STA 110B STA 110B Homework 11

12-11

  1. For the given data,
    b
    =
    b±t.025  s
      æ
    Ö

    åx2
     
    =
    800±2.02   7300
      ___
    Ö900
     
    =
    800±491.5
    308.5 < b < 1291.5

    Note that, since n = 50, the number of degrees of freedom of the t distribution is 48. This critical point can not be found in the tables, so we used d.f.=40, having in mind that this way we are being conservative and therefore we get a larger interval. Linear interpolation is also OK.

  2. Yes, since 0\not Î (308.5, 1291.5).

  3. Prediction interval:

    Y0
    =
    a+b X0 ±t.025 s     æ
    Ö

    1
    n
    + (X0-[`X]0)2
    åx2
    +1
     
    =
    1200 + 800 ·10 ±2.02 ·7300·   æ
    Ö

    1
    50
    + (10-11)2
    900
    +1
     
    =
    9200 ±14900.8
  4. No cause-effect relation between annual income and level of education can be inferred from his analysis, since the data arises from an uncontrolled observational study. Uncontrolled confounding variables can be responsible for the observed relation.

12-13

Since t = b/SE and SE = .27/t.025, we get t = 2.76. Also the hypotheses is H0: b £ 0, so 0.0025 < p-value < 0.005. Note that we used a normal standard table since the sample size is large: n = 1000.

Extra Exercise

  1. We have n = 32, so that df = 30 and t.025 = 2.05.
    b
    =
    b ±t.025 SE
    =
    102.289 ±2.05 ·24.23
    =
    102.289 ±49.43
    52.86 < b < 151.72
    a
    =
    a ±t.025 SE
    =
    -6553.57 ±2.04 ·1661.96
    =
    -6553.57 ±3390.40
    -9943.97 < a < -3163.17
  2. Yes, since the 95% confidence interval for b does not contain zero.
  3. The hypotheses to test is H0b = 100, so the t statistic is
    t = b-100
    SE
    = 102.289-100
    24.23
    = 0.094.
    We want to calculate a two-sided p-value, so, looking at the table, we recognize that p-value > 2·0.25, that is, the p-value is larger than 50%.
  4. Looking at the output tables, we see that s2 = 18761.24.


File translated from TEX by TTH, version 2.00.
On 12 Apr 1999, 14:21.