Estimate μ, Variance Unknown

Consider X a random variable with probability density function N(μ, σ2), which is the normal probability density with unknown mean μ and unknown variance σ2. To estimate μ we obtain a random sample X₁, ..., X _n from N(μ, σ2). Suppose the mean of this random sample turns out to be 4.2% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaubiaeqale % qabaGaaGOmaaqdbaGaam4Caaaakiabg2da9maaqahabaWaaeWaaeaa % daqfqaqabSqaaiaadMgaaeqaneaacaWG4baaaOGaeyOeI0Yaa0aaae % aacaWG4baaaaGaayjkaiaawMcaaaWcbaGaamyAaiabg2da9iaaigda % aeaacaWGUbaaniabggHiLdGcdaahaaWcbeqaaiaaikdaaaGccaGGVa % WaaeWaaeaacaWGUbGaeyOeI0IaaGymaaGaayjkaiaawMcaaaaa!4A85! $$\mathop s\nolimits^2 = {\sum\limits_{i = 1}^n {\left( {\mathop x\nolimits_i - \overline x } \right)} ^2}/\left( {n - 1} \right) $$, which is a crisp number, not a fuzzy number. Also, let s2 be the sample variance. Our point estimator of μ is $$\overline{x}$$. If the values of the random sample are x₁, ..., x _n then the expression we will use for s2 in this book is 4.1$${{s}^{2}}={{\sum\limits_{i=1}^{n}{({{x}_{i}}-\overline{x})}}^{2}}/(n-1).$$ We will use this form of s2, with denominator (n−1), so that it is an unbiased estimator of σ2.