STAT 110 Semester One

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STAT 110Semester One

• HYPOTHESIS TESTING

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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

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HYPOTHESIS TESTING

• In most scientific studies we set up hypotheses
  about treatments before collecting data.
• These are the focus of the study.
• A null hypothesis (H0) is a claim about a
  treatment which is assumed to be true unless the
  data collected show substantial evidence against
  it.
• The alternative hypothesis (HA) is the one which
  is adopted if there is sufficient evidence
  against H0.

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ALTERNATIVE HYPOTHESES

• Two types
• 1. Study based implies we do not know at the
  beginning of the study about the effect of a new
  treatment.
• Leads to a two sided test (two parameter values
  are not equal to each other).
• 2. Data based suggested by the collected data
  (usually suggests treatment benefit).
• Leads to a one sided test (one parameter value
  is greater than or less than the other).

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FOUR STEPS

• 1. Set up the null hypothesis (H0) about the
  population parameter of interest
• e.g. parameter current (null) value.
• 2. Propose the alternative hypothesis (HA)
• e.g. parameter ? current value.
• 3. Calculate the test statistic.
• 4. Calculate the p-value (probability of
  observing the test statistic from 3).

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STEP 3 THE TEST STATISTIC

• This is the standardised value of the sample
  parameter i.e. a z-score or t-score
• Test stat. obs. sample value null value
• est. std. error
• i.e. the no. of std. dev.s from the null value
  to the sample value.

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STEP 4 THE P-VALUE

• This is the prob. of obs. the value of the test
  statistic, or a value more extreme, calculated
  under the assumption that H0 is true.
• We draw appropriate conclusions if the test
  statistic is less than 0.05 - if the p-value is
  less than 0.05 we have significance at the 5
  level and if less than 0.01 we have significance
  at the 1 level.
• If s is unknown the p-value is found from the
  t-table.

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EXAMPLES

• From notes
• Mean Pg 198-9 (incl. notes pg 200).
• Proportion Pg 201-2.
• Comparing means Pg 203-5.
• Comparing proportions Pg 206-7.

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EXAMPLE ONE

• Suppose the resting pulse rates for young women
  are normally distributed with mean m 66 and
  std. dev. s 9.2 beats per minute. A drug for
  the treatment of a medical condition is
  administered to 100 young women and their average
  pulse rate is found to be 68 beats per minute.
  Because the drug had for a long time been
  observed to increase pulse rates, test the claim
  that the drug does in fact increase the pulse
  rates (i.e. HA is data based).

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• STEP ONE
• H0 m 66 (the null hypothesis)
• STEP TWO
• HA m gt 66 (the research hypothesis)

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• STEP THREE
• Test stat. obs. sample value null value
• est. std. error (s/vn)
• 68 66 2.174
• 9.2/v100
• STEP FOUR (diagram p199)
• Pr(Z gt 2.174) 0.5 0.4850
• 0.0150

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CONCLUSION

• Assuming the null hypothesis (H0) is true, there
  is a very small probability (0.015) of observing
  a sample mean this large or larger.
• We have observed a rare event.
• Hence little support for H0.
• Reject H0.
• Conclude the mean pulse rate has been increased
  by the treatment.

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NOTES

• Can get p-value directly from SPSS. If p-value lt
  0.05 we have significance at the 5 level (and if
  p-value lt 0.01 we have significance at the 1
  level).
• If we use s instead of s, then use t-table with
  appropriate df, v.
• e.g. if calculating Pr(t gt 2.174) then the
  p-value is between p 0.025 and 0.010.

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EXAMPLE TWO

• In a large overseas city it was estimated that
  15 of girls between the ages of 14 18 became
  pregnant. Parents and health workers introduced
  an educational programme to lower this
  percentage. After 4 years of the programme, a
  random sample of n 293 18-year-old girls showed
  that 27 had become pregnant.
• Define null and alternative hypotheses for
  investigating whether the proportion becoming
  pregnant after the educational programme has
  decreased. (Suppose HA is one sided).
• Calculate the probability value.
• State your conclusion.

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• STEP ONE
• H0 p 0.15
• STEP TWO
• HA p lt 0.15

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• STEP THREE
• Test stat. obs. sample value null value
• est. std. error vp(1- p)/n
• 0.092 0.15 -2.78
• v0.15x0.85/293
• STEP FOUR (diagram p202)
• Pr(Z lt -2.78) 0.5 0.4973
• 0.0027

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• CONCLUSION
• Very small amount of support for H0 (p-value is
  less than 0.05).
• The observation is a rare event.
• Reject H0 and accept HA.
• There is evidence that after the education
  campaign the proportion becoming pregnant has
  reduced.

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EXAMPLE THREE

• The birthweight of a baby is thought to be
  associated with the smoking habits of the mother
  during pregnancy. The means and variances of the
  individual values in the two samples of
  birthweights, one for non-smoking and the other
  for smoking mothers, are below.
• non-smoker smoker
• Sample size 100 50
• Sample mean 3.45 3.30
• Sample variance 0.36 0.32
• Investigate the claim that the mean birthweights
  are different in the two groups (assume HA is
  study driven).

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• STEP ONE
• H0 mNS - mS 0
• or H0 mNS mS
• STEP TWO
• HA mNS - mS ? 0
• or HA mNS ? mS

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• STEP THREE
• Test stat. obs. sample value null value
• est. std. error (spv1/n11/n2)
• (3.45 3.30) 0 1.47
• v 0.3468v1/1001/50
• s2p 99(0.36) 49(0.32)
• 148
• 0.3468 (use of pooling optional)

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• STEP THREE
• Test stat. obs. sample value null value
• est. std. error (vs21/n1s22/n2)
• (3.45 3.30) 0 1.5
• v0.36/1000.32/50

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• STEP FOUR
• Pr(Z gt1.47) 2(0.5 0.4292) two sided
• 2 x 0.0708
• 0.1416
• Can use z distribution (instead of t with 148 df)
  as the sample is large.

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• CONCLUSION
• Moderate amount of support for H0 (p-value is
  greater than 0.05).
• The observation is not a rare event.
• Do not reject H0.
• There is no evidence the non-smoking group has a
  different mean birthweight than the smoking group.

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EXAMPLE FOUR

• A random sample of 240 women who gave birth in
  Wellington in 2003 revealed that 96 had epidural
  anaesthesia while a similar study in Christchurch
  showed 68 women in a sample of 176 had epidural
  anaesthesia. Conduct a formal hypothesis test
  for a difference in proportions. Assume study
  based, two sided alternative.

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• STEP ONE
• H0 p1 - p2 0
• or H0 p1 p2
• STEP TWO
• HA p1 - p2 ? 0
• or HA p1 ? p2

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Pooled Sample Proportion

• Since the two proportions are assumed to be the
  same (due to null hypothesis) we can pool the two
  proportion estimates to get the pooled sample
  proportion
• p X1 X2 90 68 0.394
• n1 n2 240 176

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Estimated Standard Error

• Using this pooled sample proportion, the
  estimated standard error becomes
• vp(1-p)(1/n11/n2)
• v0.394 x (1 - 0.394) x (1/240 1/176)
• 0.0485

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• STEP THREE
• Test stat. obs. sample value null value
• est. std. error
• (0.4 0.386) 0
• 0.0485
• 0.29

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• STEP FOUR (diagram p207)
• Pr(Z gt 0.29) 2(0.5 0.1141) two sided
• 0.7718
• There is no evidence of a difference as the p
  value is much greater than 0.05.

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SUMMARY

• 1. Set up the null hypothesis (H0) about the
  population parameter of interest
• e.g. parameter current (null) value.
• 2. Propose the alternative hypothesis (HA)
• e.g. parameter ? current value.
• 3. Calculate the test statistic.
• Test stat. obs. sample value null value
• appropriate std. error
• 4. Calculate the p-value (probability of
  observing the test statistic from 3, or a value
  more extreme).