STAT 110 Lecture 15

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STAT 110Lecture 15

• CONFIDENCE INTERVAL FOR THE MEAN

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SUMMARY SAMPLE MEANS

• sX sX/vn is called the std. error of the mean.
• mX mX
• If X is normal, then X is normal for any n.
• If X is not normal, X is approx. normal for large
  n (central limit theorem).
• For random samples of size n, the sample means
  fluctuate around the pop. mean mX with std. error
  sX.
• As n increases, the distribution fluctuates less
  and approaches normality.

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CONFIDENCE INTERVAL FOR MEAN

• We need to use sample data to estimate the
  unknown population mean.
• Rather than give a single value estimate we
  calculate an interval in which we are fairly
  certain that the mean mX lies.
• This form of estimation reflects the random
  variation in the collected data.
• We use the dist. of the sample means, X.
• This is N(mX,s2X) or N(mX,s2X/n)

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CONSTRUCTING THE INTERVAL

• See diagram at bottom of pg 151.
• From this we can write
• 0.95 Pr(-1.96 lt Z lt 1.96)

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CONSTRUCTING THE INTERVAL

• See diagram at bottom of pg 151.
• From this we can write
• 0.95 Pr(-1.96 lt Z lt 1.96)
• Pr(-1.96 lt X - mX lt 1.96)
• sX/vn

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CONSTRUCTING THE INTERVAL

• See diagram at bottom of pg 151.
• From this we can write
• 0.95 Pr(-1.96 lt Z lt 1.96)
• Pr(-1.96 lt (X-mX)/sX/vn lt 1.96)
• Pr(-1.96sXlt X-mX lt 1.96 sX)
• vn
  vn

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CONSTRUCTING THE INTERVAL

• See diagram at bottom of pg 151.
• From this we can write
• 0.95 Pr(-1.96 lt Z lt 1.96)
• Pr(-1.96 lt (X-mX)/sX/vn lt 1.96)
• Pr(-1.96sX/vn lt X-mX lt 1.96 sX/vn)
• Pr(mX - 1.96sX lt X lt mX 1.96 sX)
• vn vn

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CONSTRUCTING THE INTERVAL

• See diagram at bottom of pg 151.
• From this we can write
• 0.95 Pr(-1.96 lt Z lt 1.96)
• Pr(-1.96 lt (X-mX)/sX/vn lt 1.96)
• Pr(-1.96sX/vn lt X-mX lt 1.96 sX/vn)
• Pr(mX-1.96sX/vn ltXltmX1.96 sX/vn)
• or Pr(X - 1.96sX lt mX lt X 1.96sX)
• vn vn

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CONFIDENCE INTERVAL

• We are 95 confident that the unknown population
  mean mX satisfies
• X - 1.96sX/vn lt mXlt X 1.96 sX/vn

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CONFIDENCE INTERVAL

• We are 95 confident that the unknown population
  mean mX satisfies
• x - 1.96sX lt mX lt x 1.96sX
• vn vn
• Or x 1.96 sX
• vn

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CONFIDENCE INTERVAL NOTES

• We now have an interval estimate for the
  population mean.
• A 99 C.I. replaces 1.96 with 2.58
• (use table to find Z-1(0.99/2)).
• The 99 C.I. is wider (less precise).
• As n increases the std. error of the sample mean
  (sX/vn) gets smaller and the C.I. is narrower
  (more precise)
• i.e. better estimate with larger n.

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EXAMPLE (pg 155)

• A pharmacologist is investigating the length of
  time that a sedative is effective. Eight
  patients are selected at random for a study and
  the eight times for which the sedative is
  effective have mean, x 8.4 hours. It is known
  that the standard deviation, sx 1.5 hours.
• Find 95 and 99 confidence intervals for the
  true mean number of hours mx .

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95 CONFIDENCE INTERVAL

• The 95 confidence interval is
• x 1.96 sX/vn
• 8.4 1.96 x 1.5/v8
• 8.4 1.04
• (7.36, 9.44)
• Alternatively we write
• 7.36 lt mX lt 9.44

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99 CONFIDENCE INTERVAL

• The 99 confidence interval is
• x 2.58 sX/vn
• 8.4 2.58 x 1.5/v8
• 8.4 1.37
• (7.03, 9.77)
• Alternatively we write
• 7.03 lt mX lt 9.77

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• The 99 confidence interval is
• x 2.58 sX/vn
• 8.4 2.58 x 1.5/v8
• 8.4 1.37
• (7.03, 9.77)
• Alternatively we write
• 7.03 lt mX lt 9.77
• Note that this interval is wider.

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EXAMPLE (pg 156)

• The pharmacologist is required to find the value
  of mx to within 15 minutes with 95 confidence.
• Assuming that the standard deviation, sx 1.5
  hours, find the size of the sample which must be
  taken in order to achieve this accuracy.

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SOLUTION

• Consider the confidence interval
• x 1.96 sX/vn
• The error component of this interval is
• 1.96 sX/vn
• We need this error to be within 15 minutes or
  0.25 hours.

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WORKING

• 1.96 sX/vn lt 0.25
• 1.96 x 1.5/vn lt 0.25
• 1.96 x 1.5 lt 0.25 vn
• 1.96 x 1.5 lt vn
• 0.25
• 11.76 lt vn
• 11.762 lt n
• n gt 138.2976 , smallest sample size 139

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THE STUDENTS t DISTRIBUTION

• Often the true std. dev. sX is not known.
• We estimate with the sample std. dev. sX.
• Use the t-table instead of the normal table.
• This gives larger values than 1.96 and 2.58 and
  hence wider, less precise C.I.s.
• The 95 C.I. becomes
• x tv sX/vn
• where v n 1 (degrees of freedom).

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EXAMPLE (pg 158)

• Now suppose that the pharmacologist did not know
  the value of sX and was forced to take the sample
  standard deviation from the sample of size n 8
  as the best estimate of sX, namely sX 1.5
  hours.
• Find 95 and 99 confidence intervals for mX.

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• The 95 confidence interval is
• x t7 sX/vn
• 8.4 2.365 x 1.5/v8
• 8.4 1.86
• (7.15, 9.65)
• We write
• 7.15 lt mX lt 9.65

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• The 99 confidence interval is
• x t7 sX/vn
• 8.4 3.500 x 1.5/v8
• 8.4 1.86
• (6.54, 10.26)
• We write
• 6.54 lt mX lt 10.26

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• The 99 confidence interval is
• x t7 sX/vn
• 8.4 3.500 x 1.5/v8
• 8.4 1.86
• (6.54, 10.26)
• We write
• 6.54 lt mX lt 10.26
• Both are wider than before.

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NOTES

• Be careful to use the correct column for 95
  use 2p0.05 (combined area of two tails and for
  99 use 2p0.01.
• The interval is wide when samples are small (less
  precise estimate).
• We usually need the t dist. as the pop. std. dev.
  is not known.
• Even for large n use the t-table (note last line
  gives 1.96 and 2.58 as before).

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EXAMPLE (pg 161)

• Tablets must be produced which weigh 200mg.
  Choose a sample of n 20 from the production
  line. The sample mean is 201.7 mg and the sample
  s.d., sX is 5.13 mg.
• Does this sample confirm that mX 200mg?

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• The 95 confidence interval is
• x t19 sX/vn
• 201.7 2.093 x 5.13/v20
• 201.7 2.4
• 199.3 lt mX lt 204.1

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• The 95 confidence interval is
• x t19 sX/vn
• 201.7 2.093 x 5.13/v20
• 201.7 2.4
• 199.3 lt mX lt 204.1
• The weight of 200 mg lies in this interval, hence
  200mg is an acceptable value of the mean mX with
  95 confidence.

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INTERPRETING A C.I.

• See notes pg 162-163.
• If 100 different samples are used to construct
  100 intervals, then 95 of these intervals will
  contain the population mean.
• Conversely, 5 of these intervals will miss the
  population mean.
• For 99 C.I.s only 1 out of 100 will miss.
• We say we are 95/99 confident the true mean
  lies in this interval.

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SUMMARY C.I. FOR MEAN

• Calculate an interval in which we are fairly
  certain that the mean mX lies.
• 95 C.I. is x 1.96 sX/vn
• A 99 C.I. replaces 1.96 with 2.58 (wider, less
  precise interval).
• With larger n, sX/vn is smaller and the C.I. is
  narrower (more precise) i.e. better estimate with
  larger n.
• The true std. dev. sX is usually not known, use
  sample std. dev. sX.
• Use t-table -gt larger values than 1.96 and 2.58
  and hence wider, less precise C.I.s.
• The 95 C.I. becomes x tn-1 sX/vn