 weeks? Justify your answer
First of all, we need to find the margin of error which we will call E. This will be the greatest possible distance between our point estimate and the value of the parameter, the mean, we are estimating. We need the critical z score, (or possibly critical t score) that is to say the z score corresponding to 99% confidence interval. This means 99% of the area under the standard normal curves falls within 2.576 standard deviations of the mean. Here is how you find 2.576. First, divide the confidence level by two, and then look up the area in the the Z-table and look up the z-score on the outside. .99/2 =.495 The z table or the calculator give you 2.576. (The area in one tail is 0.0050, and some people find this easier to see) Next, use your sample mean, x bar, as your point estimate. The lower limit of the confidence interval will be x bar- E and the upper limit will be x bar + E. ( By the way, x bar is used here to indicate the letter x with a bar over it which is the standard notation for the sample mean) Now we need to find E. We need the critical z score we found to do this. The notation zc is often used to denote this and is used below. If sigma, the population standard deviation is known, we use it. Otherwise if the sample size is greater than or equal to 30, we find the sample standard deviation s and use it as an estimate for sigma. (When we do this, we will not be able to use a z score if n is less than 30, we will need to use s/√n instead of s) E will equal the product of the critical z score and ( sigma divided by the square root of n) E=zc (Sigma/√n) or if we use s, E=zc (s/√n) If n is less than 30, we use a t-distribution and instead of s we use s/√n . So E will be( tc s/√n ) and once again we will add an subtract this value from x bar. There will be n-1 degrees of freedom. The critical t will be talpha/2 (It is important to understand that the true mean, MU exists. There are only two possibilities. Either it is in the interval or not. It is not accurate to state that there is a 99% probability that the actual mean is in the interval. You should say there is a 99% probability that the confidence interval contains MU.) Here is a summary of which test to use: First is sigma known? If it is not and the population is normally distributed, use the t distribution. If the population is not normally distributed and n>30, we also use the t distribution. If the population is not normally distributed and n is less than or 30 we need nonparametric methods. If sigma is known, and the population is normally distributed, use the z scores. If sigma is known and the population is not normally distributed, you can use the z distribution if n>30. If not, once again we need nonparametric methods.
Yes, the interval of a graph is the difference between any two consecutive numbers on a scale.For example, if the scale read: 2,4,6,8,10 then you could do 4-2, 6-4, etc. to find the interval. (which is 2)
You are studying the sample because you want to find out information about the whole population. If the sample you have drawn from the population does not represent the population, you will find out about the sample but will not find out about the population.
if the confidence interval is 24.4 to 38.0 than the average is the exact middle: 31.2, and the margin of error is 6.8
Generally speaking an x% confidence interval has a margin of error of (100-x)%.
The confidence interval for this problem can be calculated using the following formula: Confidence Interval = p ± z*√(p*(1-p)/n) Where: p = observed proportion (54%) n = sample size (80) z = z-score (1.96) Confidence Interval = 0.54 ± 1.96*√(0.54*(1-0.54)/80) Confidence Interval = 0.54 ± 0.07 Therefore, the confidence interval is 0.47 - 0.61, meaning that we can be 95% confident that the percentage of voters who prefer the referred candidate is between 47% and 61%.
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U find the word interval
See: http://en.wikipedia.org/wiki/Confidence_interval Includes a worked out example for the confidence interval of the mean of a distribution. In general, confidence intervals are calculated from the sampling distribution of a statistic. If "n" independent random variables are summed (as in the calculation of a mean), then their sampling distribution will be the t distribution with n-1 degrees of freedom.
Yes, if it is the closed interval. No, if it is the open interval.
The 98 percent confidence level is commonly used in statistical tests. The critical Zc refers to the amount of relation between to factors.
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what is the lunitidal interval of pireaus;
You find the the smallest and largest values. The interval is the largest minus the smallest.
You can't. You need an estimate of p (p-hat) q-hat = 1 - p-hat variance = square of std dev sample size n= p-hat * q-hat/variance yes you can- it would be the confidence interval X standard deviation / margin of error then square the whole thing