Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.
I believe you want the equation to calculate the standard deviation of a sample. The equation is: s = square root[ sum from i =1 to n of (xi- xbar)/(n-1)] where xbar is the average of values of the sample and n = size of sample.
I think, the estimate is a numerical value, wile the estimator is a function or operator, which can be generate more estimates according to some factors. For example (xbar) is estimator for (meu), which can be various when the sample size in various, the value that will be produced is an (estimate), but (xbar) is estimator.
There are two types of statistics. One is called descriptive statistics and the other is inferential statistics. Descriptive statistics is when you use numbers. Inferential statistics is when you draw conclusions or make predictions.
Psychological statistics is the application of statistics to psychology.
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Yes. The transform is z= (x-xbar)/s where x is the data value, xbar is the mean of the data and s is the standard deviation.
Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.
I believe you want the equation to calculate the standard deviation of a sample. The equation is: s = square root[ sum from i =1 to n of (xi- xbar)/(n-1)] where xbar is the average of values of the sample and n = size of sample.
I think, the estimate is a numerical value, wile the estimator is a function or operator, which can be generate more estimates according to some factors. For example (xbar) is estimator for (meu), which can be various when the sample size in various, the value that will be produced is an (estimate), but (xbar) is estimator.
s=sample standard deviation s=square root (Sum(x-(xbar))2 /(n-1) Computing formula (so you don't have to find the mean and the distance from the mean over and over): square root(Sxx /(n-1)) Sxx= Sum(x2) - ((Sum(x))2/n)
The z score is calculated from the distance of a value from the distribution center divided by the standard dev. (x-xbar)/st. dev
The standard deviation. There are many, and it's easy to construct one. The mean of a sample from a normal population is an unbiased estimator of the population mean. Let me call the sample mean xbar. If the sample size is n then n * xbar / ( n + 1 ) is a biased estimator of the mean with the property that its bias becomes smaller as the sample size rises.
import java.io.*;import java.util.*;public class LinearRegression {public static void main(String[] args) {int MAXN = 1000;int n = 0;double[] x = new double[MAXN];double[] y = new double[MAXN];// first pass: read in data, compute xbar and ybardouble sumx = 0.0, sumy = 0.0, sumx2 = 0.0;while(!StdIn.isEmpty()) {x[n] = StdIn.readDouble();y[n] = StdIn.readDouble();sumx += x[n];sumx2 += x[n] * x[n];sumy += y[n];n++;}double xbar = sumx / n;double ybar = sumy / n;// second pass: compute summary statisticsdouble xxbar = 0.0, yybar = 0.0, xybar = 0.0;for (int i = 0; i < n; i++) {xxbar += (x[i] - xbar) * (x[i] - xbar);yybar += (y[i] - ybar) * (y[i] - ybar);xybar += (x[i] - xbar) * (y[i] - ybar);}double beta1 = xybar / xxbar;double beta0 = ybar - beta1 * xbar;// print resultsSystem.out.println("y = " + beta1 + " * x + " + beta0);// analyze resultsint df = n - 2;double rss = 0.0; // residual sum of squaresdouble ssr = 0.0; // regression sum of squaresfor (int i = 0; i < n; i++) {double fit = beta1*x[i] + beta0;rss += (fit - y[i]) * (fit - y[i]);ssr += (fit - ybar) * (fit - ybar);}double R2 = ssr / yybar;double svar = rss / df;double svar1 = svar / xxbar;double svar0 = svar/n + xbar*xbar*svar1;System.out.println("R^2 = " + R2);System.out.println("std error of beta_1 = " + Math.sqrt(svar1));System.out.println("std error of beta_0 = " + Math.sqrt(svar0));svar0 = svar * sumx2 / (n * xxbar);System.out.println("std error of beta_0 = " + Math.sqrt(svar0));System.out.println("SSTO = " + yybar);System.out.println("SSE = " + rss);System.out.println("SSR = " + ssr);}}
The simple answer is no. This depends on a lot of factors such as alpha which determines the critical value and the absolute value of the difference between the claim and sample data. Mathematically speaking, all things being equal, the larger the sample size the larger the absolute value of the test statistic. The formula for the test statistic mean with sigma known is shown below. You can substitute values in and perform the mathematics. The larger the sample size, the larger the Z value; but note if the numerator is small, even a small denominator will not produce a large Z value. In fact, the numerator could be zero which would make the test statistic zero. Z = (Xbar - μxbar)/(σ/√n) (formula from Elementary Statistics by Mario F. Triola)
The two main branches of statistics is Descriptive statistics and inferential statistics.
There are two types of statistics. One is called descriptive statistics and the other is inferential statistics. Descriptive statistics is when you use numbers. Inferential statistics is when you draw conclusions or make predictions.