Exam 1

  1. Metainformation

    Tag Value
    fileInferential_Statistics_uu-Independent-samples-means-602-en_uu-Independent-samples-means-602-en
    nameuu-Independent-samples-means-602-en.Rmd
    sectionInferential Statistics/Parametric Techniques/t-test/Independent samples means
    typestring
    solutionna
    TypeCase
    LanguageEnglish

    Question

    Eight men applying to Cambridge University had a sample mean and variance on College Board tests of 1050 and 2500 respectively. The respective numbers for nine women were 1075 and 3600. Test at the .05 level of significance whether women did better than men on these tests. Use the five steps of hypothesis testing.


    1. na:

    Solution

    1. H0:μf=μmH_0: \mu_f = \mu_m and HA:μf>μmH_{A}: \mu_f > \mu_m
      1. Smale2=2500S^2_{male} = 2500 : $S^2_{female} =3600 $
      2. Spooled2=(8−1)2500+(9−1)36008+9−2=3086.7S^2_{pooled} = \frac {(8-1)2500 + (9-1)3600}{8+9-2} =3086.7
      3. Sm12=3086.78=385.8375;Sm22=3086.79=342.9667S^2_{m1} = \frac {3086.7}{8}= 385.8375;S^2_{m2} = \frac {3086.7}{9}= 342.9667
      4. Sdifference2=Sm12+Sm22=728.8042S^2_{difference}= S^2_{m1} + S^2_{m2} =728.8042
      5. Sdifference2=728.8042=26.99637S^2_{difference}= \sqrt{728.8042} = 26.99637
    3. α=.05\alpha=.05, dfTotal=8+9−2=15df_{Total} = 8+9-2=15, cut-off =−1.753=-1.753
    4. t=1050−10753086.7(1/8+1/9)=−0.9261≈−0.93t= \frac {1050-1075}{\sqrt {3086.7 (1/8 + 1/9)} } = -0.9261 \approx -0.93
    5. Do not reject H0H_0. Women’s scores were not higher than men’s scores.