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Can the margin of error be determined if only the confidence level is known?
No, more information is needed to determine the margin of error. For example, one may need to know the sample's mean, the sample size, and the standard deviations of the population and sample. Depending on the type of test one is performing, certain parameters need not be known. For example, the population standard deviation does not need to be known in a one sample T-test.
What is Confidence Intervals of Margin of Error?
The magnitude of difference between the statistic (point estimate) and the parameter (true state of nature), . This is estimated using the critical statistic and the standard error.
How does Increasing the sample size while keeping the same confidence level has what effect on the margin of error?
The margin of error is reduced.
Is the margin of error always greater than or equal to the standard of error?
Yes
99 percent confidence interval Population mean 24.4 to 38.0 find the mean sample?
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
What happens to the confidence interval as the standard deviation of a distribution increases?
The standard deviation is used in the numerator of the margin of error calculation. As the standard deviation increases, the margin of error increases; therefore the confidence interval width increases. So, the confidence interval gets wider.
What is the difference between standard deviation and margin of error?
Standard of deviation and margin of error are related in that they are both used in statistics. Level of confidence is usually shown as the Greek letter alpha when people conducting surveys allow for a margin of error - usually set at between 90% and 99%. The Greek letter sigma is used to represent standard deviation.
How do you find the sample size if you are given the confidence interval and the margin of error as well as the standard deviation?
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
In a poll of 100 adults 45 percent reported they believe in faith healing If the poll was based on 5000 adults would the confidence interval be wider or narrower?
The formula for margin of error is (Z*)*(Standard Deviation))/(sqrt(N)), so as N increases, the margin of error decreases. Here N went from 100 to 5000, so N has increased by 4900. This means the margin of error decreases. Since the confidence interval is the mean plus or minus the margin of error, a smaller margin of error means that the confidence interval is narrower.
If Web Search Results for A bank wishes to estimate the mean balances owed by their MasterCard customers within 75 The population standard deviation is estimated to be 300 If a 98 percent confidence?
A bank wishing to estimate the mean balances owed by their MasterCard customers within 75 miles with a 98 percent confidence can use the following formula to calculate the required sample size: Sample size = (Z-score)2 * population standard deviation / (margin of error)2 Where Z-score = 2.326 for 98 percent confidence Population standard deviation = 300 Margin of error = desired confidence intervalSubstituting the values into the formula the required sample size is: 2.3262 * 300 / (Confidence Interval)2 = 553.7Therefore the bank would need to have a sample size of 554 to estimate the mean balances owed by their MasterCard customers within 75 miles with a 98 percent confidence.
Can the margin of error be determined if only the confidence level is known?
No, more information is needed to determine the margin of error. For example, one may need to know the sample's mean, the sample size, and the standard deviations of the population and sample. Depending on the type of test one is performing, certain parameters need not be known. For example, the population standard deviation does not need to be known in a one sample T-test.
What is the relationship between confidence interval and standard deviation?
Short answer, complex. I presume you're in a basic stats class so your dealing with something like a normal distribution however (or something else very standard). You can think of it this way... A confidence interval re-scales margin of likely error into a range. This allows you to say something along the lines, "I can say with 95% confidence that the mean/variance/whatever lies within whatever and whatever" because you're taking into account the likely error in your prediction (as long as the distribution is what you think it is and all stats are what you think they are). This is because, if you know all of the things I listed with absolute certainty, you are able to accurately predict how erroneous your prediction will be. It's because central limit theory allow you to assume statistically relevance of the sample, even given an infinite population of data. The main idea of a confidence interval is to create and interval which is likely to include a population parameter within that interval. Sample data is the source of the confidence interval. You will use your best point estimate which may be the sample mean or the sample proportion, depending on what the problems asks for. Then, you add or subtract the margin of error to get the actual interval. To compute the margin of error, you will always use or calculate a standard deviation. An example is the confidence interval for the mean. The best point estimate for the population mean is the sample mean according to the central limit theorem. So you add and subtract the margin of error from that. Now the margin of error in the case of confidence intervals for the mean is za/2 x Sigma/ Square root of n where a is 1- confidence level. For example, confidence level is 95%, a=1-.95=.05 and a/2 is .025. So we use the z score the corresponds to .025 in each tail of the standard normal distribution. This will be. z=1.96. So if Sigma is the population standard deviation, than Sigma/square root of n is called the standard error of the mean. It is the standard deviation of the sampling distribution of all the means for every possible sample of size n take from your population ( Central limit theorem again). So our confidence interval is the sample mean + or - 1.96 ( Population Standard deviation/ square root of sample size. If we don't know the population standard deviation, we use the sample one but then we must use a t distribution instead of a z one. So we replace the z score with an appropriate t score. In the case of confidence interval for a proportion, we compute and use the standard deviation of the distribution of all the proportions. Once again, the central limit theorem tells us to do this. I will post a link for that theorem. It is the key to really understanding what is going on here!
Rolling margin of steel rod which is used for construction?
Rolling Margin is the deviation of actual unit weight to that of Standard unit weight as per IS Standards. Rolling Margin is calculated as : Sectional weight = Weight of Steel Bars dia wise / length of the bars. As per IS Standards unit weight of the Bars is calculated as dia x dia / 162 Rolling Margin is deviation of actual sectional weight to that of IS Standard unit weight. Standard Rolling Margin for different dia reinforcement bars used for construction purposes: 8mm to 10mm +- 7% 12mm to 16mm +- 5% 20mm & Above +- 3%
How do you find The margin of error for the confidence interval is?
Generally speaking an x% confidence interval has a margin of error of (100-x)%.
Why does the margin of error increase as the confidence increases?
The margin of error is dependent on the confidence interval.I'll give you examples to understand it better.We know:Confidence Interval (CI) = x(bar) ± margin of error (MOE)MOE = (z confidence)(sigma sub x bar, aka standard error of mean)When CI = 95%, MOE = (1.96)(sigma sub x bar)When CI = 90%, MOE = (1.64)(sigma sub x bar)Naturally, the margin of error will decrease as confidence level decreases.
Vitamins to analyze the variation of vitamin supplement tablets you randomly select and weigh 14 tablets the results in milligrams are shown use 90 percent level of confidence?
To analyze the variation of vitamin supplement tablets, calculate the mean and standard deviation of the weights of the 14 tablets. With a 90% confidence level, determine the margin of error using the t-distribution. This margin of error will help define the range within which the true mean weight of the tablets is likely to lie.
What is an example sentence for leeway?
The permissible margin for variation or deviation from something.
