【正文】 第四章 (二 ) 假設檢驗 Hypothesis testing 山東大學公共衛(wèi)生學院 劉靜 2 ?Example 1 The dean of a medical school states that the students of the school are a highly intelligent group with an average IQ of 135. Do you believe him? The students of our school are very clever. The average IQ of our students is up to 135. 3 Suppose ? is really 135 Random Sample sample 135?X? Population Not necessarily. Sampling error will always cause the sample mean to deviate from the population mean. 4 H0: ?=?0, then ~N(?0,?2/n) XXFig. 1 Distribution of sample means when H0 is true To suspect H0 ?0 5 H0: ?=?0 ?0 Fig. 2: Regions of rejection and nonrejection in hypothesis testing of one sample mean Reject region XNonrejection region ),(~20 nNX??area of nonrejection region, 1? area of rejection region, ? Critical value of sample mean 6 : Critical value of t in hypothesis testing for one sample mean area of nonrejection region, 1? area of rejection region, ? Critical value of t : t?,? t nsXt ???Degrees of freedom, ?=n1 7 T a b le 1 Cr iti c a l v a lu e s o f t d i str ib u tio n Pro b a b i l i t y ， P 1 t ai l 0 . 2 5 0 . 2 0 0 . 1 0 0 . 0 5 0 . 0 2 5 0 . 0 1 0 . 0 0 5 0 . 0 0 2 5 0 . 0 0 1 0 . 0 0 0 5 D eg rees o f f reed o m ? 2 t ai l 0 . 5 0 0 . 4 0 0 . 2 0 0 . 1 0 0 . 0 5 0 . 0 2 0 . 0 1 0 . 0 0 5 0 . 0 0 2 0 . 0 0 1 1 1 . 0 0 0 1 . 3 7 6 3 . 0 7 8 6 . 3 1 4 1 2 . 7 0 6 3 1 . 8 2 1 6 3 . 6 5 7 1 2 7 . 3 2 1 3 1 8 . 3 0 9 6 3 6 . 6 1 9 2 0 . 8 1 6 1 . 0 6 1 1 . 8 8 6 2 . 9 2 0 4 . 3 0 3 6 . 9 6 5 9 . 9 2 5 1 4 . 0 8 9 2 2 . 3 2 7 3 1 . 5 9 9 3 0 . 7 6 5 0 . 9 7 8 1 . 6 3 8 2 . 3 5 3 3 . 1 8 2 4 . 5 4 1 5 . 8 4 1 7 . 4 5 3 1 0 . 2 1 5 1 2 . 9 2 4 4 0 . 7 4 1 0 . 9 4 1 1 . 5 3 3 2 . 1 3 2 2 . 7 7 6 3 . 7 4 7 4 . 6 0 4 5 . 5 9 8 7 . 1 7 3 8 . 6 1 0 5 0 . 7 2 7 0 . 9 2 0 1 . 4 7 6 2 . 0 1 5 2 . 5 7 1 3. 3 6 5 4 . 0 3 2 4 . 7 7 3 5 . 8 9 3 6 . 8 6 9 6 0 . 7 1 8 0 . 9 0 6 1 . 4 4 0 1 . 9 4 3 2 . 4 4 7 3 . 1 4 3 3 . 7 0 7 4 . 3 1 7 5 . 2 0 8 5 . 9 5 9 7 0 . 7 1 1 0 . 8 9 6 1 . 4 1 5 1 . 8 9 5 2 . 3 6 5 2 . 9 9 8 3 . 4 9 9 4 . 0 2 9 4 . 7 8 5 5 . 4 0 8 8 0 . 7 0 6 0 . 8 8 9 1 . 3 9 7 1 . 8 6 0 2 . 3 0 6 2 . 8 9 6 3 . 3 5 5 3 . 8 3 3 4 . 5 0 1 5 . 0 4 1 9 0 . 7 0 3 0 . 8 8 3 1 . 3 8 3 1 . 8 3 3 2 . 2 6 2 2 . 8 2 1 3 . 2 5 0 3 . 6 9 0 4 . 2 9 7 4 . 7 8 1 10 0 . 7 0 0 0 . 8 7 9 1 . 3 7 2 1 . 8 1 2 2 . 2 2 8 2 . 7 6 4 3 . 1 6 9 3 . 5 8 1 4 . 1 4 4 4 . 5 8 7 21 0 . 6 8 6 0 . 8 5 9 1 . 3 2 3 1 . 7 2 1 2 . 0 8 0 2 . 5 1 8 2 . 8 3 1 3 . 1 3 5 3 . 5 2 7 3 . 8 1 9 22 0 . 6 8 6 0 . 8 5 8 1 . 3 2 1 1 . 7 1 7 2 . 0 7 4 2 . 5 0 8 2 . 8 1 9 3 . 1 1 9 3 . 5 0 5 3 . 7 9 2 23 0. 6 8 5 0 . 8 5 8 1 . 3 1 9 1 . 7 1 4 2 . 0 6 9 2 . 5 0 0 2 . 8 0 7 3 . 1 0 4 3 . 4 8 5 3 . 7 6 8 24 0 . 6 8 5 0 . 8 5 7 1 . 3 1 8 1 . 7 1 1 2 . 0 6 4 2 . 4 9 2 2 . 7 9 7 3 . 0 9 1 3 . 4 6 7 3 . 7 4 5 25 0 . 6 8 4 0 . 8 5 6 1 . 3 1 6 1 . 7 0 8 2 . 0 6 0 2 . 4 8 5 2 . 7 8 7 3 . 0 7 8 3 . 4 5 0 3 . 7 2 5 … … … … … … … … … … … ∞ 0 . 6 7 4 5 0 . 8 4 1 6 1 . 2 8 1 6 1 . 6 4 4 9 1 . 9 6 0 0 2 . 3 2 6 3 2 . 5 7 5 8 2 . 8 0 7 0 3 . 0 9 0 2 3 . 2 9 0 5 t t 0 8 ? A random sample of 10 students is drawn。 their IQs are as follows: 115 140 133 125 120 126 136 124 132 129 12810129132140115 ?????? ?X1/)( 22??????nnxxs9 Calculate the value of t that corresponds to the mean of the sample. 135128??????nsXt?10 t?,?= In this example, the t score () does fall in the region of rejection. Therefore, the null hypothesis should be rejected, and naturally the alternative hypothesis is accepted. Conclusion!! ?=n1=9 t 11 Steps to test a hypothesis about a mean 1. State the null and alternative hypotheses, H0 and H1. 2. Select the decision criterion ? (or “significance level”). 3. Establish the critical value(s) of the testing statistic. 4. Draw a random sample from the population, then calculate the mean and the standard deviation of that sample and estimate sampling error ( ). Xs5. Calculate the value of the test statistic t corresponding to the mean of the sample. 6. Compare the calculated value of t with the critical values of t, and then draw an inferential conclusion to reject or not to reject the null hypothesis. 12 Hypothesis testing of the current example :1 3 5:0100?????????HH In t h i s p ro b l em , ,128,10 ??? sXn 135128??????nsXt? ,8 3 , ??? Ptt. Conclusion: The null hypothesis was rejected and logically the alternative hypothesis was accepted under the significance level of . Therefore, the mean IQ of the students of that medical school was less than 135. , ?=n 1=9 13 一、假設檢驗的基本思想 反證法： 當一件事情的發(fā)生只有兩種可能 A和B，為了肯定其中的一種情況 B，但又不能直接證實 B，這時否定另一種可能 A，則間接的肯定了 B。 假設檢驗之所以可行 ,其理論背景為 “ 小概率事件原理 ” ，即小概率事件在一次抽樣中不可能發(fā)生。 14 二、假設檢驗的一般步驟 （一）建立假設和確定檢驗水準 1． 確定單、雙側檢驗 根據(jù)研究目的或專業(yè)知識確 定單側檢驗 (onesided test)或雙側檢驗 (twosided test)。 2 ． 建立假設 一種是檢驗假設（ hy pot hes i s und er t es t ）， 記為0H； 另一種是備擇假設 ( al t er nat i v e h y pot hes i s ) , 記為1H，是與0H相聯(lián)系且對立的假設。 3 ． 確定檢驗水準 亦稱顯著性水準 （ s i g ni f i ca nce l ev el ） ， 符號為 ? ，是判斷拒絕或不拒絕0H的水準，也是 允許犯 Ⅰ 型錯誤 的概率。 15 二、假設檢驗的一般步驟（續(xù)） （二）選定檢驗方法和計算統(tǒng)計量 根據(jù)研究目的、設計類型、資料類型及其分布特征等選用適當?shù)慕y(tǒng)計檢驗方法，并計算出相應的檢驗統(tǒng)計量。 （三）確定值，做出推斷結論 求出檢驗統(tǒng)計量后，查相應的統(tǒng)計用表，得出概率值。 P值是指從 H0所規(guī)定的總體中隨機抽樣時，獲得等于及 大于（負值時為等于及小于）現(xiàn)有樣本統(tǒng)計量的概率。 16 P值的含義 —— 雙側檢驗 area of nonrejection,