How Do You Know if a Populations Standard Deviation Is Known or Unkown

Population Standard Difference

From which, the population standard deviation tin can be estimated by taking the foursquare-root of the variance.

From: The Joy of Finite Mathematics , 2016

Due south

Fred Eastward. Szabo PhD , in The Linear Algebra Survival Guide, 2015

Standard Deviation of a Numerical Vector

The population standard deviation a measures the spread of a vector in ℝ ^northward . It is divers to exist the square root of the population variance of the vector. The sample standard deviation s is defined to the square root of the sample variance of the vector.

Analogy

■

The population standard deviation of a vector in ℝ ⁶

x   =   {2, 6, iii, 1, 8, 9}; mx   =   Mean [x];

$\mathsf{\sigma }=\mathsf{Sqrt}\left[\mathsf{Full}\left[\frac{\mathsf{1}}{\mathsf{half-dozen}}\mathsf{Table}\left[{\left(\mathsf{x}\left[\left[\mathsf{i}\right]\right]‐\mathsf{m}\mathsf{x}\right)}^{\mathsf{2}},\left\{\mathsf{i},\mathsf{1},\mathsf{6}\right\}\right]\right]\right]$

$\frac{\sqrt{\mathtt{329}}}{\mathtt{6}}$

■

The sample standard divergence of a vector in ℝ⁶

10   =   {2, half-dozen, three, 1, 8, ix}; mx   =   Hateful [x];

$\mathsf{s}=\mathsf{Sqrt}\left[\mathsf{Total}\left[\frac{\mathsf{1}}{\mathsf{five}}\mathsf{Tabular\; array}\left[{\left(\mathsf{x}\left[\left[\mathsf{i}\right]\right]‐\mathsf{thousand}\mathsf{10}\right)}^{\mathsf{2}},\left\{\mathsf{i},\mathsf{i},\mathsf{six}\right\}\right]\right]\right]$

$\sqrt{\frac{\mathtt{329}}{\mathtt{xxx}}}$

The Mathematica StandardDeviation part computes the sample standard deviation of a vector.

s == StandardDeviation[x]

True

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Hypothesis Testing

Gary Smith , in Essential Statistics, Regression, and Econometrics (Second Edition), 2015

Using an Estimated Standard Deviation

Our initial calculations assumed that the population standard departure is known to be viii.53. However, this is an gauge based on sample data. The preceding affiliate explains how the t distribution tin be used in identify of the normal distribution when the sample standard deviation S is used in place of the population standard deviation σ. Hither, instead of the Z statistic in Eqn (7.2), we use the t statistic:

(vii.3) $t=\frac{\overline{\mathrm{Ten}}-\mu }{S/\sqrt{\mathrm{north}}}$

The calculated t value is identical to the Z value we calculated previously:

$\begin{array}{c}t=\frac{25.53-35}{8.53/\sqrt{203}}\\ =-\mathrm{xv.82}\end{array}$

The departure is that, instead of using the normal distribution to summate the P value, we use the t distribution with n  −   one   =   203   −   ane   =   202 degrees of freedom. The probability works out to exist:

$P[t\le -15.82]=3.23\times {10}^{-37}$

so that the two-sided P value is ii(iii.23   ×   10⁻³⁷)   =   half dozen.46   ×   10⁻³⁷.

A P value calculated from the t distribution is exactly correct if the data come from a normal distribution and, considering of the power of the primal limit theorem, is an fantabulous approximation if nosotros accept at least 15 observations from a more often than not symmetrical distribution or at least 30 observations from a very asymmetrical distribution. In our poker example, the P value is minuscule and is strong evidence against the zippo hypothesis that the population hateful is 35.

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Tests on Means of Continuous Data

R.H. Riffenburgh , in Statistics in Medicine (Third Edition), 2012

The Normal Exam and the t Test Are Ii Forms of the Two-Sample Means Test

Two forms are addressed in this section: the case of known population standard deviations, or samples large enough that the sample standard deviations are not practically different from known, and the case of minor-sample estimated standard deviations. The means test uses a standard normal distribution (z distribution) in the first example and a t distribution in the second.

(Review: The means are assumed normal. Standardizing places a standard deviation in the denominator. If the standard deviation is known, it behaves as a constant and the normal distribution remains. If the standard deviation is estimated from the sample, it follows a probability distribution and the ratio of the numerator'south normal distribution to this distribution turns out to be t.)

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Computational Statistics with R

Marepalli B. Rao , Subramanyam Kasala , in Handbook of Statistics, 2014

5.1 Specifications

For sample size calculations, specify level, power, alternative population means, and within population standard deviation σ. The between population variance is divers by

Between populations variance σ _pop ²  = $\frac{{\displaystyle {\sum }_{i=1}^{\mathrm{one\; thousand}}{\left({\mu }_i-\overline{\mu }\right)}^{\mathrm{two}}}}{\mathrm{grand}}$ , where $\overline{\mu }=\left(\frac{\mathrm{i}}{k}\right){\displaystyle \sum _{i=\mathrm{i}}^{\mathrm{1000}}{\mu }_i}$   =   boilerplate of the population means.

In the context of multiple populations, the result size is defined by

$\mathrm{Effect}\mathrm{size}=\Delta =\frac{{\sigma }_{\mathrm{pop}}}{\sigma }=\mathrm{bet}.\mathrm{pop}.\mathrm{SD}/\mathrm{within}.\mathrm{pop}.\mathrm{SD}$

If the number of populations is k  =   2, the effect size works out to be $\frac{\left|{\mu }_{\mathrm{one}}-{\mu }_2\right|}{\mathrm{ii}*\sigma }$ , which is different from $\frac{\left|{\mu }_1-{\mu }_{\mathrm{ii}}\right|}{\sigma }$ .

In engineering science terminology, σ is the noise present in the populations and σ _pop is a measure of the signal. In that location are and so many signals associated with different populations we need to measure how close they are. The standard deviation σ _pop of these signals is a measure how close they are. If they are very close, i needs a large sample to detect the differences. If the ways are very close, the alternative and zippo hypotheses are very close. Let us find out how the effect size shapes in some examples.

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Confidence Intervals and Probability

ROBERT H. RIFFENBURGH , in Statistics in Medicine (2nd Edition), 2006

EXAMPLE

Using the 10 prostate volumes from Tabular array DB1.1 , what is a 95% confidence interval on the population standard difference, σ? Nosotros accept calculated s = 15.92 ml, which has n − i = 9 df. We know the probability distribution of s ⁱⁱ merely not of s; therefore, we shall find the interval on the population variance, σtwo, and take square roots: southward ⁱⁱ = 15.92² = 253.4464. Looking in Tabular array Iii or Tabular array 4.3 under right tail expanse = 0.025 for 9 df, we find xix.02. Similarly in Table IV, under left tail expanse = 0.025 for 9 df, we find 2.70. Substituting these values in the formula for confidence limits on a variance [see Eq. (4.seven)], we notice:

$\begin{array}{l}P[{\mathrm{due\; south}}^2-df/{\chi }_R^2<{\sigma }^2<s^{\mathrm{two}}\times df/{\chi }_L^{\mathrm{two}}]\hfill \\ =P[253.45\times 9/19.02<{\sigma }^2<253.45\times 9/\mathrm{2.seventy}]\hfill \\ =P[1119.93<{\sigma }^{\mathrm{ii}}<844.83]=\mathrm{0.95.}\hfill \end{array}$

Taking square roots within the brackets, we obtain

$P[10.95<\sigma <29.07]=\mathrm{0.95.}$

We note from the outset tabular array of DBl (showing ways and standard deviations) that the population standard difference (excluding known BPH to make the distribution more symmetric) is 16.35 ml, falling well within the confidence interval.

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Confidence Intervals

R.H. Riffenburgh , in Statistics in Medicine (Third Edition), 2012

Example Posed: CI on Standard Deviation of Prostate Volumes

Using the 10 prostate volumes from Table DB1.1, what is a 95% confidence interval on the population standard deviation σ? We have calculated southward=15.92   ml which has n−1=9 df. We know the probability distribution of south ² but non of s, so nosotros shall find the interval on the population variance, σ ⁱⁱ, and take square roots. s ^two=15.92²=253.4464. How do nosotros express a confidence interval?

Method for Conviction on σ ² or σ

We know (from Section 4.7) that sample variance south ² drawn randomly from a normal population is distributed equally χ ²× σ ²/df. A confidence-type statement on σ ^two from this relationship, with the additional twist that the asymmetric distribution requires the chi-square values excluding ane−α/ii in each tail to be establish separately, is given by

(7.12) $P[{\mathrm{due\; south}}^2\times df/{\chi }^{2}R<{\sigma }^{\mathrm{two}}<{\mathrm{due\; south}}^2\times df/{\chi }^2\mathrm{50}]=1-\alpha \text{,}$

where χ ⁱⁱ _R is the critical value for the correct tail found from Table Iii (see Tables of Probability Distributions) and χ ^two ₅₀ is that for the left found from Table Four (run across Tables of Probability Distributions). To detect the confidence on σ rather than σ ⁱⁱ, we just have square roots of the components within the brackets.

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The Basis of Monte Carlo

William L. Dunn , J. Kenneth Shultis , in Exploring Monte Carlo Methods, 2012

ii.eight Summary

A probability density function of a single continuous variable is a not-negative office defined on an interval and normalized and so that its integral over that interval is unity. The associated cumulative distribution office of ten is interpreted as the probability that a random sample has a value less than or equal to x. These concepts extend naturally to discrete and multidimensional random variables.

Mostly, probability distributions have well-defined population means and variances. These are explicit values that are characteristic of the distribution. Monte Carlo is a method that allows 1 to estimate the population hateful and population variance past the sample mean and sample variance. These concepts apply to functions of a discrete variable, to functions of a continuous variable, and to functions of many variables, whether detached or continuous.

The Monte Carlo estimates of the sample mean and sample variance almost surely arroyo the population hateful and population variance every bit the number of samples (usually called "histories") gets large. This of import feature is a consequence of the constabulary of big numbers, which can be stated equally follows. If

(two.69) $〈z〉={\int }_a^{b}z\left(\text{ten}\right)f\left(\text{x}\right)d\text{10}$

and

(two.70) $\overline{z}=\frac{\mathrm{i}}{N}\sum _{i=\mathrm{one}}^{N}z\left({\text{x}}_i\right),$

with the 10_i suitably sampled from f(x), and so, almost surely,

(2.71) $\underset{\mathrm{North}\to \infty }{lim}\overline{z}=〈z〉.$

The CLT and then provides a prescription for estimating the uncertainty in the sample mean and sample variance.

It should be kept in listen that the term standard deviation is ofttimes used, rather loosely, for whatsoever of the following:

•

the population standard divergence σ (z);

•

the sample standard deviation $s(z)={[N(\overline{z^2}-{\overline{z}}^{\mathrm{ii}})/(N-1)]}^{1/2}$ ;

•

the standard deviation of the sample mean $\sigma (\overline{z})=\sigma (z)/\sqrt{N}$ ; and

•

the estimation of the standard departure of the sample mean $s(\overline{z})=s(z)/\sqrt{N}={[(\overline{z^{\mathrm{two}}}-{\overline{z}}^2)/(N-1)]}^{1/2}$ .

The context must be employed to signal which interpretation is meant.

Finally, Monte Carlo can hateful a wide variety of unlike calculational methods, but all are based on the statistical power of the cardinal limit theorem that guarantees expected values can exist approximated by averages and that allows confidence intervals to be constructed for the averages. Although many dissimilar applications of Monte Carlo appear to be quite different, they all depend on this critical theorem.

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HYPOTHESIS TESTING

Rand R. Wilcox , in Applying Contemporary Statistical Techniques, 2003

five.ane.two A Two-Sided Examination: Testing for Exact Equality

One other variation of hypothesis testing needs to be described: testing the hypothesis that the mean is exactly equal to some specified value. Returning to the example regarding open-mindedness, suppose it is claimed that the average score of all adult men is exactly 50, as opposed to beingness greater than or equal to fifty. Then the nothing hypothesis is

$H_0:\mu =50.$

If the sample hateful is exactly equal to 50, y'all would not reject, considering this is consistent with H ₀. If $\overline{X}>50,$ , then the larger the sample mean happens to be, the more doubtfulness there is that μ = fifty. Similarly, if $\overline{\mathrm{10}}<\mathrm{l},$ , then the smaller the sample mean, the more doubt in that location is that μ = l. That is, now it is reasonable to refuse H ₀ if $\overline{X}$ is either too large or too pocket-size. An equivalent mode of proverb this is that you should reject if

$Z=\frac{\overline{\mathrm{Ten}}-50}{\sigma /\mathrm{due\; north}}$

is too big or also small.

Suppose you turn down H ₀ if either Z ≤ −ane.96 or Z ≥ ane.96. A more succinct way of describing this decision rule is that you lot reject if the absolute value of Z is greater than or equal to one.96. In symbols, turn down if |Z| ≥ i.96. If the null hypothesis is true and sampling is from a normal distribution, and then Z has a standard normal distribution, and then the probability of rejecting is

$P(Z\le -196)+P(Z\ge \mathrm{i.96})=.025+.025=0.5,$

which is the total area of the two shaded regions in Figure five.ii.

FIGURE five.2. Critical region for a two-sided exam such that P(Type I error) = .05.

EXAMPLE

Imagine a list of 55 pocket-sized malformations babies might have at birth. For illustrative purposes, it is assumed that the boilerplate number of malformations is 15 and the population standard departure is σ = 6. For babies built-in to diabetic women, is the average number unlike from 15? That is, can you reject the hypothesis H ₀: μ = fifteen? To find out, you sample n = sixteen babies having diabetic mothers, count the number of malformations for each, and find that the boilerplate number of malformations is $\overline{\mathrm{Ten}}=19$ . Then

$Z=\frac{\mathrm{xix}-15}{\mathrm{half-dozen}/\sqrt{16}}=\mathrm{2.67.}$

If the goal is to have the probability of a Type I error equal to .05, then the critical values are −one.96 and 1.96, for the reasons just given. Because 2.67 is greater than 1.96, y'all reject the cipher hypothesis and conclude that the boilerplate number of malformations is greater than 15.

The significance level, or p-value, can be determined when testing for exact equality, but yous must accept into business relationship that the critical region consists of both tails of the standard normal distribution.

Case

Continuing the last example, where Z = ii.67, if you had decided to reject the null hypothesis if Z ≤ −two.67 or if Z ≥ ii.67, then the probability of a Blazon I error is

$P(Z\le -\mathrm{two.67})+P(Z\ge 2.67)=.0038+.0038=\mathrm{.0076.}$

This means that the significance level is 0.0076.

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Statistical Testing, Risks, and Odds in Medical Decisions

ROBERT H. RIFFENBURGH , in Statistics in Medicine (Second Edition), 2006

THE NORMAL Examination AND THE t TEST ARE Two FORMS OF THE TWO-SAMPLE Ways TEST

This section concentrates on the type of means test nearly frequently seen in the medical literature: the test for a divergence betwixt two means. We volition look at two subclasses: the case of known population standard deviations, or samples large plenty that the sample standard deviations are not practically different from known, and the example of modest-sample estimated standard deviations. The means examination uses a standard normal distribution ( z distribution) in the first case and a t distribution in the second, for reasons discussed in Section ii.8. (Review: The means are assumed normal. Standardizing places a standard deviation in the denominator. If the standard divergence is known, it behaves as a constant and the normal distribution remains. If the standard deviation is estimated from the sample, information technology follows a probability distribution and the ratio of the numerator's normal distribution to this distribution turns out to be t.)

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Estimation

Sheldon Thousand. Ross , in Introductory Statistics (Tertiary Edition), 2010

SUMMARY

The sample mean

is an unbiased estimator of the population hateful μ. Its standard difference, sometimes referred to as the standard mistake of

as an estimator of μ, is given past

$\text{Due south}\text{D}\left(\overline{X}\right)=\frac{\sigma }{\sqrt{\mathrm{northward}}}$

where σ is the population standard divergence.

The statistic $\hat{p}$ , equal to the proportion of a random sample having a given characteristic, is the estimate of p, the proportion of the entire population with the characteristic. The standard error of the judge is

$\text{Due south}\text{D}\left(\hat{p}\right)=\sqrt{\frac{p(\mathrm{one}-p)}{n}}$

where n is the sample size. The standard error can be estimated past

$\sqrt{\frac{\hat{p}(\mathrm{ane}-\hat{p})}{\mathrm{north}}}$

The sample variance S ^two is the calculator of the population variance σ^two. Correspondingly, the sample standard deviation Southward is used to estimate the population standard deviation σ.

If 10₁, …, X_n are a sample from a normal population having a known standard difference σ,

$\overline{X}\pm z_{\alpha /{\mathrm{ii}}^{\frac{\sigma }{\sqrt{n}}}}$

is a 100(1 – α) percent confidence interval estimator of the population mean μ. The length of this interval, namely,

$\mathrm{two}{z}_{\alpha /2^{\frac{\sigma }{\sqrt{n}}}}$

will be less than or equal to b when the sample size n is such that

$n\ge (\frac{2z_{\alpha /\mathrm{ii}}\sigma }{b}{)}^2$

A 100(1 – α) lower confidence bound for μ is given past

$\overline{\mathrm{Ten}}-z_{\alpha }\frac{\sigma }{\sqrt{\mathrm{due\; north}}}$

That is, nosotros can affirm with 100(i – α) percent confidence that

$\mu >\overline{\mathrm{Ten}}+z_{\alpha }\frac{\sigma }{\sqrt{\mathrm{northward}}}$

A 100(1 – α) upper confidence bound for μ is

$\overline{X}+z_{\alpha }\frac{\sigma }{\sqrt{n}}$

That is, we can assert with 100(1 – α) percentage conviction that

$\mu <\overline{X}+z_{\alpha }\frac{\sigma }{\sqrt{n}}$

If X₁, …, X_n are a sample from a normal population whose standard deviation is unknown, a 100(1 – α) percent confidence interval estimator of μ is

$\overline{X}\pm t_{\mathrm{northward}-1,\alpha /{\mathrm{two}}^{\frac{\mathrm{Southward}}{\sqrt{n}}}}$

In the preceding, t _{n –1, α/ii} is such that

$P\{T_{n-1}>t_{n-\mathrm{one},\alpha /\mathrm{two}}\}=\frac{\alpha }{2}$

when T_{n –1} is a t random variable with n – 1 degrees of freedom.

The 100(1 – α) per centum lower and upper confidence premises for μ are, respectively, given by

$\overline{X}-t_{n-1,\alpha }\frac{S}{\sqrt{\mathrm{northward}}}$

To obtain a confidence interval judge of p, the proportion of a big population with a specific characteristic, take a random sample of size n. If $\hat{p}$ is the proportion of the random sample that has the characteristic, then an approximate 100(1 – α) per centum confidence interval calculator of p is

$\hat{p}\pm z_{\alpha /2}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}$

The length of this interval always satisfies

$\text{Length}\text{of}\text{confidence}\text{interval}\le \frac{{\text{z}}_{\alpha /2}}{\sqrt{\text{n}}}$

The distance from the center to the endpoints of the 95 percent confidence interval estimator, that is, $1.96\sqrt{\hat{p}\left(1-\hat{p}\right)/n}$ , is commonly referred to every bit the margin of error. For case, suppose a newspaper states that a new poll indicates that 64 percentage of the population consider themselves to be conservationists, with a margin of error of ±three percent. By this, the newspaper means that the results of the poll yield that the 95 percent conviction interval estimate of the proportion of the population who consider themselves to be conservationists is 0.64 ± 0.03.

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