STAT 110 Semester One

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STAT 110Semester One

• PROPORTIONS

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COMPARING TWO SAMPLES

• Unpaired t-test For LARGE SAMPLES where n1 and
  n2 30 the 95 C.I. for the differences in means
  of two populations is
• ( x1 x2) 1.96vs21/n1 s22/n2
• For SMALL SAMPLES (n1 n2 lt 30) the C.I. is
• ( x1 x2) tvspv1/n1 1/n2
• Paired t-test The C.I. is d tv sd/vn

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C.I. FOR A PROPORTION

• If X is a binomial dist. with parameters n p we
  know that mxnp and sxvnp(1-p).
• Suppose one sample produces a proportion of
  successes px/n in n trials.
• If we take many samples we get different ps.
• Using the central limit theorem, the resulting
  dist. P of these proportions is normal.

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MEAN AND STD. DEV. FOR PROPORTION

• If PX/n
• mP mX/n
• np/n
• p and
• s2P (1/n)2 Var(X)
• (1/n)2 np(1-p)
• p(1-p)/n
• sP vp(1-p)/n

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C.I. FOR A PROPORTION

• We use the sample proportion (p) to estimate the
  unknown true population proportion (p).
• The 95 C.I. for p is
• p 1.96 vp(1-p)/n
• We always use 1.96 (95) and 2.58 (99) for
  C.I.s for proportions.

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EXAMPLE (pg 183)

• A random sample of 500 Aucklanders taken in 1996
  had 173 supporting aerial spraying to eradicate
  tussock moth.
• Estimate the proportion (p) of Aucklanders who
  support this spraying.

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SOLUTION

• The 95 C.I. for p is
• p 1.96 vp(1-p)/n
• 173/500 1.96 173/500(1-173/500)
• 500
• 0.346 .042
• (0.304, 0.388)
• Or we can write 0.304 lt p lt 0.388

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CONCLUSION

• Based on the 95 confidence interval of
• 0.304 lt p lt 0.388 we can say
• we are 95 sure that between 30.4 and 38.7 of
  the Auckland population support the spraying.
• Alternatively we say
• 34.6 of the population support spraying with a
  margin of error of 4.2.
• We usually only use the margin of error if p lies
  between 0.3 and 0.7.

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EXAMPLE (pg 184)

• An epidemiologist estimates the proportion of
  women with asthma.
• Find the sample size (n) needed to give an
  estimate for this proportion with an error no
  more than 0.03 with 95 confidence.

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INVESTIGATING SAMPLE SIZE

• p p(1-p)
• 0.1 0.09
• 0.2 0.16
• 0.3 0.21
• 0.4 0.24
• 0.5 0.25
• 0.6 0.24

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• The most conservative sample size is obtained
  using p 0.5
• We solve
• 1.96 x v0.5(1-0.5)/n lt error
• for sample size n.

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• 1.96 x v0.5(1-0.5)/n lt error

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96
• 0.5(1-0.5)/n lt (0.03/1.96)2

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96
• 0.5(1-0.5)/n lt (0.03/1.96)2
• 0.5(1-0.5) lt (0.03/1.96)2 x n

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96
• 0.5(1-0.5)/n lt (0.03/1.96)2
• 0.5(1-0.5) lt (0.03/1.96)2 x n
• 0.25 lt (0.03/1.96)2 x n

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96
• 0.5(1-0.5)/n lt (0.03/1.96)2
• 0.5(1-0.5) lt (0.03/1.96)2 x n
• 0.25 lt (0.03/1.96)2 x n
• 0.25 / (0.03/1.96)2 lt n

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96
• 0.5(1-0.5)/n lt (0.03/1.96)2
• 0.5(1-0.5) lt (0.03/1.96)2 x n
• 0.25 lt (0.03/1.96)2 x n
• 0.25 / (0.03/1.96)2 lt n
• 1067.11 lt n

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• 1.96 x v0.5(1-0.5)/n lt error
• 1.96 x v0.5(1-0.5)/n lt 0.03
• v0.5(1-0.5)/n lt 0.03/1.96
• 0.5(1-0.5)/n lt (0.03/1.96)2
• 0.5(1-0.5) lt (0.03/1.96)2 x n
• 0.25 lt (0.03/1.96)2 x n
• 0.25 / (0.03/1.96)2 lt n
• 1067.11 lt n thus n 1068

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4. COMPARING PROPORTIONS

• Confidence Interval for the difference p1-p2
• (p1-p2) 1.96 v p1(1-p1)/n1 p2(1-p2)/n2

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EXAMPLE pg 186-7

• To study the effectiveness of a drug for
  arthritis, two samples of patients were randomly
  selected. One sample of 100 was injected with
  the drug, the other sample of 60 receiving a
  placebo injection. After a period of time the
  patients were asked if their arthritic condition
  had improved. Results were
• Drug Placebo
• Improved 59 22
• Not improved 41 38
• TOTAL 100 60

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SOLUTION

• The proportions improved are
• p1 59/100 0.59
• p2 22/60 0.37
• The confidence interval is
• (0.59 - 0.37)
• 1.96 v(0.59x0.41)/100 (0.37x0.63)/60
• 0.22 0.16
• (0.06, 0.38)

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• We can write
• 0.06 lt p1 p2 lt 0.38
• Since 0 is excluded from the interval and the
  interval is positive, there is evidence that the
  difference p1 p2 gt 0.
• That is, we conclude the proportion improved is
  higher when the drug is used.

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COMPARING TWO SAMPLES

• Unpaired t-test For LARGE SAMPLES where n1 and
  n2 30 the 95 C.I. for the differences in means
  of two populations is
• ( x1 x2) 1.96vs21/n1 s22/n2
• For SMALL SAMPLES (n1 n2 lt 30) the C.I. is
• ( x1 x2) tvspv1/n1 1/n2
• Paired t-test The C.I. is d tv sd/vn
• The 95 C.I. for p is
• p 1.96 vp(1-p)/n
• Interval for the difference p1-p2
• (p1-p2) 1.96 v p1(1-p1)/n1 p2(1-p2)/n2