1 / 31

Percentile Practice Problem (1)

Percentile Practice Problem (1). This question asks you to use percentile for the variable [marital]. Recall that the use of percentile is for ordinal or interval level variable. Since [marital] is a nominal level variable, it is incorrect to use percentile to answer this question.

zody
Download Presentation

Percentile Practice Problem (1)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Percentile Practice Problem (1) This question asks you to use percentile for the variable [marital]. Recall that the use of percentile is for ordinal or interval level variable. Since [marital] is a nominal level variable, it is incorrect to use percentile to answer this question.

  2. Percentile Practice Problem (2) This question asks you to use percentile for the variable [age] to answer whether a value of 55 would position in the top 10% of cases in this data set. Recall that the use of percentile is for ordinal or interval level variable. Since [age] is an interval level variable, it is good to use percentile to answer this question.

  3. Percentile in SPSS (1) To add the percentile value to each case in SPSS data set, use Transform > Rank Cases… in the Menu Bar.

  4. Percentile in SPSS (2) After selecting and moving the variable [age] to the “Variable(s):” box, click on “Rank Types…” button.

  5. Percentile in SPSS (3) In the next window, select “Fractional rank as %” option and click on “Continue” button.

  6. Percentile in SPSS (4) Click on “Ties…” button to select the rule for handling tied cases.

  7. Percentile in SPSS (5) Select “Low” option in “Rank Assigned to Ties” section. Click “Continue” button in this window and click “OK” in the next window to exit.

  8. Percentile in SPSS (6) Now, you will notice that a new variable “PERCENT of AGE” [page] was added to the data set. These values are the rank percentile values of the cases. For example, 10.67 means the age of this case is located at bottom 10.67% of the age distribution of the cases in the data set.

  9. Percentile in SPSS (7) Right click on the variable name area of the Data View will show you a pop up menu. Then select “Sort Ascending” on the list.

  10. Percentile in SPSS (8) Although there were 17 cases with tied age, the previous selection of “Low” of the “Ties…” option in the slide 7 ensured that all the respondents age of 56 positioned in the top 10% (100-90.53<10%) of the age distribution of this sample. The answer to the question is false. An age of 55 would not place a respondent in the top 10%; that would require an age of 56.

  11. Standard Scores (z-scores) Practice (1) This question asks you to use standard scores for the variable [marital]. Recall that the use of standard scores is for ordinal or interval level variable. Since [marital] is a nominal level variable, it is incorrect to use standard scores to answer this question.

  12. Standard Scores (z-scores) Practice (2) This question asks you to use standard scores for the variable [nperinhh]. Recall that the use of standard scores is for ordinal or interval level variable. Since [nperinhh] is an interval level variable, it satisfies the requirement. Also, recall that use of z-score to get the accurate probability score requires that the distribution of the variable follows a normal curve. Let’s check whether it satisfies too using SPSS.

  13. Z-Scores in SPSS (1) You can check the normality of the variable by clicking Analyze > Descriptive Statistics > Descriptives… It also provides you with standard scores that can be added to your data set.

  14. Z-Scores in SPSS (2) After select and move the variable of interest into the “Variable(s):” box, click on “Options…” button.

  15. Z-Scores in SPSS (3) In the “Descriptive: Options” window, select “Kurtosis” and “Skewness” in the “Distribution” section. Remember that Kurtosis or Skewness larger than +/- 1.00 imply that the variable is not normally distributed.

  16. Z-Scores in SPSS (4) Both Skewness (=12.749) and Kurtosis (=316.388) are way too larger than one. Because it shows that [nperinhh] is not normally distributed, you cannot use z-scores to estimate the probability value. Incorrect application of a statistic is the correct answer.

  17. Standard Scores (z-scores) Practice (3) This question asks you to use standard scores for the variable [interdur]. Recall that the use of standard scores is for ordinal or interval level variable. Since [interdur] is an ordinal level variable, it satisfies the requirement. Also, recall that use of z-score to get the accurate probability score requires that the distribution of the variable follows a normal curve.

  18. Z-Scores in SPSS (5) You can check the normality of the variable by clicking Analyze > Descriptive Statistics > Descriptives… It also provides you with standard scores that can be added to your data set.

  19. Z-Scores in SPSS (6) Since you will use standardized scores to answer the question, make sure that “Save standardized values as variables” option was selected at this time. Then click “Options…” button.

  20. Z-Scores in SPSS (7) In the “Descriptive: Options” window, select “Kurtosis” and “Skewness” in the “Distribution” section. Then, click “Continue” and “OK” buttons. Remember that Kurtosis or Skewness larger than +/- 1.00 imply that the variable is not normally distributed.

  21. Z-Scores in SPSS (8) The SPSS output shows that neither Skewness (=-0.675) nor Kurtosis (=-0.390) is greater than +/- 1, which implies that [interdur] is normally distributed. Now, you can use z-scores to estimate the probability value.

  22. Z-Scores in SPSS (9) You notice that a new variable “Zscore: First time R used Internet” [zinterdu] was created and added to the data set. Each score in this column is the standardized score, or z-score, that corresponds to each value of [interdur]. You can obtain probability values for the z-scores and add to the original data set.

  23. Z-Scores in SPSS (10) To add the probabilities, or p-values, to the data set , click on Transform > Compute…

  24. Z-Scores in SPSS (11) In the Compute Variable window, first, type a variable name you want to assign in “Target Variable” box. I usually use a “p”, followed by the first seven letters in the original variable name. Second, select CDFNORM(zvalue) from “Functions” box and move to “Numeric Expression:” box using triangle button. This function returns the corresponding p-value for the z-value of a normal distribution.

  25. Z-Scores in SPSS (12) Once you move the CDFNORM( ) function to the “Numeric Expression:” box, it will look like this. Now, you find the “Zscore: First time R used Internet” [zinterdu] and replace “?” with the variable by clicking the right arrow button.

  26. Z-Scores in SPSS (13) CDFNORM(zinterdu) calculates the probability from the left tail of the distribution up to the z-score value. Since we want the probability above the z-score value, we subtract CDFNORM(zinterdu) from 1. The 1, or 100%, represents the total probability under the normal curve. The formula for the probabilities is complete. Click on the OK button to close the dialog box.

  27. Z-Scores in SPSS (14) Now, you will notice that a new variable [pinterdu] was created. Recall that these values are the probability values corresponding to the z-values [zinterdu] under the assumption of normal distribution. To answer the question, sort the cases using right click on the [pinterdu] variable label and select “Sort Ascending”.

  28. Z-Scores in SPSS (15) You will notice that [interdur] value of 7 corresponds to the standardized score [zinterdu] of 1.09850 and probability value [pinterdu] of .14 from the top. In other words, “A value of 7, or more than 3 years ago, for the variable “first time respondent used internet” would position a survey respondent in the top 20% of cases in the data set.” The answer to the question is true with caution. The caution is added because the variable is ordinal.

  29. Steps in solving percentile problems - 1 Answering homework problems about percentiles: Is the variable ordinal or interval level? Incorrect application of a statistic No Yes Rank all cases in the data set, specifying fractional rank a %, and assigning the low value to ties. Sort the data set by the rank variable. Is 100 – percentile for specified raw score <= top % stated in the problem? False Yes True

  30. Steps in solving z-score problems - 1 Answering homework problems about standard scores: Is the variable ordinal or interval level? Incorrect application of a statistic No Yes Compute skewness and kurtosis for variable? Assumption of normality satisfied? (skew, kurtosis between -1.0 and + 1.0) Incorrect application of a statistic No Yes

  31. Steps in solving z-score problems - 2 Compute z-scores for all cases. Compute probabilities for all z-scores. Sort the data set by probabilities. Is probability for specified raw score <= top % stated in the problem? False Yes No Is the variable ordinal level? True Yes True with caution

More Related