SPSS Homework Testing Data for Normality Template

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Psychology

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Feb 20, 2024

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PSYC 354 S TEPHANIE G EORGE SPSS H OMEWORK : T ESTING D ATA FOR N ORMALITY T EMPLATE Problem Set 1: 1. For the “EATING_DRINKING” variable, using the “Explore” command, test the data for normality (as shown in the SPSS tutorial) by performing the following steps: a. Create a table of descriptive statistics for this variable. Paste the table here: (6 pts) Descriptives Statistic Std. Error EATING_DRINKIN G Mean 66.81 2.038 95% Confidence Interval for Mean Lower Bound 62.81 Upper Bound 70.81 5% Trimmed Mean 63.43 Median 60.00 Variance 2075.729 Std. Deviation 45.560 Minimum 0 Maximum 285 Range 285 Interquartile Range 60 Skewness 1.291 .109 Kurtosis 2.617 .218 b. Run a Kolomogorov-Smirnov/ Shapiro-Wilks test for this variable. Paste the table here: (6 pts) 2. Tests of Normality Kolmogorov-Smirnov a Shapiro-Wilk
PSYC 354 Statistic df Sig. Statistic df Sig. EATING_DRINKIN G .133 500 <.001 .914 500 <.001 a. Lilliefors Significance Correction a. Create a “Normal Q-Q plot” for this variable. Paste the table here: (6 pts) 3. Create a histogram for the “EATING_DRINKING” variable. Paste the histogram here: (6 pts)
PSYC 354 4. In several sentences, discuss whether the daily minutes of eating and drinking is normally distributed in this sample. Justify your answer using information from numbers 1 and 2 above. (6 pts) Based on the results of the data set given for eating and drinking it is safe to state that the scores are not normally distributed. We can first support this statement by reviewing the “SIG” box on the normality test. The score is <.001 which is way below the expected normal score of .05 or higher. Secondly, if you review the results of the Q-Q plot graph you can clearly see that most of the dots do not fall along the linear line of the graph. This further supports that the data set is not normally distributed. Problem Set 2: 1. For the “SLEEPING” variable, using the “Explore” command, test the data for normality by performing the following steps: a. Create a table of descriptive statistics for this variable. Paste the table here: (6 pts) Descriptives
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PSYC 354 Statistic Std. Error SLEEPIN G Mean 557.36 6.119 95% Confidence Interval for Mean Lower Bound 545.33 Upper Bound 569.38 5% Trimmed Mean 556.27 Median 555.00 Variance 18719.821 Std. Deviation 136.820 Minimum 0 Maximum 1120 Range 1120 Interquartile Range 150 Skewness .059 .109 Kurtosis 1.796 .218 b. Run a Kolomogorov-Smirnov/ Shapiro-Wilks test for this variable. Paste the table here: (6 pts) Tests of Normality Kolmogorov-Smirnov a Shapiro-Wilk Statistic df Sig. Statistic df Sig. SLEEPIN G .064 500 <.001 .980 500 <.001 a. Lilliefors Significance Correction c. Create a “Normal Q-Q plot” for this variable. Paste the table here: (6 pts)
PSYC 354 2. Create a histogram for the “SLEEPING” variable. Paste the histogram here: (6 pts)
PSYC 354 3. Some might argue that the amount of sleep people report is normally distributed. Based on careful evaluation of ALL of the data in numbers 4 and 5, would you agree or disagree that minutes of “Sleep” is normally distributed in this sample? Support your answer with information from the data and the reading/presentations this week. (6 pts) Although there are a few outliers present within the values given throughout this data set I would agree that the scores are distributed normally. I support this statement by reviewing the values for mean (557.36) and median (555.00). These results are close in value supporting the normality of the distribution. Also, when analyzing the Q-Q plot graph, most of the dots fall along the linear line in the graph. This also supporting the statement of normal distribution. However, the “SIG” box on the normality test falls well below the .05 or higher target coming in at <.001. This value does not support the idea of a normal distribution. I believe if the few outliers that are reported in the data set were not included this data set would produce a definite normal distribution.
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