Design and Statistics - SPSS Data - Trend Analysis

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Design and Statistics Assignment

Assignment Task

A sleep research team is interested in the effects of sleep deprivation on subjects’ ability to perform a vigilance task, such as locating objects moving on a radar screen. They house their subjects in a sleep laboratory so that they can monitor their sleeping habits. There are 4 conditions, namely, 4, 12, 20 and 28 hours without sleep. 4 subjects are randomly assigned to each of the conditions. They are scored on the number of failures to spot objects on a radar screen during a 30-minute test period. Scores are presented below:

Enter the above data into an SPSS data file using informative variable and value labels as appropriate. Append or cut and paste a screen shot of your data file
Using syntax with the glm command in SPSS, obtain descriptive statistics, perform the overall ANOVA and append the SPSS output. Report the significance of the overall test, assuming that EER is set at ?=.05
Draw a conclusion concerning the overall test.

The following refers to the above data but is a separate question. Assume that the research team had decided to test whether the amount of sleep deprivation was systematically related to missing objects on the radar screen by performing a trend analysis, controlling EER at ?=.05. Use syntax with the glm command in SPSS to perform this analysis, append the output and answer the following questions. 4. Was a trend analysis appropriate in this design? Why or why not? 5. Interpret and draw conclusions about the results of the analysis you have performed, making explicit the decision rule you are using. 6. Contrast code the deprivation condition variable so that you can replicate the trend analysis above, perform the analysis via the regression command in SPSS and append the output. Details on submitting your assignment: - Please ensure that you include the syntax used (which will appear at the top of your output) when you append your output for the different components of the question. - Please read the relevant sections of the unit outline on how to submit assignments (online via Plagiarism in iLearn). - There is no specified word or page limit, but answers won’t need to be too long, just answer the questions. - Any questions about the assignment should be posted to the discussion forum on iLearn, rather than directed at your tutor. This ensures all students have equal access to the answers to questions.

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Solution:

The data was entered in SPSS and the output is given below

2.The SPSS syntax is given below DATASET ACTIVATE DataSet0. UNIANOVA Failures BY Time /METHOD=SSTYPE(3) /INTERCEPT=INCLUDE /POSTHOC=Time(TUKEY) /PLOT=PROFILE(Time) /EMMEANS=TABLES(OVERALL) /EMMEANS=TABLES(Time) /PRINT=OPOWER HOMOGENEITY DESCRIPTIVE /PLOT=SPREADLEVEL RESIDUALS /CRITERIA=ALPHA(.05) /DESIGN=Time. 3.The SPSS output is given below

6. The SPSS syntax is given below REGRESSION /DESCRIPTIVES MEAN STDDEV CORR SIG N /MISSING LISTWISE /STATISTICS COEFF OUTS CI(95) R ANOVA COLLIN TOL /CRITERIA=PIN(.05) POUT(.10) /NOORIGIN /DEPENDENT Failures /METHOD=ENTER Twelvehour Twentyhour twentyeighthour /RESIDUALS HISTOGRAM(ZRESID) NORMPROB(ZRESID). The SPSS output is given below

The Levene’s test for equality variance test is used to check the assumption of homogeneity of variances. The value of F test statistic is F (3, 12) = 1.245 and its corresponding p – value is 0.337 > 0.05. Since the p – value of F test statistic is greater than 0.05, there is sufficient evidence to conclude that the assumption of homogeneity of variance is satisfied

The value of f test statistic is7.343 and its corresponding p – value is 0.005 < 0.05, indicating that there is a significant difference in the mean number of failures to spot objects on a radar screen during a 30-minute test period among four different time periods of sleep. Therefore, there is sufficient evidence to conclude that there is a significant effects of sleep deprivation on subjects’ ability to perform a vigilance task, such as locating objects moving on a radar screen

Conclusion The value of f test statistic is7.343 and its corresponding p – value is 0.005 < 0.05, indicating that there is a significant difference in the mean number of failures to spot objects on a radar screen during a 30-minute test period among four different time periods of sleep. Therefore, there is sufficient evidence to conclude that there is a significant effects of sleep deprivation on subjects’ ability to perform a vigilance task, such as locating objects moving on a radar screen. The post hoc test was performed to determine which time pair mean number of failures to spot objects on a radar screen differs significantly. Tukey post hoc test was used and the findings suggest that respondents who enrolled in 28 hour sleep group found more number of number of failures to spot objects on a radar screen when compared to that of respondents placed in 4 hours and 12 hours group 4.The trend analysis is not appropriate for this design, as the readings are measured on a fixed time based but on the exact time basis. When a set of measurements of a certain process are measured on a time series, trend estimation can be used to justify the readings and trends about the data. 5.The mean number of number of failures to spot objects on a radar screen by participants enrolled in 28 hour sleep group is 61.75 ± 6.133 (95% CI: 48.387 – 75.113), the mean number of number of failures to spot objects on a radar screen by participants enrolled in 20 hour sleep group is 57.5 ± 6.133 (95% CI: 44.137 – 70.863), the mean number of number of failures to spot objects on a radar screen by participants enrolled in 12 hour sleep group is 37.75 ± 6.133 (95% CI: 24.387 – 51.113), and the mean number of number of failures to spot objects on a radar screen by participants enrolled in 4 hour sleep group is 26.5 ± 6.133 (95% CI: 13.137 – 39.863). Thus, we see that the respondents who enrolled in 28 hour sleep group found more number of number of failures to spot objects on a radar screen when compared to that of respondents placed in 4 hours and 12 hours group 6.The SPSS syntax is given below REGRESSION /DESCRIPTIVES MEAN STDDEV CORR SIGN /MISSING LISTWISE /STATISTICS COEFF OUTS CI(95) R ANOVA COLLIN TOL /CRITERIA=PIN(.05) POUT(.10) /NOORIGIN /DEPENDENT Failures /METHOD=ENTER Twelvehour Twentyhour twentyeighthour /RESIDUALS HISTOGRAM(ZRESID) NORMPROB(ZRESID). The SPSS output is given below

The regression equation is No of Failure = 26.5 + 11.25 * Twelve Hour + 31 * Twenty Hour + 35.25 * Twenty eight Hour The value of F test statistic is 7.343 and its corresponding p – value is 0.005 < 0.05, indicating that the estimated regression model is good fit in predicting number of failure spot objects on a radar screen The coefficient of determination is 0.647. This indicates that the regression model which was constructed to predict future values of number of failure spot objects on a radar screen explains 64.7% of the variation in the dependent variable number of failure spot objects on a radar screen The coefficient of the independent variable twelve hour is 11.25. This indicates that, when the participant is from 12 hour sleep group, then the number of failure spot objects on a radar screen increases by 11.25 units, provided the other independent variables held constant, but statistically insignificant (t test statistic = 1.297, p – value = 0.219 > 0.05) The coefficient of the independent variable twenty hour is 31. This indicates that, when the participant is from 20 hour sleep group, then the number of failure spot objects on a radar screen increases by 31 units, provided the other independent variables held constant and statistically significant (t test statistic = 3.574, p – value = 0.004 < 0.05) The coefficient of the independent variable twenty eight hour is 35.25. This indicates that, when the participant is from 28 hour sleep group, then the number of failure spot objects on a radar screen increases by 35.25 units, provided the other independent variables held constant, but statistically insignificant (t test statistic = 4.064, p – value = 0.002 < 0.05)

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