IBMメーカーSPSS 19の使用説明書/サービス説明書
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1 Using IBM SPSS 19* Descriptive Statistics SPSS Help. SPSS has a good online help system. Once SPSS is up and running, you can nd it by going to Help>T opics in the menu bar , i.e., click Help in the menu bar and then click T opics in the drop window that opens.
2 Y ou can then open any of the books comprising the tutorial by clicking on the + to get to the various subtopics. Once in a subtopic is open, you can just keep clicking on the right and left arrows to move through it page by page. I suggest going through the entire Overview booklet.
3 Sorting the Data . From the menu, choose Data>Sort Cases… , click the right arrow to move protein to the Sort by box, make sure Ascending is chosen, and click OK . Our data column is now in ascending order . However , the rst thing that come up is an output page telling you what has happened.
4 Click the Statistics... button, then make sure Descriptives and Percentiles are checked. W e will use 95% for Condence Interv al for Mean . Click Continue . Then click Plots... . Under Bo xplots , select F actor levels together , and under Descriptiv e , choose both Stem-and-leaf and Histogr am .
5 Then click OK . This opens an output window with two frames. The frame on the left contains an outline of the data on the right..
6 The Standard Error of the Mean is a measure of how much the value of the mean may vary from repeated samples of the same size taken from the same distribution. The 95% Condence Interv al for Mean are two numbers that we would expect 95% of the means from repeated samples of the same size to fall between.
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8 Now click on a number on the horizontal axis and then click on Number Format . In the diagram to the left below , we see that we have 2 decimal places. The values in this window can be changed as desired. Next, click on one of the bars and then Binning in the Properties window .
9 Next choose P ercentiles from either output frame. The following comes up. Obviously , there are two dif ferent methods at work here. The formulas are given in the SPSS Algorithms Manual . T ypically , use the W eighed A verage . T ukey’ s Hinges was designed by T ukey for use with the boxplot.
10 Then click back to Data View . From the menu, choose T ransform>Compute V ariable... . When the Compute V ariable window comes up, click Reset , and type cum_bin in the box labeled T arget V ariable . Scroll down the Function group: window to CDF & Noncentral CDF to select it, then scroll to and select Cdf .
1 1 Poisson Distribution. Let us assume that l =.5. W e will rst nd P(X ≤ x | .5)for x = 0, ..., 15, i.e., the cumula- tive probabilities. First put the numbers 0 through 15 in a column of a worksheet. (W e have already done this above. Again, you only need to enter the numbers whose cumulative probability you desire.
12 cumulative Poisson probabilities are now found in the column cum_pois. Now we want to put the individual Poisson probabilities into the column pois_pro . Do basically the same as above, except make the T arget V ariable “ pois_pro ,” and the Numeric Expression “ CDF .
13 The probability is now found in the column cum_norm . Staying with the normal distribution with mean 100 and standard deviation 20, suppose we with to nd P(90 ≤ X ≤135). Do as above except make the T arget V ariable “ int_norm ,” and the Numeric Expression “ CDF .
14 Condence Intervals and Hypothesis T esting Using t A Single Population Mean . W e found earlier that the sample mean of the data given on page 2, which you may have saved under the name protein.
15 SPSS gives us the basic descriptives in the rst table. In the second table, we are given that the t -value for our test is 1.110 . The p -value (or Sig. (2-tailed) ) is given as .272 . Thus the p -value for our one-tailed test is one- half of that or .
16 and again press Add . Then hit OK and complete the V ariable View as follows. Returning to Data View gives a window whose beginning looks like that below . Now we wish to test the hypotheses H 0 : m 1 - m 2 = 0 H a : m 1 - m 2 ≠ 0 where m 1 refers to the population mean for the non-smokers and m 2 refers to the population mean for the smokers.
17 Then click Continue . As before, click Options... , enter 95 (or any other number) for Condence Inter- val , and again click Continue followed by OK .
18 discount this hypothesis, so we will take our results from the Equal V ariances Assumed column. W e see that, with 30 degrees of freedom, we have t =-2.468 and p =.020, so we reject the null hypothesis H 0 : m 1 - m 2 = 0 at the a =.05 level of signicance.
19 The rst output table gives the descriptives and a second (not shown here) gives a correlation coefcient. From the third table, which has been pivoted to interchange rows and columns, we see that we have a t -score of 12.740. The fact that Sig.
20 H 0 : m N = m F = m C H a : Not all of m N , m F , and m C are equal. From the menu we choose Analyze>Compare Means>One- W ay ANOV A... . In the window that opens, place volume under Dependent List and Smok er[smoking] u nder F actor . Then click P ost Hoc.
21 Then we click options and choose Descriptive , Homogeneity of variance test , and Means plot . The Homogeneity of v ariance test calculates the Levene statistic to test for the equality of group variances. This test is not dependent on the assumption of normality .
22 The results of the T est of Homogeneity of V ariances is nonsignicant since we have a p value of .974 , showing that there is no reason to believe that the variances of the three groups are different from one another . This is reassuring since both ANOV A and T ukey's HSD have equal variance assumptions.
23 Simple Linear Regression and Corr elation W e will use the following 109 x-y data pairs for simple linear regression and correlation. The x 's are waist circumferences (cm) and the y 's are measurements of deep abdominal adipose tissue gathered by CA T scans.
24 Then click OK to get the following scatter plot, which leads us to suspect that there is a signicant linear relation- ship. Regression. T o explore this relationship, choose Analyze>R egression>Linear ... from the menu, select and move y under Dependent and x under Independent(s) .
25 Then click Statistics... , and in the window that opens with Estimates and Model t already checked, also check Condence interv als and Descriptives .
26 Then click Continue followed by OK to get the output. W e rst see the mean and the standard deviation for the two variables in the Descriptive Statistics . In the Model Summary , we see that the bivariate correlation coefcient r ( R ) is .819, indicating a strong positive linear relationship between the two variables.
27 reject the null hypothesis of b =0. W e now return to the scatter plot. Double click on the plot to bring up the Chart Editor and choose Options>Y Axis R eference Line from the menu. In the window that opens, select Refernce Line and, from the drop- down menue for Set to: , choose Mean and then click Apply .
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29 for the mean value m y|74.5 is ( 32.41572, 52.72078) , corresponding to the limits of the inner bands at x=74.5 in the scatter plot, and the 95% condence interval for the individual value y I (74.5)is (-23.7607,108.8972), correspond- ing to the limits of the outer bands at x =74.
30 W e see again that the Pearson Correlation r is .819, and from the Sig. of .000, we know that the p -value is less than .001 and so we would reject a null hypothesis of r =0. Multiple Regr ession W e will use the following data set for multiple linear regression.
31 Choose Analyze>R egression>Linear ... from the menu, select and move minutes under Dependent and ram , input , and output , in that order , under Independent(s) . Then ll in the options for Statistics , Plots , and Save exactly as you did for simple linear regression.
32 1.049x 3 . From the last two rows of numbers in the table, one gets that 95% condence intervals are (-.694,2.645) for a , (.061,.138) for b 1 , (.000,.487) for b 2 , and (.692,1.407) for b 3 . The t test is used for testing the various null hypotheses b i =0.
33 Finally , consider the residual plot below . On the horizontal axis are the standardized y values from the data points, and on the vertical axis are the standardized residuals for each such y .
34 Then click back to Data View . From the menu, choose T ransform>Compute V ariable... . When the Compute V ariable window comes up, click R eset , then type lny in the box labeled T arget V ariable .
35 Choosing a Model using Curve Estimation. T o nd an appropriate model for a given data set, such as the one in the previous section, choose Analyz e>Regression>Curve Estimation.
36 Finally , click OK . W e show below the output for the Quadratic model. The regression equation is ŷ=336.790- 693.691x+295.521x 2 . The other data, although arranged differently , is similar to that for linear and multiple regression. W e do note that the Standard Error is 1 1 1.
37 Chi-Squar e T est of Independence For data, we will use a survey of a sample of 300 adults in a certain metropolitan area where they indicated which of three policies they favored with respect to smoking in public places. W e wish to test if there is a relationship between education level and attitude to- ward smoking in public places.
38 This is not very well documented, but the rst thing we need to do for c 2 is to tell SPSS which column contains the frequency counts. Choose Data>W eight Cases... from the menu, and in the window that opens, choose W eight cases by and move the variable count under Frequency V ariable .
39 Check Observed and Expected under Counts , followed by Continue and OK . The rst table of output simply provides a table of the Counts and the Expected Counts if the variables are independent. From the second table, the Pearson Chi-Square statistic is 22.
40 From the menu, choose Analyze>Nonpar ametric T ests>Legacy Dialogs>2 Related Samples... . In the window that opens, rst click output followed by the arrow to make it V ariable 1 for Pair 1 , then con- stant followed by the arrow to make it V ariable 2 .
41 The Z in the second table is the standardized normal approximation to the test statistic, and the Asymp. Sig (2-tailed) of .140, which we will use as our p -value, is estimated from the normal approximation. Because of the size of this p -value, we will not reject the null hypothesis at any of the usual levels of signicance.
42 Put 1 in the box for Group 1 and 2 in the box for Group 2 . Then click Continue . Y ou may click Options... if you want the output to include descriptive statistics and/or quartiles.
43 T o create the control chart(s), click Analyze>Quality Control>Control Charts... from the menu bar , and in the window that opens, select X -Bar , R, s under V ariable Charts and make sure Cases are units is checked under Data Organization .
44 Click Options , and enter 2 for Number of Sigmas . After clicking Continue, since we have specications for the mean, we click Statistics... , and in the window that opens, based on our specied mean and standard deviation, enter 50.756 for Upper and 49.
45 Control Charts for the Proportion. T o illustrate control charts for the proportion, we use the number of defec- tives in samples of size 100 from a production process for twenty days in August.
46 Now click Options, and enter 3 for Number of Sigmas . Then click Continue followed by OK to get the control chart, which is again pretty much self-explanatory . W e see that the process is out of control on August 24 and 25, although it is hard to call too few defectives out of control.
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ですが、ユーザガイドが果たす重要な役割の一つは、IBM SPSS 19に関する問題の解決を支援することです。そこにはほとんどの場合、トラブルシューティング、すなわちIBM SPSS 19デバイスで最もよく起こりうる故障・不良とそれらの対処法についてのアドバイスを見つけることができるはずです。たとえ問題を解決できなかった場合でも、説明書にはカスタマー・サービスセンター又は最寄りのサービスセンターへの問い合わせ先等、次の対処法についての指示があるはずです。