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Spss activity assessing population, Summaries of Statistics

analyzing data looking at two groups of individuals. Given placebo pill to reduce stress. Given stress reduction pill

Typology: Summaries

2017/2018

Uploaded on 10/19/2023

sara-parchami
sara-parchami 🇨🇦

1 document

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Questions (for Part 1)
1. For each response variable (i.e., Attention to Detail and Social Desirability), report the number
of individuals in each group (which might differ slightly for each variable due to missing values)
along with the sample means and sample standard deviations for each group
Group Statistics
What is your sex? N Mean Std. Deviation Std. Error Mean
AQAttDet Male 107 24.86 4.475 .433
Female 223 26.35 4.371 .293
SDSTot Male 108 16.97 2.211 .213
Female 224 17.05 1.873 .125
2. For each response variable, is it safe to use the “equal variances” version of the t-test and why
or why not?
It is safe to use the equal variances version for both variables because the difference in the output
that the Levenès test shows us in the chart, is;
AQATTDet equal variances assumed is .581 compared to SDSTot which is 3.231 which is
significantly higher and because of this big difference between the data it is not ok to rely on the
'equal variances' for this portion.
3. For each response variable, report the value of the mean difference between the two groups
and its standard error (for the version of the test that is consistent with the previous part ii).
mean difference for AQATTDet is -1.494 & standard error difference is .518 (for equal
variances assumed)
.522 (for equal variances not
assumed)
mean difference for SDSTot is -.077 & standard error difference is .233 (Equal variances
assumed)
.247 (Equal variances not assumed)
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Questions (for Part 1)

  1. For each response variable (i.e., Attention to Detail and Social Desirability), report the number of individuals in each group (which might differ slightly for each variable due to missing values) along with the sample means and sample standard deviations for each group Group Statistics What is your sex? N Mean Std. Deviation Std. Error Mean AQAttDet Male 107 24.86 4.475. Female 223 26.35 4.371. SDSTot Male 108 16.97 2.211. Female 224 17.05 1.873.
  2. For each response variable, is it safe to use the “equal variances” version of the t-test and why or why not? It is safe to use the equal variances version for both variables because the difference in the output that the Levenès test shows us in the chart, is; AQATTDet equal variances assumed is .581 compared to SDSTot which is 3.231 which is significantly higher and because of this big difference between the data it is not ok to rely on the 'equal variances' for this portion.
  3. For each response variable, report the value of the mean difference between the two groups and its standard error (for the version of the test that is consistent with the previous part ii). mean difference for AQATTDet is -1.494 & standard error difference is .518 (for equal variances assumed) .522 (for equal variances not assumed) mean difference for SDSTot is -.077 & standard error difference is .233 (Equal variances assumed) .247 (Equal variances not assumed)
  1. For each response variable, report the results of the t-test (i.e., the value of t, the degrees of freedom, and the P-value). In addition, for each test, show how the value for the degrees of freedom was obtained. For AQATTDet: Value of t; -2.885 (Equal variances assumed), -2.861 (Equal variances not assumed) degrees of freedom; 328 (Equal variances assumed), 204.755 (not assumed) P-value; .004 (Equal variances assumed), .005 (Equal variances not assumed) How the value for the degrees of freedom obtained? Df= N − 1 where: Df=degrees of freedom N =sample size N1 +N2= 107+223=331 330-1= For SDSTot ; Value of t; -.330 (Equal variances assumed), -.312 (Equal variances not assumed) degrees of freedom; 330 (Equal variances assumed), 183.359 (Equal variances not assumed) P-value; .742 (Equal variances assumed), .756 (Equal variances not assumed) How the value for the degrees of freedom obtained? Df= N − 1 where: Df=degrees of freedom N =sample size N1 +N2=108+224= 332 332-1=
  2. Given a 2-tailed significance level of .05, what conclusion can be made for each test (please be complete by referring to the groups and variables themselves here)? A sample mean with a z-score less than or equal to the critical value of -1. is significant at the 0.05 level. Therefore the significance level of .05 is a true Null hypothesis, it will not be rejected. .05 and lower significance level means it will not be rejected. the. significance level means that there is a 95% confidence interval, which shows barely any error, or error that can be accepted. The confidence interval for the Lower is -2.513 and -2.524 and the upper -.475 and -.465 for AQATTDet,
  1. For each response variable, report the quantity labeled Std. Deviation under “Paired Samples Test” and state what this quantity represents (i.e., it is the standard deviation of what?). Pair 1. Std. Deviation 2. Pair 2. Std. Deviation 1. Standard deviation is the measure of dispersion of a set of data from its mean. It measures the absolute variability of a distribution; the higher the dispersion or variability, the greater is the standard deviation and greater will be the magnitude of the deviation of the value from their mean. Standard deviations are important here because the shape of a normal curve is determined by its mean and standard deviation. CV >= 1 indicates a relatively high variation, while a CV < 1 can be considered low.
  2. For each response variable, report the value of the mean difference between the two groups and its standard error (for the version of the test that is consistent with the previous part ii). Paired Samples Test Paired Differences Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the Difference Lower Upper Pair 1 AQAttDet - AQAttDet2 .35976 2.38468 .13167 .10073. Pair 2 SDSTot - SDSTot2 .10241 1.39758 .07670 -.04848.
  3. For each response variable, report the results of the t-test (i.e., the value of t, the degrees of freedom, and the P-value). In addition, for each test, show how the value for the degrees of freedom was obtained. Pair 1 ; t= 2.732 df= 327 P-value=. Pair 2 ; t= 1.335 df= 331 P-value =. Pair 1 df = 328-1= 327 Pair 2 df = 332-1=
  1. Given a 2-tailed significance level of .05, what conclusion can be made fro each test (please be complete by referring to the variables and the groups themselves here) The significance level for both pairs are 0 therefore "reject the null hypothesis" or "fail to reject the null hypothesis" is the conclusion.
  2. What is the point of making these plots (i.e., why should we be looking at them)? Are there any potential issues suggested by these plots (please indicate why or why not)? The histograms allow to show continuous data rather than categorical data. Box plots show differences in distributions and especially the mean. Box plots show the five-number summary of a set of data: including the minimum score, first (lower) quartile, median, third (upper) quartile, and maximum score. It shows the frequency of data. but according to SPSS, it is missing the total and list wise. It does not suggest a complete analysis of the variables. DATASET ACTIVATE DataSet2. COMPUTE AQdiff=AQAttDet. EXECUTE. COMPUTE AQdiff=AQAttDet-AQAttDet2. EXECUTE. COMPUTE SDSdiff=SDSTot-SDSTot2. EXECUTE. EXAMINE VARIABLES=AQdiff SDSdiff /PLOT BOXPLOT HISTOGRAM /COMPARE GROUPS /STATISTICS DESCRIPTIVES /CINTERVAL 95 /MISSING LISTWISE /NOTOTAL.