Confidence interval for the mathematical expectation of a normal sample

The case of known variance

Let be $X_{1},\ldots ,X_{n}\sim \mathrm {N} (\mu ,\sigma ^{2})$ ${\ displaystyle X_ {1}, \ ldots, X_ {n} \ sim \ mathrm {N} (\ mu, \ sigma ^ {2})}$ Is an independent sample from the normal distribution , where $\sigma ^{2}$ ${\ displaystyle \ sigma ^ {2}}$ - known dispersion . Define an arbitrary $\alpha \in [0,1]$ ${\ displaystyle \ alpha \ in [0,1]}$ and build the confidence interval for the unknown average $\mu$ ${\ displaystyle \ mu}$ .

Statement. Random value

Z={\frac {{\bar {X}}-\mu }{\sigma /{\sqrt {n}}}}

{\ displaystyle Z = {\ frac {{\ bar {X}} - \ mu} {\ sigma / {\ sqrt {n}}}}}

has a standard normal distribution $\mathrm {N} (0,1)$ ${\ displaystyle \ mathrm {N} (0,1)}$ . Let be $z_{\alpha }$ ${\ displaystyle z _ {\ alpha}}$ - $\alpha$ ${\ displaystyle \ alpha}$ is the quantile of the standard normal distribution . Then, due to the symmetry of the latter, we have:

\mathbb {P} \left(-z_{1-{\frac {\alpha }{2}}}\leq Z\leq z_{1-{\frac {\alpha }{2}}}\right)=1-\alpha

{\ displaystyle \ mathbb {P} \ left (-z_ {1 - {\ frac {\ alpha} {2}}} \ leq Z \ leq z_ {1 - {\ frac {\ alpha} {2}}} \ right) = 1- \ alpha}

.

After substituting the expression for $Z$ ${\ displaystyle Z}$ and simple algebraic transformations we get:

\mathbb {P} \left({\bar {X}}-z_{1-{\frac {\alpha }{2}}}{\frac {\sigma }{\sqrt {n}}}\leq \mu \leq {\bar {X}}+z_{1-{\frac {\alpha }{2}}}{\frac {\sigma }{\sqrt {n}}}\right)=1-\alpha

{\ displaystyle \ mathbb {P} \ left ({\ bar {X}} - z_ {1 - {\ frac {\ alpha} {2}}} {\ frac {\ sigma} {\ sqrt {n}}} \ leq \ mu \ leq {\ bar {X}} + z_ {1 - {\ frac {\ alpha} {2}}} {\ frac {\ sigma} {\ sqrt {n}}} \ right) = 1 - \ alpha}

.

The Case of Unknown Variance

Let be $X_{1},\ldots ,X_{n}\sim \mathrm {N} (\mu ,\sigma ^{2})$ ${\ displaystyle X_ {1}, \ ldots, X_ {n} \ sim \ mathrm {N} (\ mu, \ sigma ^ {2})}$ Is an independent sample from the normal distribution, where $\mu ,\sigma ^{2}$ ${\ displaystyle \ mu, \ sigma ^ {2}}$ - unknown constants. We construct a confidence interval for an unknown average $\mu$ ${\ displaystyle \ mu}$ .

Statement. Random value

T={\frac {{\bar {X}}-\mu }{S/{\sqrt {n}}}}

{\ displaystyle T = {\ frac {{\ bar {X}} - \ mu} {S / {\ sqrt {n}}}}}

,

has a student distribution with $n-1$ ${\ displaystyle n-1}$ degrees of freedom $\mathrm {t} (n-1)$ ${\ displaystyle \ mathrm {t} (n-1)}$ where $S$ ${\ displaystyle S}$ - unbiased sample standard deviation. Let be $t_{\alpha ,n-1}$ ${\ displaystyle t _ {\ alpha, n-1}}$ - $\alpha$ ${\ displaystyle \ alpha}$ - quantiles of student distribution . Then, due to the symmetry of the latter, we have:

\mathbb {P} \left(-t_{1-{\frac {\alpha }{2}},n-1}\leq T\leq t_{1-{\frac {\alpha }{2}},n-1}\right)=1-\alpha

{\ displaystyle \ mathbb {P} \ left (-t_ {1 - {\ frac {\ alpha} {2}}, n-1} \ leq T \ leq t_ {1 - {\ frac {\ alpha} {2 }}, n-1} \ right) = 1- \ alpha}

.

After substituting the expression for $T$ ${\ displaystyle T}$ and simple algebraic transformations we get:

\mathbb {P} \left({\bar {X}}-t_{1-{\frac {\alpha }{2}},n-1}{\frac {S}{\sqrt {n}}}\leq \mu \leq {\bar {X}}+t_{1-{\frac {\alpha }{2}},n-1}{\frac {S}{\sqrt {n}}}\right)=1-\alpha

{\ displaystyle \ mathbb {P} \ left ({\ bar {X}} - t_ {1 - {\ frac {\ alpha} {2}}, n-1} {\ frac {S} {\ sqrt {n }}} \ leq \ mu \ leq {\ bar {X}} + t_ {1 - {\ frac {\ alpha} {2}}, n-1} {\ frac {S} {\ sqrt {n}} } \ right) = 1- \ alpha}

.

Confidence interval for the mathematical expectation of a normal sample

The case of known variance

The Case of Unknown Variance

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