<p>AMS312 Spring_2007 </p><p>Apr 10th</p><p>1、 student ’s t-test (continued: hypothesis test) Example (Page 490): Pica is a children’s disorder characterized by a craving for nonfood substances such as clay, plaster, and paint. Anyone affected runs the risk of ingesting high levels of lead, which can result in kidney damage and neurological dysfunction. Checking a child’s blood lead level is a standard procedure for diagnosing the condition. Among children between the ages of six months and five years, blood lead levels of 16.0mg/l are considered “normal”. Recently, a random sample of twelve children enrolled in Molly’s Mighty Bear Nursery had their blood lead levels checked. The resulting sample mean and sample standard derivation were 18.65 and 5.049, respectively. Can it be concluded that children at this particular facility tend to have higher lead levels? At thea = 0.05level, is the increase from 16.0 to 18.65 statistically significant?</p><p>H :m = 16 Solution: 0 Average lead level for all children in Milly’s daycare Ha :m > 16</p><p>(X- X )2 X =18.65 , s=5.049 ( s = i )a = 0.05 n -1</p><p>Assuming the blood lead distribution for the population is normal</p><p>X - m 18.65- 16.0 T =0 = =1.82 Test statistic: 0 s 5.049 n 12</p><p>Ata = 0.05, we will reject the null hypothesis if T0> tn- 1,a = t 11,0.05 =1.7959</p><p>Now 1.82>1.7959 hence we reject the null hypothesis. 2、 student ’s t-test (continued: confidence interval) Confidence interval for population mean based on student t-distribution.</p><p>2 2 Let X1, X 2 ,..., Xn : N (m , s ) i.i.d. X and s are known, then: </p><p>1 AMS312 Spring_2007 </p><p>X - m T= : t s n-1 n</p><p>P(- t < T < t ) = 1 -a We have: n-1,a n - 1, a 2 2</p><p>X - m P(< t < = t ) - 1 a � n-1,a n - 1, a 2s 2 n P( X< ts <m + X = t s ) - 1 a � n-1,a n - 1, a 2n 2 n</p><p>[X- ts , X + t s ] Hence the confidence interval is: n-1,a n - 1, a 2n 2 n 3、 likelihood ratio test (LRT):</p><p>H0 :q w c Ha :q w</p><p>Data: X1, X 2 ,..., X n : i.i.d. follow p.d.f. f (x;q )</p><p> n</p><p>Likelihood function: L= f( xi ;q ) i=1</p><p> max L l = q w 1 max L q wc</p><p>If Ho is true, then l ∐1 so we should reject Ho if l �c 1</p><p>P(l� c | H 0 ) a</p><p>2 2 Example: Let X1, X 2 ,..., Xn : N (m , s ) ,s is unknown. Please derive the LRT for m .</p><p>2 2 H0: m= m 0 w= {( m0 , s ), s positive }</p><p>2 2 Ha : m m0 w= {( m , s ), m m0 & s positive }</p><p> c 2 2 W =wU w = {( m , s ), mR , s positive }</p><p>2 AMS312 Spring_2007 </p><p>2 2 (X -m ) n (Xi -m ) n n -i - 1 2- 2 L= f( x ;m , s2 ) = e2s = (2 ps 2 ) 2 e 2 s 照 i 2 i=1 i = 1 2ps</p><p> max L m= m, s 2 > 0 l = 0 MLE’s for m ands 2 in w max L m�R, s 2 0</p><p>For the numerator:</p><p> n 2 n n 2 (X i - m0 ) max2 lnL0 = max 2 [ - ln(2p ) - ln( s ) - ] (*) m= m0, s > 0 m = m 0 , s > 0 2 2 2 i=1 s</p><p>2 d(*) n 1 邋(Xi-m0 )2 ( X i - m 0 ) Then = - + = 0 � sˆ ds22 s 2 2 s 4 0 n</p><p> n2 n n -2p (X - m ) - - Hence maxL= (2psˆ 2 ) e2 = [i 0 ] 2 e 2 q w 0 n</p><p>Similarly, we can derive: For the denominator: </p><p>2 2 (Xi - X ) max2 L� mˆ = X , s m�R, s 0 n</p><p> n2 n n -2p (X- X ) - - maxL= (2ps 2 ) e2 = [i ] 2 e 2 n</p><p>2 n (X - m ) - i 0 2 Therefore, l = [2 ] (Xi - X )</p><p>At the significance level a , we reject the null hypothesis if l �c 1 C is the value such that: </p><p>P(l� c | H0 ) a</p><p>2 n (X - m ) - 蓿 i 0 2 P([2 ] c | H0 :m= m 0 ) = a (Xi - X ) (X - m )2 蕹 i 0 * P(2 c | H0 :m= m 0 ) = a (Xi - X )</p><p>3 AMS312 Spring_2007 </p><p> n n n n 2 2 2 2 2 And, 邋(Xi-m0 ) = ( X i - X ) + 邋 ( X - m 0 ) = ( X i - X ) + n ( X - m 0 ) i=1 i = 1 i = 1 i = 1</p><p>2 2 2 (X i - m0 ) n( X-m0 )n ( X - m 0 ) n 2 n=1 + n = 1 + n = 1 + (t0 ) 2 2n-1 2 n - 1 邋(Xi- X ) ( X i - X ) ( X i - X ) i=1 i = 1 i = 1 n -1</p><p>Hence the above formula becomes: </p><p>2 ** P(( t0 )� c | H 0 :m= m 0 ) a Given H0: T 0: tn- 1 ** ** 蕹 P( T0 c | H 0 )+ P ( T 0 � = c | H 0 ) a</p><p>Symmetric, therefore we reject Ho at significance levela if:</p><p>(T )2 ( t ) 2 | T |> t 0驰 n-1,a 0 n - 1, a 2 2</p><p>4</p>