From d65a5f231c2bf35aab218febf0b9454db9265467 Mon Sep 17 00:00:00 2001 From: Richard Conrardy Date: Tue, 12 Nov 2024 20:21:27 +0100 Subject: [PATCH] last changes Zukunftsforum --- docs/index.html | 2 +- docs/search.json | 21 ++++-- docs/site_libs/revealjs/dist/theme/quarto.css | 4 +- .../2024_11_21_Zukunftsforum.html | 74 +++++++++++-------- .../2024_11_21_Zukunftsforum.qmd | 48 ++++++------ 5 files changed, 83 insertions(+), 66 deletions(-) diff --git a/docs/index.html b/docs/index.html index f305d16..9913050 100644 --- a/docs/index.html +++ b/docs/index.html @@ -676,7 +676,7 @@

Lerngelegenheiten

12. GEBF-Tagung - + 15.11.2024 diff --git a/docs/search.json b/docs/search.json index 075d7a9..7aefbf6 100644 --- a/docs/search.json +++ b/docs/search.json @@ -333,7 +333,7 @@ "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#empirische-forschung", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "Empirische Forschung", - "text": "Empirische Forschung\n\n\n\n\nMetastudie\n\n\n\n\ng\nSE\n\n\n\n\nOverall effect\n0.37\n0.025\n\n\nSchool-age children\n0.68\n0.025\n\n\nMath\n0.26\n0.067\n\n\n\n\\[n=173\\]\n\n\n\n\n4F Modell\n\n\nFail\nFlip\n\n\n\n\nFix\nFeed\n\n\n\n\n(Kapur et al., 2022, S. 14)" + "text": "Empirische Forschung\n\n\n\n\nMetastudie\n\n\n\n\ng\nSE\n\n\n\n\nOverall effect\n0.37\n0.025\n\n\nSchool-age children\n0.68\n0.025\n\n\nMath\n0.26\n0.067\n\n\n\nn=173\n\n\n\n\n4F Modell\n\n\nFail\nFlip\n\n\n\n\nFix\nFeed\n\n\n\n\n(Kapur et al., 2022, S. 14)" }, { "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#productive-failure-pf", @@ -375,7 +375,7 @@ "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#untersuchungsdesign", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "Untersuchungsdesign", - "text": "Untersuchungsdesign\n\n\n\nMathematikunterricht auf der Sekundarstufe 1\nThema: Mittlere absolute Abweichung\nExperimentelles Forschungsdesign\n\\(n=220\\)\n\n\n\n\n\n\n\n4F\nALT\n\n\n\n\nFail\nin class\nout of class\n\n\nFlip\nout of class\nin class\n\n\nFix\nin class\nin class\n\n\nFeed\nin class\nin class" + "text": "Untersuchungsdesign\n\n\n\nMathematikunterricht in der Sekundarstufe 1\nThema: Mittlere absolute Abweichung\nExperimentelles Forschungsdesign (randomisiert innerhalb der Klassen)\n220 Lernende aus 12 Klassen\n\n\n\n\n\n\n\n\n\n4F\nALT\n\n\n\n\nFail\nin class\nout of class\n\n\nFlip\nout of class\nin class\n\n\nFix\nin class\nin class\n\n\nFeed\nin class\nin class" }, { "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#forschungsfragen", @@ -389,14 +389,14 @@ "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#engagement-is-highest-during-in-class-activities-irrespective-of-variant.", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "1. Engagement is highest during in-class activities, irrespective of variant.", - "text": "1. Engagement is highest during in-class activities, irrespective of variant.\n\nResultatTestConsoleInstrument\n\n\n\n\n\n\n\n\n\nWilcoxon\np\n\n\n\n\neng_1\n0.405\n\n\neng_2\n0.691\n\n\n4F\n0.605\n\n\nAlt\n0.812\n\n\n\n\n\n\n## 3. Deskriptive Vergleiche 4F und ALT\nmean(calcdata$engagement_1[calcdata$FF_Alt==\"4F\"],na.rm=TRUE)\nsd(calcdata$engagement_1[calcdata$FF_Alt==\"4F\"],na.rm=TRUE)\n\nmean(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\nsd(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\n\n## 4. Auf Normalverteilung prüfen (mit Shapiro-Wilk Test)\nshapiro.test(calcdata$engagement_1)\nshapiro.test(calcdata$engagement_2)\n\n## 4. Hypothesentest\n## a) Gruppenunterschiede zwischen 4F und Alt zum gleichen Zeitpunkt\nwilcox.test(calcdata$engagement_1 ~ calcdata$FF_Alt)\nwilcox.test(calcdata$engagement_2 ~ calcdata$FF_Alt)\n\n## b) Gruppenunterschiede innerhalb der 4F resp. ALT Gruppe zu t1 vs. t2 (fail vs. flip)\n\n#Untergruppen der Daten erstellen\ncalcdata_4f <- subset(calcdata, FF_Alt == \"4F\")\ncalcdata_alt <- subset(calcdata,FF_Alt == \"ALT\")\n\nwilcox.test(calcdata_4f$engagement_1,calcdata_4f$engagement_2,paired=TRUE)\nwilcox.test(calcdata_alt$engagement_1,calcdata_alt$engagement_2,paired=TRUE)\n\n\n> mean(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\n[1] 16.67391\n> sd(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\n[1] 6.988413\n\n> shapiro.test(calcdata$engagement_1)\n\n Shapiro-Wilk normality test\n\ndata: calcdata$engagement_1\nW = 0.8085, p-value = 9.904e-16\n\n> shapiro.test(calcdata$engagement_2)\n\n Shapiro-Wilk normality test\n\ndata: calcdata$engagement_2\nW = 0.82215, p-value = 3.876e-15\n\n> wilcox.test(calcdata$engagement_1 ~ calcdata$FF_Alt)\n\n Wilcoxon rank sum test with continuity correction\n\ndata: calcdata$engagement_1 by calcdata$FF_Alt\nW = 6369, p-value = 0.4051\nalternative hypothesis: true location shift is not equal to 0\n\n> wilcox.test(calcdata$engagement_2 ~ calcdata$FF_Alt)\n\n Wilcoxon rank sum test with continuity correction\n\ndata: calcdata$engagement_2 by calcdata$FF_Alt\nW = 6166, p-value = 0.6911\nalternative hypothesis: true location shift is not equal to 0\n\n> wilcox.test(calcdata_4f$engagement_1,calcdata_4f$engagement_2,paired=TRUE)\n\n Wilcoxon signed rank test with continuity correction\n\ndata: calcdata_4f$engagement_1 and calcdata_4f$engagement_2\nV = 1563, p-value = 0.605\nalternative hypothesis: true location shift is not equal to 0\n\n> wilcox.test(calcdata_alt$engagement_1,calcdata_alt$engagement_2,paired=TRUE)\n\n Wilcoxon signed rank test with continuity correction\n\ndata: calcdata_alt$engagement_1 and calcdata_alt$engagement_2\nV = 1773, p-value = 0.8123\nalternative hypothesis: true location shift is not equal to 0\n\n\n5 Fragen mit einer 5-Punkte Likert Skala. Übernommen aus Kapur (2014), \\(\\alpha = 0.79\\).\n\nDiese Lektion hat mich dazu gebracht, dass ich mich aktiv beteiligen möchte.\n\nIch war während der Lektion fokussiert.\n\nIch war aufmerksam während der Lektion.\n\nIch habe an den Aktivitäten der Lektion teilgenommen.\n\nIch habe mich während der Lektion konzentriert.\n\n\nstimme überhaupt nicht zu\n\nstimme nicht zu\n\nneutral\n\nstimme zu\n\nstimme voll und ganz zu\n\n\n\n\n\n\nVerworfen" + "text": "1. Engagement is highest during in-class activities, irrespective of variant.\n\nResultatTestConsoleInstrument\n\n\n\n\n\n\n\n\n\nWilcoxon\np\n\n\n\n\neng_1\n0.405\n\n\neng_2\n0.691\n\n\n4F\n0.605\n\n\nAlt\n0.812\n\n\n\n\n\n\n## 3. Deskriptive Vergleiche 4F und ALT\nmean(calcdata$engagement_1[calcdata$FF_Alt==\"4F\"],na.rm=TRUE)\nsd(calcdata$engagement_1[calcdata$FF_Alt==\"4F\"],na.rm=TRUE)\n\nmean(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\nsd(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\n\n## 4. Auf Normalverteilung prüfen (mit Shapiro-Wilk Test)\nshapiro.test(calcdata$engagement_1)\nshapiro.test(calcdata$engagement_2)\n\n## 4. Hypothesentest\n## a) Gruppenunterschiede zwischen 4F und Alt zum gleichen Zeitpunkt\nwilcox.test(calcdata$engagement_1 ~ calcdata$FF_Alt)\nwilcox.test(calcdata$engagement_2 ~ calcdata$FF_Alt)\n\n## b) Gruppenunterschiede innerhalb der 4F resp. ALT Gruppe zu t1 vs. t2 (fail vs. flip)\n\n#Untergruppen der Daten erstellen\ncalcdata_4f <- subset(calcdata, FF_Alt == \"4F\")\ncalcdata_alt <- subset(calcdata,FF_Alt == \"ALT\")\n\nwilcox.test(calcdata_4f$engagement_1,calcdata_4f$engagement_2,paired=TRUE)\nwilcox.test(calcdata_alt$engagement_1,calcdata_alt$engagement_2,paired=TRUE)\n\n\n> mean(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\n[1] 16.67391\n> sd(calcdata$engagement_1[calcdata$FF_Alt==\"ALT\"],na.rm=TRUE)\n[1] 6.988413\n\n> shapiro.test(calcdata$engagement_1)\n\n Shapiro-Wilk normality test\n\ndata: calcdata$engagement_1\nW = 0.8085, p-value = 9.904e-16\n\n> shapiro.test(calcdata$engagement_2)\n\n Shapiro-Wilk normality test\n\ndata: calcdata$engagement_2\nW = 0.82215, p-value = 3.876e-15\n\n> wilcox.test(calcdata$engagement_1 ~ calcdata$FF_Alt)\n\n Wilcoxon rank sum test with continuity correction\n\ndata: calcdata$engagement_1 by calcdata$FF_Alt\nW = 6369, p-value = 0.4051\nalternative hypothesis: true location shift is not equal to 0\n\n> wilcox.test(calcdata$engagement_2 ~ calcdata$FF_Alt)\n\n Wilcoxon rank sum test with continuity correction\n\ndata: calcdata$engagement_2 by calcdata$FF_Alt\nW = 6166, p-value = 0.6911\nalternative hypothesis: true location shift is not equal to 0\n\n> wilcox.test(calcdata_4f$engagement_1,calcdata_4f$engagement_2,paired=TRUE)\n\n Wilcoxon signed rank test with continuity correction\n\ndata: calcdata_4f$engagement_1 and calcdata_4f$engagement_2\nV = 1563, p-value = 0.605\nalternative hypothesis: true location shift is not equal to 0\n\n> wilcox.test(calcdata_alt$engagement_1,calcdata_alt$engagement_2,paired=TRUE)\n\n Wilcoxon signed rank test with continuity correction\n\ndata: calcdata_alt$engagement_1 and calcdata_alt$engagement_2\nV = 1773, p-value = 0.8123\nalternative hypothesis: true location shift is not equal to 0\n\n\n5 Fragen mit einer 5-Punkte Likert Skala. Übernommen aus Kapur (2014), \\(\\alpha = 0.79\\).\n\nDiese Lektion hat mich dazu gebracht, dass ich mich aktiv beteiligen möchte.\n\nIch war während der Lektion fokussiert.\n\nIch war aufmerksam während der Lektion.\n\nIch habe an den Aktivitäten der Lektion teilgenommen.\n\nIch habe mich während der Lektion konzentriert.\n\n\nstimme überhaupt nicht zu\n\nstimme nicht zu\n\nneutral\n\nstimme zu\n\nstimme voll und ganz zu" }, { "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#engagement-predicts-performance.", "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#engagement-predicts-performance.", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "2. Engagement predicts performance.", - "text": "2. Engagement predicts performance.\n\nResultatTestConsoleInstrument\n\n\n\nLineare Regression: \\(\\quad R^2=0.297 \\quad p=2.2\\cdot 10^{-16}\\)\n\n\n# Engagement als Prädiktor an Tag 1\nlm_engagement_day1 <- lm(Lernkontrolle1_tot ~ engagement_tag1, data = calcdata)\nsummary(lm_engagement_day1)\n# Engagement als Prädiktor an Tag 2\nlm_engagement_day2 <- lm(Lernkontrolle2_tot ~ engagement_tag2, data = calcdata)\nsummary(lm_engagement_day2)\n# Engagement als Prädiktor an Tag 2 (inklusive Engagement am 1. TAg)\nlm_engagement_day2_2 <- lm(Lernkontrolle2_tot ~ engagement_tag1 + engagement_tag2, data = calcdata)\nsummary(lm_engagement_day2_2) #nur das direkt preceding engagement relevant\n# Total engagement mit total Lernkontrolle\n# totale Lernzunahme berechnen\ncalcdata$Lernkontrolle_tot <- calcdata$Lernkontrolle1_tot + calcdata$Lernkontrolle2_tot\n# lineare regression: \nlm_engagement_tot <- lm(Lernkontrolle_tot ~ engagement_tot, data = calcdata)\nsummary(lm_engagement_tot)\n\n\n> # Engagement als Prädiktor an Tag 1\n> lm_engagement_day1 <- lm(Lernkontrolle1_tot ~ engagement_tag1, data = calcdata)\n> summary(lm_engagement_day1)\n\nCall:\nlm(formula = Lernkontrolle1_tot ~ engagement_tag1, data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.9723 -0.8754 -0.0512 1.0909 4.7330 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.53850 0.31880 1.689 0.0926 . \nengagement_tag1 0.11369 0.01176 9.671 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 1.725 on 218 degrees of freedom\nMultiple R-squared: 0.3002, Adjusted R-squared: 0.297 \nF-statistic: 93.53 on 1 and 218 DF, p-value: < 2.2e-16\n\n> # Engagement als Prädiktor an Tag 2\n> lm_engagement_day2 <- lm(Lernkontrolle2_tot ~ engagement_tag2, data = calcdata)\n> summary(lm_engagement_day2)\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ engagement_tag2, data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.3717 -1.3722 -0.4974 1.1274 5.2502 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.50400 0.32823 1.536 0.126 \nengagement_tag2 0.12476 0.01301 9.591 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.004 on 218 degrees of freedom\nMultiple R-squared: 0.2967, Adjusted R-squared: 0.2935 \nF-statistic: 91.98 on 1 and 218 DF, p-value: < 2.2e-16\n\n> # Engagement als Prädiktor an Tag 2 (inklusive Engagement am 1. TAg)\n> lm_engagement_day2_2 <- lm(Lernkontrolle2_tot ~ engagement_tag1 + engagement_tag2, data = calcdata)\n> summary(lm_engagement_day2_2) #nur das direkt preceding engagement relevant\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ engagement_tag1 + engagement_tag2, \n data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.333 -1.336 -0.229 1.149 5.204 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.19928 0.41221 0.483 0.629 \nengagement_tag1 0.01798 0.01474 1.220 0.224 \nengagement_tag2 0.11827 0.01404 8.422 5.08e-15 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.002 on 217 degrees of freedom\nMultiple R-squared: 0.3015, Adjusted R-squared: 0.2951 \nF-statistic: 46.84 on 2 and 217 DF, p-value: < 2.2e-16\n\n> # Total engagement mit total Lernkontrolle\n> # totale Lernzunahme berechnen\n> calcdata$Lernkontrolle_tot <- calcdata$Lernkontrolle1_tot + calcdata$Lernkontrolle2_tot\n> # lineare regression: \n> lm_engagement_tot <- lm(Lernkontrolle_tot ~ engagement_tot, data = calcdata)\n> summary(lm_engagement_tot)\n\nCall:\nlm(formula = Lernkontrolle_tot ~ engagement_tot, data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-7.8733 -2.0002 -0.3149 1.8721 9.8729 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.7576 0.6287 1.205 0.229 \nengagement_tot 0.1249 0.0123 10.150 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 3.074 on 218 degrees of freedom\nMultiple R-squared: 0.3209, Adjusted R-squared: 0.3178 \nF-statistic: 103 on 1 and 218 DF, p-value: < 2.2e-16\n\n\nAufgaben von Kapur (2014), gleichmässig aufgeteilt nach procedural knowledge, conceptual knowledge und transfert knowledge.\nAufgabe 1\nBerechne die mittlere absolute Abweichung der folgenden Noten in einer Statistikprüfung:\n\n30, 60, 50, 40, 70\n10, 12, 20, 22\n\nEin anderer Schüler, Schüler A, verwendet die folgende Methode, um Teilaufgabe a. dieser Frage zu beantworten:\n\\(|30 - 70| + |60 - 70| + |50 - 70| + |40 - 70| + |70 - 70| = 100\\)\n\\(100 \\div 5 = 20\\) Die Lösung ist 20.\nWie unterscheidet sich die Methode von Schüler A von der Berechnung der mittleren absoluten Abweichung? Welche Methode ist besser? Begründe deine Antwort.\nAufgabe 2\nBeachte die folgenden sechs Datensätze (ein Datensatz pro Zeile):\n\nA: 1, 5, 6, 10\nB: 4, 4, 4, 4\nC: 101, 102, 103, 104\nD: 7, 8, 9, 10\nE: 1, 2, 9, 10\nF: 1, 2, 3, 4\n\nWelcher der Datensätze hat die kleinste mittlere absolute Abweichung?\n\nDatensatz A\nDatensatz B\nDatensatz C\nDatensatz D\nDatensatz E\nDatensatz F\n\nWelcher der Datensätze hat die grösste mittlere absolute Abweichung?\n\nDatensatz A\nDatensatz B\nDatensatz C\nDatensatz D\nDatensatz E\nDatensatz F\n\nAufgabe 3\nEin aus fünf Zahlen bestehender Datensatz hat den Mittelwert ( M = 7 ) und die mittlere absolute Abweichung = 4.\nVerwende diese Informationen, um die Fragen a. und b. zu beantworten.\n\nWie verändern sich Mittelwert und mittlere absolute Abweichung, wenn die Zahlen 1 und 13 zu dem Datensatz hinzugefügt werden, sodass ein Datensatz mit sieben Zahlen entsteht?\n\n\nMittelwert (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nMittlere absolute Abweichung (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\n\n\nWie verändern sich Mittelwert und mittlere absolute Abweichung, wenn die Zahlen 3 und 13 zu dem Datensatz hinzugefügt werden, sodass ein Datensatz mit sieben Zahlen entsteht?\n\n\nMittelwert (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nMittlere absolute Abweichung (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\n\nAufgabe 4\nEine gleiche Anzahl von Schülern nahm am Finale eines Sportwettbewerbs über 100 m Sprint und 100 m Schwimmen teil. Die Zeiten (in Sekunden) der Sieger des 100m Sprints und des 100m Schwimmens sind unten aufgeführt, ebenso wie die Durchschnittszeiten und die Standardabweichung der Finalisten der beiden Wettbewerbe.\n\n\n\n\n\n\n\n\n\n100m Sprint\n100m Schwimmen\n\n\n\n\nGewinner\n11s\n40s\n\n\nMittelwert der Finalisten, ( M )\n12s\n45s\n\n\nMittlere absolute Abweichung der Finalisten\n1s\n10s\n\n\n\nUnter der Annahme, dass alles andere gleich ist, wer von den beiden Gewinnern ist der bessere Sportler?\n\nDer Sprint-Gewinner\nDer Schwimm-Gewinner\nBeide\nEs gibt nicht genügend Informationen, um dies zu entscheiden\n\nAufgabe 1\n\nBei der Berechnung der mittleren absoluten Abweichung, erkläre, warum es wichtig ist, den Absolutbetrag der Differenzen zum Mittelwert zu rechnen.\nBei der Berechnung der mittleren absoluten Abweichung, erkläre, warum es wichtig ist, die Summe durch die Anzahl der Werte zu teilen.\n\nAufgabe 2\nDie Eigentümer von zwei Kinos, A und B, argumentieren, dass ihr jeweiliges Kino eine gleichmäßigere Besucherzahl aufweist. Sie erfassten die täglichen Besucherzahlen ihrer Kinos an 11 zufälligen Tagen. Die Ergebnisse ihrer Datenerhebung sind im Folgenden dargestellt:\n\n\n\nKino A\nKino B\n\n\n\n\nMittelwert, M\n72\n\n\nMittlere absolute Abweichung\n10\n\n\n\nWelches Kino hat Ihrer Meinung nach eine gleichmäßigere Anwesenheit?\n\nKino A\nKino B\nBeide hatten eine ähnlich gleichmäßige Besucheranzahl\nKeine der oben genannten Lösungen\n\nAufgabe 3\nBeachte die folgenden sechs Datensätze (ein Datensatz pro Zeile):\n\n\n\nA\n1\n5\n6\n10\n\n\n\n\nB\n4\n4\n4\n4\n\n\nC\n101\n102\n103\n104\n\n\nD\n7\n8\n9\n10\n\n\nE\n1\n2\n9\n10\n\n\nF\n1\n2\n3\n4\n\n\n\nWelche Datensätze haben die gleiche mittlere absolute Abweichung? (1 Antwort)\n\nDatensätze A, E, F\nDatensätze C, D, F\nDatensätze A, B\nDatensätze C, F\nDatensätze B, D, F\nDatensätze B, E, F\n\nAufgabe 4\nEin aus fünf Zahlen bestehender Datensatz hat den Mittelwert M = 7 und die mittlere absolute Abweichung = 4. Verwende diese Informationen, um die Fragen a. bis c. zu beantworten.\n\nWenn jede der fünf Zahlen um 2 erhöht wird, wie lauten der neue Mittelwert M und die neue mittlere absolute Abweichung?\n\nM = 7, Mittlere absolute Abweichung = 4\nM = 9, Mittlere absolute Abweichung = 4\nM = 7, Mittlere absolute Abweichung = 6\nM = 9, Mittlere absolute Abweichung = 6\n\nWenn jede der fünf Zahlen mit 5 multipliziert wird, wie lauten der neue Mittelwert und die neue mittlere absolute Abweichung?\n\nM = 7, Mittlere absolute Abweichung = 4\nM = 35, Mittlere absolute Abweichung = 4\nM = 7, Mittlere absolute Abweichung = 20\nM = 35, Mittlere absolute Abweichung = 20\n\nWie verändern sich Mittelwert und mittlere absolute Abweichung, wenn die Zahlen 5 und 9 zu dem Datensatz hinzugefügt werden, sodass ein Datensatz mit sieben Zahlen entsteht?\n\nMittelwert (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nMittlere absolute Abweichung (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nAufgabe 5\nDavids Ergebnisse für Mathematik, Physik und Chemie in den Abschlussprüfungen sind unten aufgeführt. Die Leistungen seiner Klasse in diesen drei Fächern sind ebenfalls unten aufgeführt:\n\n\n\nFach\nMathematik\nPhysik\nChemie\n\n\n\n\nDavids Punkte\n95\n90\n85\n\n\nKlassenmittelwert\n80\n80\n80\n\n\nMittlere absolute Abweichung der Klasse\n15\n5\n4\n\n\n\n\nIn welchem Fach hat David im Vergleich zu seiner Klasse am besten abgeschnitten?\n\nMathematik\nPhysik\nChemie\n\nIn welchem Fach hat David im Vergleich zu seiner Klasse am schlechtesten abgeschnitten?\n\nMathematik\nPhysik\nChemie\n\n\n\n\n\n\n\nBestätigt" + "text": "2. Engagement predicts performance.\n\nResultatTestConsoleInstrument\n\n\n\nLineare Regression: \\(\\quad R^2=0.297 \\quad p=2.2\\cdot 10^{-16}\\)\n\n\n# Engagement als Prädiktor an Tag 1\nlm_engagement_day1 <- lm(Lernkontrolle1_tot ~ engagement_tag1, data = calcdata)\nsummary(lm_engagement_day1)\n# Engagement als Prädiktor an Tag 2\nlm_engagement_day2 <- lm(Lernkontrolle2_tot ~ engagement_tag2, data = calcdata)\nsummary(lm_engagement_day2)\n# Engagement als Prädiktor an Tag 2 (inklusive Engagement am 1. TAg)\nlm_engagement_day2_2 <- lm(Lernkontrolle2_tot ~ engagement_tag1 + engagement_tag2, data = calcdata)\nsummary(lm_engagement_day2_2) #nur das direkt preceding engagement relevant\n# Total engagement mit total Lernkontrolle\n# totale Lernzunahme berechnen\ncalcdata$Lernkontrolle_tot <- calcdata$Lernkontrolle1_tot + calcdata$Lernkontrolle2_tot\n# lineare regression: \nlm_engagement_tot <- lm(Lernkontrolle_tot ~ engagement_tot, data = calcdata)\nsummary(lm_engagement_tot)\n\n\n> # Engagement als Prädiktor an Tag 1\n> lm_engagement_day1 <- lm(Lernkontrolle1_tot ~ engagement_tag1, data = calcdata)\n> summary(lm_engagement_day1)\n\nCall:\nlm(formula = Lernkontrolle1_tot ~ engagement_tag1, data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.9723 -0.8754 -0.0512 1.0909 4.7330 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.53850 0.31880 1.689 0.0926 . \nengagement_tag1 0.11369 0.01176 9.671 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 1.725 on 218 degrees of freedom\nMultiple R-squared: 0.3002, Adjusted R-squared: 0.297 \nF-statistic: 93.53 on 1 and 218 DF, p-value: < 2.2e-16\n\n> # Engagement als Prädiktor an Tag 2\n> lm_engagement_day2 <- lm(Lernkontrolle2_tot ~ engagement_tag2, data = calcdata)\n> summary(lm_engagement_day2)\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ engagement_tag2, data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.3717 -1.3722 -0.4974 1.1274 5.2502 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.50400 0.32823 1.536 0.126 \nengagement_tag2 0.12476 0.01301 9.591 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.004 on 218 degrees of freedom\nMultiple R-squared: 0.2967, Adjusted R-squared: 0.2935 \nF-statistic: 91.98 on 1 and 218 DF, p-value: < 2.2e-16\n\n> # Engagement als Prädiktor an Tag 2 (inklusive Engagement am 1. TAg)\n> lm_engagement_day2_2 <- lm(Lernkontrolle2_tot ~ engagement_tag1 + engagement_tag2, data = calcdata)\n> summary(lm_engagement_day2_2) #nur das direkt preceding engagement relevant\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ engagement_tag1 + engagement_tag2, \n data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.333 -1.336 -0.229 1.149 5.204 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.19928 0.41221 0.483 0.629 \nengagement_tag1 0.01798 0.01474 1.220 0.224 \nengagement_tag2 0.11827 0.01404 8.422 5.08e-15 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.002 on 217 degrees of freedom\nMultiple R-squared: 0.3015, Adjusted R-squared: 0.2951 \nF-statistic: 46.84 on 2 and 217 DF, p-value: < 2.2e-16\n\n> # Total engagement mit total Lernkontrolle\n> # totale Lernzunahme berechnen\n> calcdata$Lernkontrolle_tot <- calcdata$Lernkontrolle1_tot + calcdata$Lernkontrolle2_tot\n> # lineare regression: \n> lm_engagement_tot <- lm(Lernkontrolle_tot ~ engagement_tot, data = calcdata)\n> summary(lm_engagement_tot)\n\nCall:\nlm(formula = Lernkontrolle_tot ~ engagement_tot, data = calcdata)\n\nResiduals:\n Min 1Q Median 3Q Max \n-7.8733 -2.0002 -0.3149 1.8721 9.8729 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 0.7576 0.6287 1.205 0.229 \nengagement_tot 0.1249 0.0123 10.150 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 3.074 on 218 degrees of freedom\nMultiple R-squared: 0.3209, Adjusted R-squared: 0.3178 \nF-statistic: 103 on 1 and 218 DF, p-value: < 2.2e-16\n\n\nAufgaben von Kapur (2014), gleichmässig aufgeteilt nach procedural knowledge, conceptual knowledge und transfert knowledge.\nAufgabe 1\nBerechne die mittlere absolute Abweichung der folgenden Noten in einer Statistikprüfung:\n\n30, 60, 50, 40, 70\n10, 12, 20, 22\n\nEin anderer Schüler, Schüler A, verwendet die folgende Methode, um Teilaufgabe a. dieser Frage zu beantworten:\n\\(|30 - 70| + |60 - 70| + |50 - 70| + |40 - 70| + |70 - 70| = 100\\)\n\\(100 \\div 5 = 20\\) Die Lösung ist 20.\nWie unterscheidet sich die Methode von Schüler A von der Berechnung der mittleren absoluten Abweichung? Welche Methode ist besser? Begründe deine Antwort.\nAufgabe 2\nBeachte die folgenden sechs Datensätze (ein Datensatz pro Zeile):\n\nA: 1, 5, 6, 10\nB: 4, 4, 4, 4\nC: 101, 102, 103, 104\nD: 7, 8, 9, 10\nE: 1, 2, 9, 10\nF: 1, 2, 3, 4\n\nWelcher der Datensätze hat die kleinste mittlere absolute Abweichung?\n\nDatensatz A\nDatensatz B\nDatensatz C\nDatensatz D\nDatensatz E\nDatensatz F\n\nWelcher der Datensätze hat die grösste mittlere absolute Abweichung?\n\nDatensatz A\nDatensatz B\nDatensatz C\nDatensatz D\nDatensatz E\nDatensatz F\n\nAufgabe 3\nEin aus fünf Zahlen bestehender Datensatz hat den Mittelwert ( M = 7 ) und die mittlere absolute Abweichung = 4.\nVerwende diese Informationen, um die Fragen a. und b. zu beantworten.\n\nWie verändern sich Mittelwert und mittlere absolute Abweichung, wenn die Zahlen 1 und 13 zu dem Datensatz hinzugefügt werden, sodass ein Datensatz mit sieben Zahlen entsteht?\n\n\nMittelwert (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nMittlere absolute Abweichung (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\n\n\nWie verändern sich Mittelwert und mittlere absolute Abweichung, wenn die Zahlen 3 und 13 zu dem Datensatz hinzugefügt werden, sodass ein Datensatz mit sieben Zahlen entsteht?\n\n\nMittelwert (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nMittlere absolute Abweichung (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\n\nAufgabe 4\nEine gleiche Anzahl von Schülern nahm am Finale eines Sportwettbewerbs über 100 m Sprint und 100 m Schwimmen teil. Die Zeiten (in Sekunden) der Sieger des 100m Sprints und des 100m Schwimmens sind unten aufgeführt, ebenso wie die Durchschnittszeiten und die Standardabweichung der Finalisten der beiden Wettbewerbe.\n\n\n\n\n\n\n\n\n\n100m Sprint\n100m Schwimmen\n\n\n\n\nGewinner\n11s\n40s\n\n\nMittelwert der Finalisten, ( M )\n12s\n45s\n\n\nMittlere absolute Abweichung der Finalisten\n1s\n10s\n\n\n\nUnter der Annahme, dass alles andere gleich ist, wer von den beiden Gewinnern ist der bessere Sportler?\n\nDer Sprint-Gewinner\nDer Schwimm-Gewinner\nBeide\nEs gibt nicht genügend Informationen, um dies zu entscheiden\n\nAufgabe 1\n\nBei der Berechnung der mittleren absoluten Abweichung, erkläre, warum es wichtig ist, den Absolutbetrag der Differenzen zum Mittelwert zu rechnen.\nBei der Berechnung der mittleren absoluten Abweichung, erkläre, warum es wichtig ist, die Summe durch die Anzahl der Werte zu teilen.\n\nAufgabe 2\nDie Eigentümer von zwei Kinos, A und B, argumentieren, dass ihr jeweiliges Kino eine gleichmäßigere Besucherzahl aufweist. Sie erfassten die täglichen Besucherzahlen ihrer Kinos an 11 zufälligen Tagen. Die Ergebnisse ihrer Datenerhebung sind im Folgenden dargestellt:\n\n\n\nKino A\nKino B\n\n\n\n\nMittelwert, M\n72\n\n\nMittlere absolute Abweichung\n10\n\n\n\nWelches Kino hat Ihrer Meinung nach eine gleichmäßigere Anwesenheit?\n\nKino A\nKino B\nBeide hatten eine ähnlich gleichmäßige Besucheranzahl\nKeine der oben genannten Lösungen\n\nAufgabe 3\nBeachte die folgenden sechs Datensätze (ein Datensatz pro Zeile):\n\n\n\nA\n1\n5\n6\n10\n\n\n\n\nB\n4\n4\n4\n4\n\n\nC\n101\n102\n103\n104\n\n\nD\n7\n8\n9\n10\n\n\nE\n1\n2\n9\n10\n\n\nF\n1\n2\n3\n4\n\n\n\nWelche Datensätze haben die gleiche mittlere absolute Abweichung? (1 Antwort)\n\nDatensätze A, E, F\nDatensätze C, D, F\nDatensätze A, B\nDatensätze C, F\nDatensätze B, D, F\nDatensätze B, E, F\n\nAufgabe 4\nEin aus fünf Zahlen bestehender Datensatz hat den Mittelwert M = 7 und die mittlere absolute Abweichung = 4. Verwende diese Informationen, um die Fragen a. bis c. zu beantworten.\n\nWenn jede der fünf Zahlen um 2 erhöht wird, wie lauten der neue Mittelwert M und die neue mittlere absolute Abweichung?\n\nM = 7, Mittlere absolute Abweichung = 4\nM = 9, Mittlere absolute Abweichung = 4\nM = 7, Mittlere absolute Abweichung = 6\nM = 9, Mittlere absolute Abweichung = 6\n\nWenn jede der fünf Zahlen mit 5 multipliziert wird, wie lauten der neue Mittelwert und die neue mittlere absolute Abweichung?\n\nM = 7, Mittlere absolute Abweichung = 4\nM = 35, Mittlere absolute Abweichung = 4\nM = 7, Mittlere absolute Abweichung = 20\nM = 35, Mittlere absolute Abweichung = 20\n\nWie verändern sich Mittelwert und mittlere absolute Abweichung, wenn die Zahlen 5 und 9 zu dem Datensatz hinzugefügt werden, sodass ein Datensatz mit sieben Zahlen entsteht?\n\nMittelwert (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nMittlere absolute Abweichung (kreuze eine Antwort an):\n\nunverändert\nnimmt zu\nnimmt ab\n\nAufgabe 5\nDavids Ergebnisse für Mathematik, Physik und Chemie in den Abschlussprüfungen sind unten aufgeführt. Die Leistungen seiner Klasse in diesen drei Fächern sind ebenfalls unten aufgeführt:\n\n\n\nFach\nMathematik\nPhysik\nChemie\n\n\n\n\nDavids Punkte\n95\n90\n85\n\n\nKlassenmittelwert\n80\n80\n80\n\n\nMittlere absolute Abweichung der Klasse\n15\n5\n4\n\n\n\n\nIn welchem Fach hat David im Vergleich zu seiner Klasse am besten abgeschnitten?\n\nMathematik\nPhysik\nChemie\n\nIn welchem Fach hat David im Vergleich zu seiner Klasse am schlechtesten abgeschnitten?\n\nMathematik\nPhysik\nChemie" }, { "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#the-4f-model-does-not-outperform-the-alt-model.", @@ -410,7 +410,7 @@ "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#mental-load-does-not-predict-performance.", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "4. Mental load does not predict performance.", - "text": "4. Mental load does not predict performance.\n\nResultatTestConsoleInstrument\n\n\n\nANOVA: \\(\\quad p=0.626\\)\n\n\n\n## 5. Testen, ob das Modell mit oder ohne Mental Effort den Leistungszuwachs besser vorhersagt\n# nur Daten behalten, die nirgendwo NA haben (weil sonst ANOVA nicht gerechnet werden kann)\ncalcdata_me <- calcdata[complete.cases(calcdata[,c(\"mental_effort_tag1\",\"mental_effort_tag2\",\"Lernkontrolle1_tot\",\"Lernkontrolle2_tot\")]),]\n\n# Tag 1\nmodel1_day1 <- lm(Lernkontrolle1_tot ~ mental_effort_tag1, data = calcdata_me)\nsummary(model1_day1)\nmodel2_day1 <- lm(Lernkontrolle1_tot ~ 1, data = calcdata_me)\nsummary(model2_day1)\n\nanova(model1_day1,model2_day1)\nAIC(model1_day1)\nAIC(model2_day1)\n\n# Tag 2\nmodel1_day2 <- lm(Lernkontrolle2_tot ~ mental_effort_tag2, data = calcdata_me)\nsummary(model1_day2)\nmodel2_day2 <- lm(Lernkontrolle2_tot ~ 1, data = calcdata_me)\nsummary(model2_day2)\n\nanova(model1_day2,model2_day2)\nAIC(model1_day2)\nAIC(model2_day2)\n\n# Insgesamt \ncalcdata_me$mental_effort_tot <- calcdata_me$mental_effort_tag1 + calcdata_me$mental_effort_tag2\n\nmodel1_tot <- lm(Lernkontrolle_tot ~ mental_effort_tot, data = calcdata_me)\nmodel2_tot <- lm(Lernkontrolle_tot ~ 1, data = calcdata_me)\n\nanova(model1_tot,model2_tot)\nAIC(model1_tot)\nAIC(model2_tot)\n\n\n> ## 5. Testen, ob das Modell mit oder ohne Mental Effort den Leistungszuwachs besser vorhersagt\n> # nur Daten behalten, die nirgendwo NA haben (weil sonst ANOVA nicht gerechnet werden kann)\n> calcdata_me <- calcdata[complete.cases(calcdata[,c(\"mental_effort_tag1\",\"mental_effort_tag2\",\"Lernkontrolle1_tot\",\"Lernkontrolle2_tot\")]),]\n> \n> # Tag 1\n> model1_day1 <- lm(Lernkontrolle1_tot ~ mental_effort_tag1, data = calcdata_me)\n> summary(model1_day1)\n\nCall:\nlm(formula = Lernkontrolle1_tot ~ mental_effort_tag1, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-3.8770 -0.8787 0.1247 1.1247 5.1213 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 3.861468 0.508998 7.586 3.95e-12 ***\nmental_effort_tag1 0.001727 0.061862 0.028 0.978 \n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 1.861 on 142 degrees of freedom\nMultiple R-squared: 5.486e-06, Adjusted R-squared: -0.007037 \nF-statistic: 0.0007791 on 1 and 142 DF, p-value: 0.9778\n\n> model2_day1 <- lm(Lernkontrolle1_tot ~ 1, data = calcdata_me)\n> summary(model2_day1)\n\nCall:\nlm(formula = Lernkontrolle1_tot ~ 1, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-3.875 -0.875 0.125 1.125 5.125 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 3.8750 0.1545 25.07 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 1.854 on 143 degrees of freedom\n\n> \n> anova(model1_day1,model2_day1)\nAnalysis of Variance Table\n\nModel 1: Lernkontrolle1_tot ~ mental_effort_tag1\nModel 2: Lernkontrolle1_tot ~ 1\n Res.Df RSS Df Sum of Sq F Pr(>F)\n1 142 491.75 \n2 143 491.75 -1 -0.0026979 8e-04 0.9778\n> AIC(model1_day1)\n[1] 591.5081\n> AIC(model2_day1)\n[1] 589.5089\n> \n> # Tag 2\n> model1_day2 <- lm(Lernkontrolle2_tot ~ mental_effort_tag2, data = calcdata_me)\n> summary(model1_day2)\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ mental_effort_tag2, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.1609 -1.7240 -0.1694 1.4534 5.6036 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 4.59772 0.53518 8.591 1.4e-14 ***\nmental_effort_tag2 -0.10921 0.06905 -1.582 0.116 \n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.264 on 142 degrees of freedom\nMultiple R-squared: 0.01731, Adjusted R-squared: 0.01039 \nF-statistic: 2.502 on 1 and 142 DF, p-value: 0.1159\n\n> model2_day2 <- lm(Lernkontrolle2_tot ~ 1, data = calcdata_me)\n> summary(model2_day2)\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ 1, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-3.8056 -1.8056 -0.3056 1.1944 5.1944 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 3.8056 0.1896 20.07 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.276 on 143 degrees of freedom\n\n> \n> anova(model1_day2,model2_day2)\nAnalysis of Variance Table\n\nModel 1: Lernkontrolle2_tot ~ mental_effort_tag2\nModel 2: Lernkontrolle2_tot ~ 1\n Res.Df RSS Df Sum of Sq F Pr(>F)\n1 142 727.73 \n2 143 740.56 -1 -12.822 2.5019 0.1159\n> AIC(model1_day2)\n[1] 647.9519\n> AIC(model2_day2)\n[1] 648.4669\n> \n> # Insgesamt \n> calcdata_me$mental_effort_tot <- calcdata_me$mental_effort_tag1 + calcdata_me$mental_effort_tag2\n> \n> model1_tot <- lm(Lernkontrolle_tot ~ mental_effort_tot, data = calcdata_me)\n> model2_tot <- lm(Lernkontrolle_tot ~ 1, data = calcdata_me)\n> \n> anova(model1_tot,model2_tot)\nAnalysis of Variance Table\n\nModel 1: Lernkontrolle_tot ~ mental_effort_tot\nModel 2: Lernkontrolle_tot ~ 1\n Res.Df RSS Df Sum of Sq F Pr(>F)\n1 142 1826.2 \n2 143 1829.3 -1 -3.0746 0.2391 0.6256\n> AIC(model1_tot)\n[1] 780.4426\n> AIC(model2_tot)\n[1] 778.6848\n\n\n9-Punkte Likert Skala genutzt von Kapur (2014) aus Paas (1992).\n\n[…] nine-point rating scale that is commonly used in the cognitive load literature as a measure of cognitive load (Paas, 1992).\n\n\nKapur (2014)\n\nIn dieser Lektion hatte ich eine:\n\nsehr, sehr geringe mentale Anstrengung\n\nsehr geringe mentale Anstrengung\n\ngeringe mentale Anstrengung\n\neher geringe mentale Anstrengung\n\nweder hohe noch niedrige mentale Anstrengung\n\neher hohe mentale Anstrengung\n\nhohe mentale Anstrengung\n\nsehr hohe mentale Anstrengung\n\nsehr, sehr hohe mentale Anstrengung\n\n\n\n\n\n\nBestätigt" + "text": "4. Mental load does not predict performance.\n\nResultatTestConsoleInstrument\n\n\n\nANOVA: \\(\\quad p=0.626\\)\n\n\n\n## 5. Testen, ob das Modell mit oder ohne Mental Effort den Leistungszuwachs besser vorhersagt\n# nur Daten behalten, die nirgendwo NA haben (weil sonst ANOVA nicht gerechnet werden kann)\ncalcdata_me <- calcdata[complete.cases(calcdata[,c(\"mental_effort_tag1\",\"mental_effort_tag2\",\"Lernkontrolle1_tot\",\"Lernkontrolle2_tot\")]),]\n\n# Tag 1\nmodel1_day1 <- lm(Lernkontrolle1_tot ~ mental_effort_tag1, data = calcdata_me)\nsummary(model1_day1)\nmodel2_day1 <- lm(Lernkontrolle1_tot ~ 1, data = calcdata_me)\nsummary(model2_day1)\n\nanova(model1_day1,model2_day1)\nAIC(model1_day1)\nAIC(model2_day1)\n\n# Tag 2\nmodel1_day2 <- lm(Lernkontrolle2_tot ~ mental_effort_tag2, data = calcdata_me)\nsummary(model1_day2)\nmodel2_day2 <- lm(Lernkontrolle2_tot ~ 1, data = calcdata_me)\nsummary(model2_day2)\n\nanova(model1_day2,model2_day2)\nAIC(model1_day2)\nAIC(model2_day2)\n\n# Insgesamt \ncalcdata_me$mental_effort_tot <- calcdata_me$mental_effort_tag1 + calcdata_me$mental_effort_tag2\n\nmodel1_tot <- lm(Lernkontrolle_tot ~ mental_effort_tot, data = calcdata_me)\nmodel2_tot <- lm(Lernkontrolle_tot ~ 1, data = calcdata_me)\n\nanova(model1_tot,model2_tot)\nAIC(model1_tot)\nAIC(model2_tot)\n\n\n> ## 5. Testen, ob das Modell mit oder ohne Mental Effort den Leistungszuwachs besser vorhersagt\n> # nur Daten behalten, die nirgendwo NA haben (weil sonst ANOVA nicht gerechnet werden kann)\n> calcdata_me <- calcdata[complete.cases(calcdata[,c(\"mental_effort_tag1\",\"mental_effort_tag2\",\"Lernkontrolle1_tot\",\"Lernkontrolle2_tot\")]),]\n> \n> # Tag 1\n> model1_day1 <- lm(Lernkontrolle1_tot ~ mental_effort_tag1, data = calcdata_me)\n> summary(model1_day1)\n\nCall:\nlm(formula = Lernkontrolle1_tot ~ mental_effort_tag1, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-3.8770 -0.8787 0.1247 1.1247 5.1213 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 3.861468 0.508998 7.586 3.95e-12 ***\nmental_effort_tag1 0.001727 0.061862 0.028 0.978 \n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 1.861 on 142 degrees of freedom\nMultiple R-squared: 5.486e-06, Adjusted R-squared: -0.007037 \nF-statistic: 0.0007791 on 1 and 142 DF, p-value: 0.9778\n\n> model2_day1 <- lm(Lernkontrolle1_tot ~ 1, data = calcdata_me)\n> summary(model2_day1)\n\nCall:\nlm(formula = Lernkontrolle1_tot ~ 1, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-3.875 -0.875 0.125 1.125 5.125 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 3.8750 0.1545 25.07 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 1.854 on 143 degrees of freedom\n\n> \n> anova(model1_day1,model2_day1)\nAnalysis of Variance Table\n\nModel 1: Lernkontrolle1_tot ~ mental_effort_tag1\nModel 2: Lernkontrolle1_tot ~ 1\n Res.Df RSS Df Sum of Sq F Pr(>F)\n1 142 491.75 \n2 143 491.75 -1 -0.0026979 8e-04 0.9778\n> AIC(model1_day1)\n[1] 591.5081\n> AIC(model2_day1)\n[1] 589.5089\n> \n> # Tag 2\n> model1_day2 <- lm(Lernkontrolle2_tot ~ mental_effort_tag2, data = calcdata_me)\n> summary(model1_day2)\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ mental_effort_tag2, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-4.1609 -1.7240 -0.1694 1.4534 5.6036 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 4.59772 0.53518 8.591 1.4e-14 ***\nmental_effort_tag2 -0.10921 0.06905 -1.582 0.116 \n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.264 on 142 degrees of freedom\nMultiple R-squared: 0.01731, Adjusted R-squared: 0.01039 \nF-statistic: 2.502 on 1 and 142 DF, p-value: 0.1159\n\n> model2_day2 <- lm(Lernkontrolle2_tot ~ 1, data = calcdata_me)\n> summary(model2_day2)\n\nCall:\nlm(formula = Lernkontrolle2_tot ~ 1, data = calcdata_me)\n\nResiduals:\n Min 1Q Median 3Q Max \n-3.8056 -1.8056 -0.3056 1.1944 5.1944 \n\nCoefficients:\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 3.8056 0.1896 20.07 <2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\nResidual standard error: 2.276 on 143 degrees of freedom\n\n> \n> anova(model1_day2,model2_day2)\nAnalysis of Variance Table\n\nModel 1: Lernkontrolle2_tot ~ mental_effort_tag2\nModel 2: Lernkontrolle2_tot ~ 1\n Res.Df RSS Df Sum of Sq F Pr(>F)\n1 142 727.73 \n2 143 740.56 -1 -12.822 2.5019 0.1159\n> AIC(model1_day2)\n[1] 647.9519\n> AIC(model2_day2)\n[1] 648.4669\n> \n> # Insgesamt \n> calcdata_me$mental_effort_tot <- calcdata_me$mental_effort_tag1 + calcdata_me$mental_effort_tag2\n> \n> model1_tot <- lm(Lernkontrolle_tot ~ mental_effort_tot, data = calcdata_me)\n> model2_tot <- lm(Lernkontrolle_tot ~ 1, data = calcdata_me)\n> \n> anova(model1_tot,model2_tot)\nAnalysis of Variance Table\n\nModel 1: Lernkontrolle_tot ~ mental_effort_tot\nModel 2: Lernkontrolle_tot ~ 1\n Res.Df RSS Df Sum of Sq F Pr(>F)\n1 142 1826.2 \n2 143 1829.3 -1 -3.0746 0.2391 0.6256\n> AIC(model1_tot)\n[1] 780.4426\n> AIC(model2_tot)\n[1] 778.6848\n\n\n9-Punkte Likert Skala genutzt von Kapur (2014) aus Paas (1992).\n\n[…] nine-point rating scale that is commonly used in the cognitive load literature as a measure of cognitive load (Paas, 1992).\n\n\nKapur (2014)\n\nIn dieser Lektion hatte ich eine:\n\nsehr, sehr geringe mentale Anstrengung\n\nsehr geringe mentale Anstrengung\n\ngeringe mentale Anstrengung\n\neher geringe mentale Anstrengung\n\nweder hohe noch niedrige mentale Anstrengung\n\neher hohe mentale Anstrengung\n\nhohe mentale Anstrengung\n\nsehr hohe mentale Anstrengung\n\nsehr, sehr hohe mentale Anstrengung" }, { "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#bibliographie", @@ -4082,7 +4082,7 @@ "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#the-alt-model-outperforms-the-4f-model.", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "3. The Alt model outperforms the 4F model.", - "text": "3. The Alt model outperforms the 4F model.\n\nResultatTestConsole\n\n\n\nWilcoxon: \\(\\quad p=0.804\\)\n\n\n## 2. TOST rechnen: Two-Sided Significance Testing\ntost_result <- TOSTtwo(\n m1 = mean(FF$Lernkontrolle1_tot, na.rm = TRUE),\n m2 = mean(ALT$Lernkontrolle1_tot, na.rm = TRUE),\n sd1 = sd(FF$Lernkontrolle1_tot, na.rm = TRUE),\n sd2 = sd(ALT$Lernkontrolle1_tot, na.rm = TRUE), \n n1 = length(FF),\n n2 = length(ALT),\n low_eqbound_d = -0.5, # maximal 0.5 Punkte unter\n high_eqbound_d = +0.5, # maximal 0.5 Punkte über\n alpha = 0.05 # Konfidenz: 95%\n)\n\n\n> ## 2. TOST rechnen: Two-Sided Significance Testing\n> tost_result <- TOSTtwo(\n+ m1 = mean(FF$Lernkontrolle1_tot, na.rm = TRUE),\n+ m2 = mean(ALT$Lernkontrolle1_tot, na.rm = TRUE),\n+ sd1 = sd(FF$Lernkontrolle1_tot, na.rm = TRUE),\n+ sd2 = sd(ALT$Lernkontrolle1_tot, na.rm = TRUE), \n+ n1 = length(FF),\n+ n2 = length(ALT),\n+ low_eqbound_d = -0.5, # maximal 0.5 Punkte unter\n+ high_eqbound_d = +0.5, # maximal 0.5 Punkte über\n+ alpha = 0.05 # Konfidenz: 95%\n+ )\nTOST results:\nt-value lower bound: 4.44 p-value lower bound: 0.000009\nt-value upper bound: -1.77 p-value upper bound: 0.040\ndegrees of freedom : 150.83\n\nEquivalence bounds (Cohen's d):\nlow eqbound: -0.5 \nhigh eqbound: 0.5\n\nEquivalence bounds (raw scores):\nlow eqbound: -1.0284 \nhigh eqbound: 1.0284\n\nTOST confidence interval:\nlower bound 90% CI: -0.105\nupper bound 90% CI: 0.992\n\nNHST confidence interval:\nlower bound 95% CI: -0.212\nupper bound 95% CI: 1.098\n\nEquivalence Test Result:\nThe equivalence test was significant, t(150.83) = -1.765, p = 0.0398, given equivalence bounds of -1.028 and 1.028 (on a raw scale) and an alpha of 0.05.\n\nNull Hypothesis Test Result:\nThe null hypothesis test was non-significant, t(150.83) = 1.337, p = 0.183, given an alpha of 0.05.\n\nNHST: don't reject null significance hypothesis that the effect is equal to 0 \nTOST: reject null equivalence hypothesis\n\n\n\n\n\nVerworfen und …\n\n\n\nTOST-Äquivalenztest (vgl. Walker & Nowacki, 2010) - Statistically equivalent and not different." + "text": "3. The Alt model outperforms the 4F model.\n\nResultatTestConsole\n\n\n\nWilcoxon: \\(\\quad p=0.804\\)\n\n\n## 2. TOST rechnen: Two-Sided Significance Testing\ntost_result <- TOSTtwo(\n m1 = mean(FF$Lernkontrolle1_tot, na.rm = TRUE),\n m2 = mean(ALT$Lernkontrolle1_tot, na.rm = TRUE),\n sd1 = sd(FF$Lernkontrolle1_tot, na.rm = TRUE),\n sd2 = sd(ALT$Lernkontrolle1_tot, na.rm = TRUE), \n n1 = length(FF),\n n2 = length(ALT),\n low_eqbound_d = -0.5, # maximal 0.5 Punkte unter\n high_eqbound_d = +0.5, # maximal 0.5 Punkte über\n alpha = 0.05 # Konfidenz: 95%\n)\n\n\n> ## 2. TOST rechnen: Two-Sided Significance Testing\n> tost_result <- TOSTtwo(\n+ m1 = mean(FF$Lernkontrolle1_tot, na.rm = TRUE),\n+ m2 = mean(ALT$Lernkontrolle1_tot, na.rm = TRUE),\n+ sd1 = sd(FF$Lernkontrolle1_tot, na.rm = TRUE),\n+ sd2 = sd(ALT$Lernkontrolle1_tot, na.rm = TRUE), \n+ n1 = length(FF),\n+ n2 = length(ALT),\n+ low_eqbound_d = -0.5, # maximal 0.5 Punkte unter\n+ high_eqbound_d = +0.5, # maximal 0.5 Punkte über\n+ alpha = 0.05 # Konfidenz: 95%\n+ )\nTOST results:\nt-value lower bound: 4.44 p-value lower bound: 0.000009\nt-value upper bound: -1.77 p-value upper bound: 0.040\ndegrees of freedom : 150.83\n\nEquivalence bounds (Cohen's d):\nlow eqbound: -0.5 \nhigh eqbound: 0.5\n\nEquivalence bounds (raw scores):\nlow eqbound: -1.0284 \nhigh eqbound: 1.0284\n\nTOST confidence interval:\nlower bound 90% CI: -0.105\nupper bound 90% CI: 0.992\n\nNHST confidence interval:\nlower bound 95% CI: -0.212\nupper bound 95% CI: 1.098\n\nEquivalence Test Result:\nThe equivalence test was significant, t(150.83) = -1.765, p = 0.0398, given equivalence bounds of -1.028 and 1.028 (on a raw scale) and an alpha of 0.05.\n\nNull Hypothesis Test Result:\nThe null hypothesis test was non-significant, t(150.83) = 1.337, p = 0.183, given an alpha of 0.05.\n\nNHST: don't reject null significance hypothesis that the effect is equal to 0 \nTOST: reject null equivalence hypothesis\n\n\n\n\nTOST-Äquivalenztest (vgl. Walker & Nowacki, 2010) - Statistically equivalent and not different." }, { "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#fragen", @@ -4096,7 +4096,7 @@ "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#danke-für-ihre-aufmerksamkeit", "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", "section": "Danke für Ihre Aufmerksamkeit", - "text": "Danke für Ihre Aufmerksamkeit\n\n\n\n\n\nPH Heidelberg, Richard Conrardy" + "text": "Danke für Ihre Aufmerksamkeit\n\n\n\n\n\n4F\nALT\n\n\n\n\nFail\nin class\nout of class\n\n\nFlip\nout of class\nin class\n\n\nFix\nin class\nin class\n\n\nFeed\nin class\nin class" }, { "objectID": "lerngelegenheiten/LG_digi-datenschutz/slides_datenschutz.html#section", @@ -4181,5 +4181,12 @@ "title": "Digitalität und Datenschutz", "section": "Bibliographie", "text": "Bibliographie\n\n\n\n\nPHBern, Institut Sekundarstufe 1, Richard Conrardy, Dozent\n\n\n\n\nKülling, C., Waller, G., Suter, L., Bernath, J., Skirgaila, P., Streule, P., & Süss, D. (2022). JAMES – Jugend, Aktivitäten, Medien – Erhebung Schweiz. Zürich: Zürcher Hochschule für Angewandte Wissenschaften." + }, + { + "objectID": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#untersuchungsdesign-1", + "href": "tagungsunterlagen/2024_11_21_Zukunftsforum.html#untersuchungsdesign-1", + "title": "Empirische Untersuchung eines Vier-Phasen-Unterrichtsmodells in Schweizer Sekundarklassen", + "section": "Untersuchungsdesign", + "text": "Untersuchungsdesign\n\n\n\n\nPH Heidelberg, Richard Conrardy" } ] \ No newline at end of file diff --git a/docs/site_libs/revealjs/dist/theme/quarto.css b/docs/site_libs/revealjs/dist/theme/quarto.css index bee35d7..dd23d09 100644 --- a/docs/site_libs/revealjs/dist/theme/quarto.css +++ b/docs/site_libs/revealjs/dist/theme/quarto.css @@ -1,8 +1,8 @@ -@import"./fonts/source-sans-pro/source-sans-pro.css";:root{--r-background-color: #fff;--r-main-font: sans-serif;--r-main-font-size: 40px;--r-main-color: #222;--r-block-margin: 12px;--r-heading-margin: 0 0 12px 0;--r-heading-font: sans-serif;--r-heading-color: 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!important}.quarto-float-caption-bottom.quarto-float-caption.quarto-float-fig{text-align:center}/*# sourceMappingURL=f95d2bded9c28492b788fe14c3e9f347.css.map */ diff --git a/docs/tagungsunterlagen/2024_11_21_Zukunftsforum.html b/docs/tagungsunterlagen/2024_11_21_Zukunftsforum.html index 44532e5..1dba8dc 100644 --- a/docs/tagungsunterlagen/2024_11_21_Zukunftsforum.html +++ b/docs/tagungsunterlagen/2024_11_21_Zukunftsforum.html @@ -570,7 +570,7 @@

Metastudie

-

\[n=173\]

+

n=173

@@ -731,14 +731,16 @@

Forschungsfragen

Untersuchungsdesign

-
+
    -
  • Mathematikunterricht auf der Sekundarstufe 1
    • +
  • Mathematikunterricht in der Sekundarstufe 1
  • Thema: Mittlere absolute Abweichung
    • -
  • Experimentelles Forschungsdesign
    • -
  • \(n=220\)
    • +
  • Experimentelles Forschungsdesign (randomisiert innerhalb der Klassen)
    • +
  • 220 Lernende aus 12 Klassen
-
+
+ +
@@ -949,11 +951,6 @@

1. Engagement is highest during in-class activities, irrespective of variant -
-
-

Verworfen

-
-

2. Engagement predicts performance.

@@ -1363,11 +1360,6 @@

2. Engagement predicts performance.

-
-
-

Bestätigt

-
-

3. The Alt model outperforms the 4F model.

@@ -1438,11 +1430,6 @@

3. The Alt model outperforms the 4F model.

-
-

Verworfen und …

-
-
-

TOST-Äquivalenztest (vgl. Walker & Nowacki, 2010) - Statistically equivalent and not different.

@@ -1654,11 +1641,6 @@

4. Mental load does not predict performance.

-
-
-

Bestätigt

-
-

Zentrale Erkenntnisse

@@ -1732,14 +1714,48 @@

Bibliographie

Danke für Ihre Aufmerksamkeit

-

+
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
4FALT
Failin classout of class
Flipout of classin class
Fixin classin class
Feedin classin class
+
+
+
+

Untersuchungsdesign

+ -
+

PH Heidelberg, Richard Conrardy

-
diff --git a/tagungsunterlagen/2024_11_21_Zukunftsforum.qmd b/tagungsunterlagen/2024_11_21_Zukunftsforum.qmd index 0d71284..9b09caa 100644 --- a/tagungsunterlagen/2024_11_21_Zukunftsforum.qmd +++ b/tagungsunterlagen/2024_11_21_Zukunftsforum.qmd @@ -97,7 +97,7 @@ SOL: [@hilbe_selbst_2016; @ammann-tinguely_selbst_2020]; reichhaltige Aufgaben: | School-age children | 0.68 | 0.025 | | Math | 0.26 | 0.067 | -$$n=173$$ +n=173 ::: @@ -192,15 +192,16 @@ $$n=173$$ ## Untersuchungsdesign :::{.columns} -::: {.column width="40%"} -* Mathematikunterricht auf der Sekundarstufe 1 +::: {.column width="43%"} +* Mathematikunterricht in der Sekundarstufe 1 * Thema: Mittlere absolute Abweichung -* Experimentelles Forschungsdesign -* $n=220$ +* Experimentelles Forschungsdesign (randomisiert innerhalb der Klassen) +* 220 Lernende aus 12 Klassen +::: +::: {.column width="4%"} ::: - -::: {.column width="50%"} +::: {.column width="43%"} ::: {.fragment .fade-in} | | 4F | ALT | |-----------------------|---------------------------|--------------| @@ -343,11 +344,7 @@ alternative hypothesis: true location shift is not equal to 0 ::: -:::{.fragment .fade-up} -::: {style="color: #cc3333; font-size: 2em; font-weight: bold; text-align: center;"} -Verworfen -::: -::: + ## 2. Engagement predicts performance.{style="font-size:0.6em;"} @@ -673,11 +670,7 @@ Davids Ergebnisse für Mathematik, Physik und Chemie in den Abschlussprüfungen ::: -:::{.fragment .fade-up} -::: {style="color: #0c710b; font-size: 2em; font-weight: bold; text-align: center;"} -Bestätigt -::: -::: + ## 3. The Alt model outperforms the 4F model. {style="font-size:0.6em;"} @@ -755,11 +748,7 @@ TOST: reject null equivalence hypothesis ::: -:::{.fragment .fade-up} -::: {style="color: #cc3333; font-size: 2em; font-weight: bold; text-align: center;"} -Verworfen und ... -::: -::: + :::{.fragment .fade-up} @@ -976,11 +965,7 @@ In dieser Lektion hatte ich eine: ::: -:::{.fragment .fade-up} -::: {style="color: #0c710b; font-size: 2em; font-weight: bold; text-align: center;"} -Bestätigt -::: -::: + ## Zentrale Erkenntnisse @@ -1007,4 +992,13 @@ Bestätigt . . . +| | 4F | ALT | +|-----------------------|---------------------------|--------------| +| **Fail** | in class | out of class | +| **Flip** | out of class | in class | +| **Fix** | in class | in class | +| **Feed** | in class | in class | + +## Untersuchungsdesign + ![](2024_11_21_Zukunftsforum/aufbau.jpg)