EXAMINATION PAPER. Exam in: STA-3300 Applied statistics 2 Date: Wednesday, November 25th 2015 Time: Kl 09:00 13:00 Place: Teorifagb.

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1 EXAMINATION PAPER Exam in: STA-3300 Applied statistics 2 Date: Wednesday, November 25th 2015 Time: Kl 09:00 13:00 Place: Teorifagb.,hus 1, plan 3 Approved aids: Calculator All printed and written The exam contains 9 pages including this cover page Contact person: Elinor Ytterstad Phone: NB! sheets It is not allowed to submit rough paper along with the answer 1

2 Sta-3300 November 2015, page 2 of 9 More SPSS outputs are presented at the end of the exam paper. Some numbers may be deleted from the SPSS tables. If not specified, use 5 % significance level. A brief translation to Norwegian starting at page 8 The exam paper may be answered in Norwegian Problem 1 (SPSS outputs from page 5) A study was carried out to explore the relationship between Aggression and several potential factors in 666 children that had older siblings. The independent variables were: Y = Aggression (high score = more aggression in the child) X 1 = Television (high score = more time spent watching television) X 2 = Computer Games (high value = more time spent playing computer games) X 3 = Sibling Aggression (high score = more aggression seen in their older sibling) X 4 = Diet (high score = the child has a good diet low in E-numbers) X 5 = Parenting Style (high score = bad parenting practices) Data from Discovering statistics using SPSS, Andy Field. SAGE Publications Ltd 2009 Descriptive Statistics Y Aggression X1 Time spent watching television. X3 Sibling Aggression Valid N (listwise) N Minimum Maximum Mean Std. Deviation 666-1,30 1,18 -,0050, ,46,98 -,0276, ,15 1,62,0104, ,43 1,10,0083, ,28 1,22,0116, ,46 3,99,0000 1, Zero order correlations: r Y,X1 = 0.159, r Y,X2 = 0.186, r Y,X3 = 0.129, r Y,X4 = and r Y,X5 = Model A: Y = β 0 + β 3 X 3 + E, A linear model for Aggression explained by Sibling Aggression Model B: Y = β 0 + β 5 X 5 + E, A linear model for Aggression explained by Parenting Style a) Use the zero order correlations above, and choose the variable that seems to be the most important in explaining the aggression in a child. Explain why you chose this variable. Calculate the estimated slope in model A, ˆβ 3, from the information in the descriptive statistics table and the correlations above.

3 Sta-3300 November 2015, page 3 of 9 b) Model B. You may use that Ŷ = X 5 (from the SPSS output) Make a drawing of this regression line. Test the hypothesis H 0 : β 5 = 0 versus H 1 : β 5 0, use 5 % significant level. What is the conclusion for this particular data set? Model C: Y = β 0 + β 5 X 5 + β 3 X 3 + E Model D: Y = β 0 + β 5 X 5 + β 3 X 3 + β 1 X 1 + β 2 X 2 + β 4 X 4 + E Previous research have shown that parenting style (X 5 ) and aggression in older siblings (X 3 ) are important factors in explaining child aggression. We will also study whether television (X 1 ), computer games (X 2 ) and diet (X 4 ) are related to aggression in children. c) Test whether there in addition to parenting style and sibling aggression, is a significant effect of television, computer games and diet. State the null hypothesis, and perform the test on 5% significance level. Are any of the extra variables (X 1, X 2, X 4 ) non-significant as added last to model D? If yes, which variable(s). We will split the data in two groups, children who spent more time on television (X 1 > 0) in one group, and children who spent less time on television (X 1 0)in the other group. The regression model D is run separately for both groups, and the results are shown on page 6. d) Write down the two estimated regression models: Ŷ T V + = for the group of children who spent more time on television, and ŶT V = for the group of children who spent less time on television. Compare these two estimated regression models, and explain what the differences means for the two television groups. We continue with the extended model E: Y = β 0 + β 5 X 5 + β 3 X 3 + β 1 X 1 + β 2 X 2 + β 4 X 4 + β 6 X 1 X 3 + E e) What kind of variable is this new independent variable X 1 X 3 (the product of X 1 and X 3 )? Use the SPSS printout and write down the estimated regression model E: Ŷ = Use model E and find (calculate) the estimated regression model when X 1 = 0.3. Do the same for X 1 = 0.3. (These estimated regression models should contain ˆβ 3 = when X 1 = 0.3, and ˆβ 3 = when X 1 = 0.3). Compare this result to your answer in d) f) Calculate a 95% confidence interval for β 6 in model E. Use this interval to test the hypothesis: H 0 : β 6 = 0 versus H 1 : β 6 0 on a 5% significant level. (If you are not able to find the confidence interval, then test the hypotheses by a t-test.)

4 Sta-3300 November 2015, page 4 of 9 Some SPSS results for model E: The largest Cook s distance is d i = The largest leverage is h i = of the Jackknife residuals are greater than 1.96 and 16 are less than 1.96 g) Why is Jackknife residuals better than studentized residuals in detecting outliers? Use the information above on Cook s distance, leverage and Jackknife residuals, and discuss whether there are any outliers in the data. Confirm your conclusion with what you see in the residual scatter plot on page 7 Problem 2 (SPSS outputs on page 7) Three battery types are tested on four flashlights to see whether the battery types have different expected life times. Y = life time is measured in hours Flashlight Battery type A B C Model: Y ij = µ + α i + β j + E. i = 1, 2, 3 (battery type) and j = 1, 2, 3, 4 (flashlight) a) Calculate and fill in the three missing numbers for flash in the ANOVA table on page 7. Use the information in the SPSS output and test the hypotesis H 0 : α 1 = α 2 = α 3 = 0 on 5% significance level. What does the test result tell us about battery life time? Use the given SPSS output and write down the ANOVA-table for the one-way model Y ij = µ + α i + E, where i = battery type.

5 Sta-3300 November 2015, page 5 of 9 Problem 1 SPSS results Model A: Y = β 0 + β 3 X 3 + E Model B: Y = β 0 + β 5 X 5 + E Model C: Y = β 0 + β 5 X 5 + β 3 X 3 + E Model D: Y = β 0 + β 5 X 5 + β 3 X 3 + β 1 X 1 + β 2 X 2 + β 4 X 4 + E Model E: Y = β 0 + β 5 X 5 + β 3 X 3 + β 1 X 1 + β 2 X 2 + β 4 X 4 + β 6 X 1 X 3 + E Model B Coefficients a Model B Unstandardized Coefficients Standardized Coefficients B Std. Error Beta t Sig. -,005,012 -,414,679,067,012,211 Model C and D, Coefficients a Model C D X3 Sibling Aggression X3 Sibling Aggression X1 Time spent watching television. Unstandardized Coefficients Standardized Coefficients B Std. Error Beta t Sig. -,006,012 -,479,632,062,012,194 5,057,000,093,038,096 2,491,013 -,005,012 -,416,677,057,015,177 3,891,000,082,039,084 2,106,036,033,046,032,715,475,142,037,152 3,851,000 -,109,038 -,115-2,864,004 Model C and D ANOVA a Model C D Regression Residual Total Regression Residual Total Sum of Squares df Mean Square F Sig. 3, ,806 18,644,000 b 64, ,097 67, , ,120 11,882,000 c 62, ,094 67, b. Predictors:, X3 Sibling Aggression, c. Predictors:, X3 Sibling Aggression,,,, X1 Time spent watching television. Page 1 Page 1

6 Sta-3300 November 2015, page 6 of 9 Model D Coefficients a Unstandardized Coefficients TVgroup less time, X1<0 more time, X1>=0 X3 Sibling Aggression X1 Time spent watching television. X3 Sibling Aggression X1 Time spent watching television. B Std. Error t Sig.,040,025 1,593,112,078,021 3,671,000,152,058 2,623,009,128,085 1,510,132,089,054 1,653,099 -,151,056-2,707,007 -,010,024 -,400,690,030,020 1,512,132 -,003,051 -,064,949,039,091,429,668,186,050 3,695,000 -,051,051-1,011,313 Model E Coefficients a Model E X3 Sibling Aggression X1 Time spent watching television. X1*X3 Unstandardized Coefficients Standardized Coefficients B Std. Error Beta t Sig.,001,012,057,954,054,015,170 3,742,000,068,039,070 1,754,080,027,046,026,592,554,148,037,158 4,009,000 -,109,038 -,115-2,876,004 -,254,095 -,101-2,669,008 Page 1

7 Sta-3300 November 2015, page 7 of 9 Model E: The largest Cook s distance is d i = The largest leverage is h i = of the Jackknife residuals are greater than 1.96 and 16 are less than 1.96 Jackknife residuals (Studentized Deleted) Model E 4,00000 Studentized Deleted Residual 2,00000, , , , , ,20000,00000,20000,40000 Unstandardized Predicted Value Problem 2 SPSS results Tests of Between-Subjects Effects Dependent Variable: Source Corrected Model Intercept batt flash Error Total Corrected Total Y life time Type III Sum of Squares df Mean Square F Sig. 100,417 a 5 20,083 14,176, , , ,471,000 2, ,083,765,506 98,250,001 8, , , , a. R Squared =,922 (Adjusted R Squared =,857) Page 1

8 Sta-3300 November 2015, page 8 of 9 Norwegian translation Oppgave 1 En studie er gjennomført for å undersøke aggresjon hos barn relatert til flere potensielle faktorer. Materialet inneholder data fra 666 barn som har eldre søsken. a) Bruke de oppgitte nullte-ordens korrelasjonene og velg variabelen som ser ut til å best forklare aggresjon hos barn. Forklar hvorfor du har valgt denne variabelen. Regn ut estimert stigningskoeffisient i modell A, ˆβ 3, fra opplysningene i tabell descriptive statistics og de oppgitte korrelasjonene. b) Modell B Bruk gjerne at Ŷ = X 5 (fra SPSS utskriften) Tegn denne regresjonslinjen i en figur. Test hypotesen H 0 : β 5 = 0 mot H 1 : β 5 0, bruk 5 % signifikansnivå. Hva blir konklusjonen for dette datamaterialet? Model C: Y = β 0 + β 5 X 5 + β 3 X 3 + E Model D: Y = β 0 + β 5 X 5 + β 3 X 3 + β 1 X 1 + β 2 X 2 + β 4 X 4 + E Tidligere forskning har vist at oppdragelse (parenting style) (X 5 ) og aggressjon hos eldre søsken (X 3 ) er viktige faktorer i forklaring av aggresjon hos barn. Vi skal undersøke om også television (X 1 ), computer games (X 2 ) og diet (X 4 ) ytterligere forklarer aggresjon hos barn. c) Test om det er signifikant effekt av television, computer games og diet i modell som allerede inneholder parenting style og aggresjon hos eldre søsken. Skriv opp nullhypotesen og utfør testen på 5% signifikansnivå.. Er noen av tilleggsvariablene (X 1, X 2, X 4 ) ikke-signifikant som added last til model D? Hvis ja, hvilke variable(r). Vi skal nå splitte datamaterialet i to grupper, den ene gruppen består av barn som ser mye på TV (X 1 > 0) og den andre gruppen er barn som ser lite på TV (X 1 0). Vi kjører separate regresjonsanalyser for hver av de to gruppene og modell D, og resultatene finner du på side 6. d) Skriv ned de to estimerte regresjonsmodellene: Ŷ T V + = for barn som ser mye på TV og Ŷ T V = for barn som ser lite på TV. Sammenlign de to estimerte regresjonsmodellene og forklar betydningen av forskjellene relatert til de to TV gruppene. Vi fortsetter med en utvidet model E: Y = β 0 + β 5 X 5 + β 3 X 3 + β 1 X 1 + β 2 X 2 + β 4 X 4 + β 6 X 1 X 3 + E

9 Sta-3300 November 2015, page 9 of 9 e) Hvilken type variabel er denne nye uavhengige variabelen X 1 X 3 (produktet av X 1 og X 3 )? Bruk SPSS utskriften og skriv ned den estimerte regresjonsmodellen modell E: Ŷ = Bruk modell E og regn ut estimert regresjonsmodell når X 1 = 0.3. Gjenta for X 1 = 0.3. (Resultatet skal da bli at ˆβ 3 = når X 1 = 0.3, og ˆβ 3 = når X 1 = 0.3). Sammenlign dette med svaret du kom frem til i d) f) Regn ut et 95% konfidensintervall for β 6 i modell E. Bruk intervallet til å teste hypotesen: H 0 : β 6 = 0 mot H 1 : β 6 0 på 5% signifikansnivå. (I tilfelle du ikke kommer frem til et konfidensintervall, skal du gjøre en t-test av hypotesene) g) Hvorfor er Jackknife bedre enn studentized residualer til å identifisere utenforliggerobservasjoner? Bruk de gitte opplysningene om Cook s distance, leverage og Jackknife residualer og diskuter om det er utenforliggerobservasjoner i materialet. Bruk spredningsplottet av residualer side 7 som bekreftelse på din konklusjon her. Oppgave 2 Tre batterityper testes i fire lommelykter for å undersøke om batteritypene har ulik forventet levetid. Y = levetid i timer. Modell: Y ij = µ + α i + β j + E. i = 1, 2, 3 (batteritype) og j = 1, 2, 3, 4 (lommelykt) a) Regn ut og fyll inn de tre tallene som mangler for flash i ANOVA-tabellen side 7. Bruk SPSS-utskriftene og test hypotesen H 0 : α 1 = α 2 = α 3 = 0 på 5% signifikansnivå. Hva forteller testresultatet oss om batterilevetider? Bruk opplysninger fra SPSS-utskriften og skriv ned ANOVA-tabellen for en enveis modell Y ij = µ + α i + E der i = batteritype.