error


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error - difference between an observed value and the true value of a measurement.

a confidence interval for a population mean has a margin of error of 0.081
a confidence interval for a population mean has a margin of error of 0.081

a confidence interval for a population mean has a margin of error of 0.081
Margin of error = Interval Size/2 0.081 = Interval Size/2 Cross Multiply: Interval Size = 0.081 * 2 Interval Size = [B]0.162[/B]

A professor wants to test all possible pairwise comparisons among three means. If we need to maintai
A professor wants to test all possible pairwise comparisons among three means. If we need to maintain an experiment-wise alpha of 0.05, what is the error rate per comparison after applying Bonferroni correction? We are given: [LIST] [*]? = 0.05 [*]n = 3 [/LIST] Bonferroni Correction = ?/n Bonferroni Correction = 0.05/3 Bonferroni Correction = [B]0.01666666667[/B]

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find
A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places) Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get [B]58.89 < u < 63.11[/B]

A researcher posed a null hypothesis that there was no significant difference between boys and girls
A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 100 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 5.0. What is the standard error of the difference in means? [B]0.707106781187[/B] using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+100&xbar1=70&stdev1=5&n2=+100&xbar2=65&stdev2=5&conf=+99&pl=Hypothesis+Test']difference of means calculator[/URL]

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find t
A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating at the 99% level of confidence

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find t
A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating at the 99% level of confidence

A student hypothesized that girls in his class had the same blood pressure levels as boys. The proba
A student hypothesized that girls in his class had the same blood pressure levels as boys. The probability value for his null hypothesis was 0.15. So he concluded that the blood pressures of the girls were higher than boys'. Which kind of mistake did he make? a. Type I error b. Type II error c. Type I and Type II error d. He did not make any mistake [B]d. He did not make any mistake[/B] [I]p value is high, especially using a significance level of 0.05[/I]

based on a sample of size 41, a 95% confidence interval for the mean score of all students,, on an
based on a sample of size 41, a 95% confidence interval for the mean score of all students,, on an aptitude test is from 60 to 66. Find the margin of error

Basic Statistics
Given a number set, and an optional probability set, this calculates the following statistical items:
Expected Value
Mean = μ
Variance = σ2
Standard Deviation = σ
Standard Error of the Mean
Skewness
Mid-Range
Average Deviation (Mean Absolute Deviation)
Median
Mode
Range
Pearsons Skewness Coefficients
Entropy
Upper Quartile (hinge) (75th Percentile)
Lower Quartile (hinge) (25th Percentile)
InnerQuartile Range
Inner Fences (Lower Inner Fence and Upper Inner Fence)
Outer Fences (Lower Outer Fence and Upper Outer Fence)
Suspect Outliers
Highly Suspect Outliers
Stem and Leaf Plot
Ranked Data Set
Central Tendency Items such as Harmonic Mean and Geometric Mean and Mid-Range
Root Mean Square
Weighted Average (Weighted Mean)
Frequency Distribution
Successive Ratio

Confidence Interval for the Mean
Calculates a (90% - 99%) estimation of confidence interval for the mean given a small sample size using the student-t method with (n - 1) degrees of freedom or a large sample size using the normal distribution Z-score (z value) method including Standard Error of the Mean. confidence interval of the mean

Confidence Interval of a Proportion
Given N, n, and a confidence percentage, this will calculate the estimation of confidence interval for the population proportion π including the margin of error. confidence interval of the population proportion

Confidence Interval/Hypothesis Testing for the Difference of Means
Given two large or two small distriutions, this will determine a (90-99)% estimation of confidence interval for the difference of means for small or large sample populations.
Also performs hypothesis testing including standard error calculation.

Find Necessary Sample Size
The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

If power is big, you can assume:
If power is big, you can assume: a. The difference between the means is more likely to be detected b. The significance level set by the researcher must be high c. We increase the probability of type I error d. Your study result will be more likely to be inconclusive [B]b. The significance level set by the researcher must be high[/B]

Imagine that a researcher wanted to know the average weight of 5th grade boys in a high school. He r
Imagine that a researcher wanted to know the average weight of 5th grade boys in a high school. He randomly sampled 5 boys from that high school. Their weights were: 120 lbs., 99 lbs, 101 lbs, 87 lbs, 140 lbs. What's the [B][U]standard error of the mean[/U][/B]? 9.29839 using our [URL='http://www.mathcelebrity.com/statbasic.php?num1=120%2C99%2C101%2C87%2C140&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics#standard_error_of_the_mean']statistics calculator[/URL]

Joanie multiplied 0.78 times 0.34 and got the product 26.52. What error did she make
Joanie multiplied 0.78 times 0.34 and got the product 26.52. What error did she make [B]She didn't move the decimal point over 2 spots[/B]: 0.78 * 0.34 = 0.2652

MAPE - MPE - MAPD
Given a time series of actual and forecasted values, this determines the following:
* Mean Absolute Percentage Error (MAPE) also known as the Mean Absolute Percentage Deviation (MAPD)
* Symmetric Mean Absolute Percentage Error (sMAPE)
* Mean Absolute Percentage Error (MPE)

Margin of Error from Confidence Interval
Given a confidence interval, this determines the margin of error and sample mean.

Percent Error
Percentage error is the difference between an experimental measured value and a theoretical actual value

Point Estimate and Margin of Error
Given an upper bound and a lower bound and a sample size, this calculate the point estimate, margin of error.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the pho
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, [U][B]the Type I error is[/B][/U]: a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
d. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher [B]b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same [/B] [I]A Type I error is when you reject the null hypothesis when it is in fact true[/I]

Relative Error
Relative error is the absolute error divided by quantity

Six Years ago, 12.2% of registered births were to teenage mothers. A sociologist believes that the
Six Years ago, 12.2% of registered births were to teenage mothers. A sociologist believes that the percentage has decreased since then. (a) Which of the following is the hypothesis to be conducted? A. H0: p = 0.122, H1 p > 0.122 B. H0: p = 0.122, H1 p <> 0.122 C. H0: p = 0.122, H1 p < 0.122 (b) Which of the following is a Type I error? A. The sociologist rejects the hypothesis that the percentage of births to teenage mothers is 12.2%, when the true percentage is less than 12.2% B. The sociologist fails to reject the hypothesis that the percentage of births to teenage mothers is 12.2%, when the true percentage is less than 12.2% C. The sociologist rejects the hypothesis that the percentage of births to teenage mothers is 12.2%, when it is the true percentage. c) Which of the following is a Type II error? A. The sociologist rejects the hypothesis that the percentage of births to teenage mothers is 12.2%, when it is the true percentage B. The sociologist fails to reject the hypothesis that the percentage of births to teenage mothers is 12.2%, when it is the true percentage C. The sociologist fails to reject the hypothesis that the percentage of births to teenage mothers is 12.2%, when the true percentage is less than 12.2% (a) [B]C H0: p = 0.122, H1: p < 0.122[/B] because a null hypothesis should take the opposite of what is being assumed. So the assumption is that nothing has changed while the hypothesis is that the rate has decreased. (b) [B]C.[/B] The sociologist rejects the hypothesis that the percentage of births to teenage mothers is 12.2%, when it is the true percentage. Type I Error is rejecting the null hypothesis when it is true c) [B]C.[/B] The sociologist fails to reject the hypothesis that the percentage of births to teenage mothers is 12.2%, when the true percentage is less than 12.2% Type II Error is accepting the null hypothesis when it is false.

Solve Problem
based on a sample of size 41, a 95% confidence interval for the mean score of all students,, on an aptitude test is from 60 to 66. Find the margin of error

Solve Problem
A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating at the 99% level of confidence

Solve Problem
[URL]http://www.mathcelebrity.com/marginoferror.php?num=60%2C66&pl=Calculate+Margin+of+Error+and+Sample+Mean[/URL]

Solve the problem
a confidence interval for a population mean has a margin of error of 0.081. Determine the length of the confidence interval

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and
standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and
Standard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: [B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognit
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the bran. How many adults must he survey in order to be 90% confident that his estimate is within seven percentage points of the true population percentage? [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5 1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5 margin of error (E) = 0.07 At 90% confidence level the t is, alpha = 1 - 90% alpha = 1 - 0.90 alpha = 0.10 alpha / 2 = 0.10 / 2 = 0.05 Zalpha/2 = Z0.05 = 1.645 sample size = n = (Z[IMG]https://ci4.googleusercontent.com/proxy/mwhpkw3aM19oMNA4tbO_0OdMXEHt9juW214BnNpz4kjXubiVJgwolO7CLbmWXXoSVjDPE_T0CGeUxNungBjN=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Calpha[/IMG] / 2 / E )2 * [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] * (1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] ) = (1.645 / 0.07)^2 *0.5*0.5 23.5^2 * 0.5 * 0.5 552.25 * 0.5 * 0.5 = 138.06 [B]sample size = 138[/B] [I]He must survey 138 adults in order to be 90% confident that his estimate is within seven percentage points of the true population percentage.[/I]

The monthly earnings of a group of business students are are normally distributed with a standard de
The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

The monthly earnings of a group of business students are are normally distributed with a standard de
The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

Typing Speed
Solves for words per minute, number of words typed, errors, or number of minutes typing based on user entry.

Which of the following could reduce the rate of Type I error? a. Making the significant level from
Which of the following could reduce the rate of Type I error? a. Making the significant level from 0.01 to 0.05 b. Making the significant level from 0.05 to 0.01 c. Increase the Β level d. Increase the power [B]a. Making the significant level from 0.01 to 0.05[/B] [I]This widens the space under the graph and makes the test less strict.[/I]

You choose an alpha level of .01 and then analyze your data.
(a) What is the probability that
You choose an alpha level of .01 and then analyze your data. (a) What is the probability that you will make a Type I error given that the null hypothesis is true? (b) What is the probability that you will make a Type I error given that the null hypothesis is false. [B](a) 0.01. Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error.[/B] [B](b) Impossible Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error.[/B]

___is the probability of a Type II error; and ___ is the probability of correctly rejecting a false
___is the probability of a Type II error; and ___ is the probability of correctly rejecting a false null hypothesis. a. 1 - ?; ? b. ?; 1 - ?; c. ?; ?; d. ?; ? [B]b. ?; 1 - ?;[/B] [LIST] [*]H0 is true = Correct Decision 1 - ? Confidence Level = Size of a Test ? = Type I Error [*]Ho is false = Type II Error ? = Correct Decision 1 - ? = Power of a Test [/LIST]