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Dr. Nafez M. Barakat Lecture no.1 Descriptive measures Measure of center o Mean Definition: the mean of the data set is the sum of the observations divided by the number of observation symbolized n by X x i 1 i n Example1 : ( exam scores) let x1 = 88, x2 = 75, x3=95, and x4=100 Compute the mean n Solution : X x i 1 i n 358 89.5 4 o Median Definition: median of a data set Arrange the data in increasing order. -if the number of observation is odd then the median is the observation exactly in the middle of the ordered lest - if the number of observation is even then the median is the mean of the two middle observations in the ordered list. Example2 : weekly salaries 1 Dr. Nafez M. Barakat Find the median for the tow weekly salaries data Table (1.1) Data set I( n=13) odd 300 300 300 940 300 300 400 300 400 450 800 450 1050 Data set II( n= 10)even 300 300 940 450 400 400 300 300 1050 300 SOLUTION : Median for data set (I) = 400$, and for data set (II) = 350$ o Mode Definition: the mode is the value that occurs most frequently in a data set Example : find the mode in data set (I) in table 1.1 Solution: the frequency distribution of the data shown in table 1.2 Table 1.2 Salary 300 400 450 800 Frequency 6 2 2 1 2 940 1050 1 1 Dr. Nafez M. Barakat NOTE: Median< mean Median =mean Right-skewed symmetric mean< Median left-skewed Measure of variation (measure of spread) o Range Definition: the range of the data set is the difference between its maximum and minimum observations : Range = Max - Min o Standard deviation Definition : Standard deviation equal to the square root of the arithmetic mean of the squares of the deviations from the arithmetic mean denoted by S. S (X X ) n 1 2 or S x 2 ( x) 2 / n n 1 o Variance Definition : the variance equal the square of Standard deviation 3 Dr. Nafez M. Barakat Example : the height, in inches of five players on team II are 67, 72, 76, 76 and 84. Obtain the Standard deviation Of these height Solution : X x 75 n X x x 67 -8 72 -3 76 1 76 1 84 9 S (X X ) n 1 2 156 6.2 4 Try to use the formula (x x)2 64 9 1 1 81 156 inches S x 2 ( x) 2 / n n 1 to commute Standard deviation o Inter quartile range Definition : inter quartile range or ( IQR), is the difference between the first and third quartiles, that is Standard deviation IQR = Q3 – Q1 Example : find the IQR fore these data 25 41 27 32 43 66 35 31 15 5 34 26 32 38 16 30 38 30 20 21 Solution : Q1 = 23 , Q3 = 36.5 4 Dr. Nafez M. Barakat IQR = 36.5- - 23 = 13.5 Hypothesis Test for One Population Mean Definition : Null hypotheses and Alternative hypothesis Null hypotheses : a hypothesis to be tested, We use the symbol H0 to represent the null hypothesis. Alternative hypothesis: a hypothesis to be conceder as alternative to null hypothesis, We use the symbol Ha to represent the alternative hypothesis. One sample t test for population Mean We present two step by step procedure for performing a one sample t-test. Procedure (I) covers the critical-value approach, and Procedure (II) covers the p-value approach. One sample t test for population Mean (critical-value approach) Assumptions 1. Normal population or large sample 2. unknown 5 Dr. Nafez M. Barakat Step 1: the null hypothesis is H O : O and the alternative hypothesis is H a : O (two tailed ) H O : O (left tailed ) H O : O ( right tailed ) Step 2 : decide on the significance level, Step 3: compute the value of the test statistic T X O S/ n Step 4: the critical value (s) are t / 2 (two tailed ) or t (left tailed ) or t (right tailed ) with degrees of freedom (df= n-1) 6 Dr. Nafez M. Barakat Step 5 : if the value of the t test statistics falls in the rejection region, reject HO ; otherwise, fail to reject H0 Step 6 : interpret the results of the hypothesis test. Example : table below show the pH levels for 15 lakes; test if the lakes has pH greater than 6 at 5% significant level.( use the critical value approach) Solution : Step 1: state the null and alternative hypotheses 7 Dr. Nafez M. Barakat H O : 6 ( mean PH Level is not greater than 6) H a : 6 (mean PH Level is greater than 6) Step 2 : decide on the significance level, 0.05 Step 3: compute the value of the test statistic T X O 6.6 6 3.458 S / n 0.672 / 15 Step 4: the critical value for a right-tailed test is t t 0.05 1.761 (from table) with df = 15-1 = 14 Step 5: the value of the test statistic, found in step 3 is T=3.458 fail in the rejection region. Consequently , we reject HO One sample t test for population Mean (P-Value Approach) 8 Dr. Nafez M. Barakat Assumptions 3. Normal population or large sample 4. unknown Step 1: the null hypothesis is H O : O and the alternative hypothesis is H a : O (two tailed ) H O : O (left tailed ) H O : O ( right tailed ) Step 2 : decide on the significance level, Step 3: compute the value of the test statistic T X O S/ n Step 4: find the p-value by using table with degrees of freedom (df= n-1) 9 Dr. Nafez M. Barakat Step 5 : if the P- value less than or equal , ( p value ), reject HO ; otherwise, fail to reject H0 Step 6 : interpret the results of the hypothesis test. 10 Dr. Nafez M. Barakat Example : table below show the pH levels for 15 lakes; test if the lakes has pH greater than 6 at 5% significant level. ( use the p-value Approach) Solution : Step 1: state the null and alternative hypotheses H O : 6 ( mean PH Level is not greater than 6) H a : 6 (mean PH Level is greater than 6) Step 2 : decide on the significance level, 0.05 Step 3: compute the value of the test statistic T X O 6.6 6 3.458 S / n 0.672 / 15 Step 4: the p-value = p ( t>= 3.458) = 0.00192 (with df = 15-1 = 14 ) 11 Dr. Nafez M. Barakat Step 5: p value < 0.05) so we reject HO use SPSS program use the SPSS program to perform the hypothesis in previous example STEP 1: Enter The Data As Shown Below 12 Dr. Nafez M. Barakat Step 3 : the result shown below One-Sample Statistics N PH 15 Mean 6.600 Std. Deviation .6719 Std. Error Mean .1735 One-Sample Test Test Value = 6 PH t 3.459 df 14 Sig. (2-tailed) .004 13 Mean Difference .600 95% Confidenc e Int erval of t he Difference Lower Upper .228 .972