Introduction to Formal Inference

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Chapter 6

• Introduction to Formal Inference
• Part III One- and Two- Sample Inference for Means

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6.3 Inference for Means

• Recap
• When we have large sample size (large-n) we can
  compute (1-a)100 confidence intervals as
• if s is known
• -OR- if s is unknown
• This is because of the CLT, for large enough n,

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6.3 Inference for Means

• What happens when we dont have large enough
  sample size?
• If s known, generally use the same interval we
  did for large-n
• If s unknown, then can we still say
• ?
• Unfortunately, NO you CANT!!

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6.3 Inference for Means

• ???

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6.3 Inference for Means

• Note Formulas (presented in this class) for
  confidence intervals based on small samples are
  valid ONLY for
• iid random variables
• That follow a Normal distribution
• In other wordswe still must assume that the
  random variables are normal BUT we cant use a
  normal distribution to compute the confidence
  interval because the sample mean is no longer
  normal

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6.3 Inference for Means

• Fact If X1, , Xn are iid N(µ, s2) then the
  random variable
• follows a student t-distribution with n-1
  degrees of freedom
• Note A t-distribution with 8 degrees of freedom
  corresponds to the standard normal distribution.
  Just like the normal distribution, we will use a
  table to obtain its quantiles (Table B.4)

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6.3 Inference for Means

• t-distribution

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6.3 Inference for Means

• t-distribution depends on the degrees of freedom
  (often labeled ?) each row of the table gives
  values associated with t? for a given ? and the
  columns give the quantile, Q(p)
• Recall Q(p) P(T? lt Q(p)) p
• Example if ?7, then PT7 1.895 0.95, where
  T7 has the t-distribution with 7 degrees of
  freedom
• t-distribution is symmetric so probabilities in
  one tail are the same as in the other tail and we
  can use the same tricks as with the normal
  distribution
• Example for the t7 distribution, we can also say
  
• PT7 -1.895 PT7 gt 1.895 1 PT7 lt
  1.895 0.05
• or
• PT7 gt 1.895 1 - PT7 lt 1.895 0.05
• so 1.895 t7,a/2 when a.10

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6.3 Inference for Means

• T-table
• This table gives probability of being less than t
  in terms of quantiles..ie PTv lt t
• So to look up ta/2 up we will need to use the
  fact that
• a 1-p for any of the quantiles Q(p)
• The table contains only positive values to find
  probabilities associated with a negative value,
  we will need to convert it to a positive value
  (like on previous example)

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6.3 Inference for Means

• When constructing two-sided confidence intervals
  we will be interested in finding t?, a/2, where
  t?,a/2 is such that PT gt t?,a/2 a /2
• It may be easier to think in terms of 1- a /2
• PT gt t?,a/2 1 - PT gt t?,1-a/2 PT lt
  t?,1-a/2
• Example, if ? 15 and a .05, then we find a
  /2.025. So, we want t15, .975 and look in row 15
  under column Q(.975) giving t15, .975
  2.131.
• For one-sided intervals, we use t?,1-a and thus
  we dont divide the value of a in half.

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6.3 Inference for Means

• Use Table B.4 to answer the following questions
• What value would you use if you have a sample of
  10 values and want to construct a two-sided 95
  CI?
• t9,0.025 (-t9,0.975 ) 2.262
• What if you have a sample of 20 values?
• t19,0.025 (-t19,0.975) 2.093
• A sample of 40 values?
• t39,0.025 (-t39,0.975 ) 2.022

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6.3 Inference for Means

• Using Table B.4 to get p-values
• Recall a p-value is the probability of being as
  extreme or more extreme than the value we got
  from our observed data (i.e. p-value is a
  probability)
• The t-table only has a few probabilities on
  itQ(.9), Q(.95), Q(.975), Q(.99), Q(.995),
  Q(.999), Q(.9995)
• These correspond to the a values 0.10, 0.05,
  0.025, 0.01, 0.005, 0.001, 0.0005
• P-values for the t-distribution will be in the
  form of inequalities, i.e. 0.05 lt p-value lt 0.10

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6.3 Inference for Means

• P-values Find p-values for the following test
  statistics given the associated sample sizes
• t 2.25 and n 10 so we know ? 9, and
• 1.833 lt 2.25 lt 2.262 Q(.95) lt 2.25 lt Q(.975)
• (IT FLIPS!) .025 lt p-value lt .05
• t 1.15 and n 20 so we know ? 19
• 1.15 lt 1.328 1.15 lt Q(.9)
• (IT FLIPS!) p-value gt .1
• t 3.15 and n 35 so we know ? 34 (use 30 or
  40)
• Using 30 2.750 lt 3.15 lt 3.385 Q(.995) lt 2.25
  lt Q(.999)
• (IT FLIPS!) .001 lt p-value lt .005

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6.3 Inference for Means

• P-values Notes
• P-values for the t-distribution will always be in
  the form of an inequality
• If you have a two-sided hypothesis, you still
  need to multiply by 2
• The interval you find in terms of quantiles (ie.
  Q(.95) to Q(.975)) will need to flip when you put
  it in terms of probabilities (ie. .025 to .05)
  since p 1-a
• The table goes backwards across the top so
  bigger p-values for smaller test-statistics and
  vice versa

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6.3 Inference for Means

• Small-n Two-Sided CI for µ with Unknown s2
• If X1, , Xn are iid N(µ, s2) and n is small, a
  (1-a)100 CI for µ is
• Small-n Test Statistics for Testing H0 µ µ0
• If X1, , Xn are iid N(µ, s2) and n is small, a
  (1-a)100 CI for µ is

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6.3 Inference for Means

• Example 4 An engineer who works for Consumer
  Reports is interested in the performance of
  various cars. Suppose one way to measure
  reliability is to determine the total cost of
  maintenance/repairs for a particular make and
  model during the first four years of operation.
  The engineer randomly selects 20 people that have
  a particular make and model and determines the
  average cost of maintenance/repairs is 2,300 and
  the standard deviation is 400.
• Give a 99 CI for the true average cost
• Test the claim that the true average cost has
  changed from 2,200 using a .01.

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6.3 Inference for Means

• A 99 CI for the true average cost
• a .01 so a/2 .005
• n 20, not large enough so we must use the
• t-distribution with 19 df
• t.005,19 2.861
• We are 99 confident that the true average cost
  of maintenance/repairs is between 2,044.10 and
  2,555.90.

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6.3 Inference for Means

• Test the claim that the true average cost has
  changed from 2,200.
• Step 1 H0 µ 2,200 vs HA µ ? 2,200
• Step 2 n 20 so we have
• where t t19
• Step 3 p-value 2PT19 gt 1.118
• 2(p-value gt .10) p-value gt .20
• Step 4 p-value .20 gt a .01 so we FTR H0
• Step 5 There is not enough sufficient evidence
  to conclude that the average cost is not 2,200.

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6.3 Inference for Means

• Example A special type of hybrid corn was
  planted on eight different plots. The plots
  produced yield values (in bushels) of 140, 70,
  39, 110, 134, 104, 100 and 125. Assume the
  yields follow a normal distribution.
• Note the sample mean is 102.75 and the sample
  variance is 1151.071.
• a) Give a 95 CI for the true average yield of
  this type of hybrid corn.
• b) Test H0 100 vs Ha ? 100 and give the
  p-value.

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6.3 Inference for Means

• Give a 95 CI for the true average yield of this
  type of hybrid corn
• I am 95 confident that the true average yield
  of this type of hybrid corn is between 74.381
  bushels and 131.119 bushels.

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6.3 Inference for Means

• Test H0 100 vs Ha ? 100 and give the
  p-value.
• Step 1 H0 µ 100 vs HA µ ? 100
• Step 2 n 8 so we have
• where t t7
• Step 3 p-value 2PT7 gt 0.22926
• 2(p-value gt .10) p-value gt .20
• Step 4 p-value .20 gt a .05 so we FTR H0
• Step 5 There is not enough sufficient evidence
  to conclude that the average yield is not 100
  bushels

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6.3 Inference for Means

• Example Suppose the weights of 10 newly minted
  U.S. pennies were recorded. The table below
  contains the data.
• Note the sample mean is 3.00, the sample variance
  is 0.000175

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6.3 Inference for Means

• Assuming the weights follow a normal
  distribution, give a 90 CI for the true average
  weight of newly minted penny.
• We are 90 confident that the true average
  weight of a newly minted penny is between 2.9919g
  and 3.0081g.

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6.3 Inference for Means

• Now suppose the weights of 40 newly minted U.S.
  pennies were measured. The table below contains
  the data.

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6.3 Inference for Means

• Give a 90 CI for the true average weight of a
  newly minted penny.
• We are 90 confident that the true average
  weight of a newly minted penny is between 2.9949
  g and 3.0051 g.

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6.3 Comparison of Two Means

• Up until now, we have only considered a single
  population at a time
• Often we want to compare(consider the difference
  between) two population means i.e. µ1 µ2
• E.g. suppose the engineer of Consumer Reports
  wants to compare average repair costs between the
  Honda Civic and Toyota Corolla
• Must consider two cases independent populations
  and dependent populations

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6.3.2 Comparison of Two Means (Dependent)

• If two populations are dependent, then we cant
  consider them individually
• Called Paired Data
• Before and After Studies
• Studies on Twins (twin A and twin B) or couples
• Two measurements on one item
• Combine observations by looking at the
  differences within pairs of observations and do
  analysis on the average difference,
• Uses the same methodologies (large/small sample
  sizes) as before with

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6.3.2 Comparison of Two Means (Dependent)

• A group of engineering students took resistance
  measurements on n 5 different resistors using
  two different resistance meters

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6.3.2 Comparison of Two Means (Dependent)

• Want to compare mean of resistance of meter 1 to
  that of meter 2
• Data are clearly dependent (since each
  measurement is on the same resistor) paired
  data
• Take the difference between the two measurements
  for each object, this is our random quantity of
  interest
• Now measurements on different subjects are
  independent (since resistors are different) so we
  can proceed as we have before

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6.3.2 Comparison of Two Means (Dependent)
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6.3.2 Comparison of Two Means (Dependent)

• n lt 25 so we have to use small-sample methods
• What is a 90 CI for the mean difference in
  resistance measurement for meter 2 and meter 1?

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6.3.2 Comparison of Two Means (Dependent)

• How do we interpret this?
• We are 90 confident that the true mean
  difference in resistance measurement for meter 2
  and meter 1 is between 11.878 and 12.922.
• We estimate that the 2nd meter tends to read
  between 11.88 and 12.92 units higher than the 1st
  meter, on average.

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6.3.2 Comparison of Two Means (Dependent)

• Note If n is large, we proceed with our large
  sample methodology. i.e. we appeal to the CLT
  to get that the distribution of is
  approximately normal with mean µd and variance
  sd2. Using this, we can build confidence
  intervals and test hypotheses in the same fashion
  as for a single mean,

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6.3.3 Comparison of Two Means (Independent)

• If your samples are independent, we can consider
  each population separately and compare the means
  of the two groups
• What could we use to estimate µ1 µ2?
• Note that
• What is the variance of our estimator?

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6.3.3 Comparison of Two Means (Independent)

• Large-Sample Two-Sided CI for Difference of Means
  From Two INDEPENDENT Populations (n1 25 and n2
  25)
• If X11, , Xn1 are iid with mean µ1 and variance
  s21 and
• X12, , Xn2 are iid with mean µ2 and variance
  s22
• then a (1-a)100 CI for µ1 µ2 is

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6.3.3 Comparison of Two Means (Independent)

• Large-Sample Test Statistic for Testing
• H0 µ1-µ2 with Two INDEPENDENT Populations
  (n1 25 and n2 25)
• If X11, , Xn1 are iid with mean µ1 and variance
  s12 and
• X12, , Xn2 are iid with mean µ2 and variance
  s22
• then

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6.3.3 Comparison of Two Means (Independent)

• Similar to the previous large sample formulas
  since both populations are assumed to be large
• In each formula, s1 and s2 can be replaced with
  s1 and s2 when the population standard deviations
  are unknown

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6.3.3 Comparison of Two Means (Independent)

• Example Two brands of golf balls, Brand A and
  Brand B, are to be compared with respect to
  driving distance. Suppose we randomly sample 25
  balls from each brand and test them using an
  automatic driving device known to give normally
  distributed distances with a standard deviation
  of 15 yards. The mean distance for golf ball A
  is 300 yards and the mean distance for golf ball
  B is 320 yards.

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6.3.3 Comparison of Two Means (Independent)

• Construct a two-sided 99 CI for the difference
  in mean driving distances for the two types of
  golf balls.

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6.3.3 Comparison of Two Means (Independent)

• How do we interpret this interval?
• We are 99 confident that that true difference
  in mean driving distances in golf ball B and golf
  ball A is between 9.07 and 30.93 yards.
• -or-
• We are 99 confident that, on average, golf ball
  B travels between 9.07 and 30.93 yards farther
  than golf ball A.

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6.3.3 Comparison of Two Means (Independent)

• Given a 1 level of significance, (i.e. a.01),
  do the two brands of golf balls differ?
• Note this is equivalent to asking is µB µA
  0? (should look like a hypothesis test!)
• If we test H0 µB µA 0 vs HA µB µA ? 0,
  since 0 is not inside the 99 CI from a), 0 is
  not a plausible value for the difference in means
  
• p-value lt a0.01
• we would reject H0 and conclude that
  the mean driving distances differ for brands A
  and B (i.e. the mean is not 0)

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6.3.3 Comparison of Two Means (Independent)

• Example Two varieties of apples, Granny Smith
  and Macintosh, were analyzed for their potassium
  content in milligrams. A sample of 100 Granny
  Smith apples resulted in a mean of 0.30 mg and a
  sd of 0.07 mg. A sample of 150 Macintosh apples
  resulted in a mean of 0.27 mg and a sd of 0.05
  mg. Is there clear evidence to conclude that one
  variety has more potassium than another variety?
  If so, which variety contains more potassium?

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6.3.3 Comparison of Two Means (Independent)

• Step 1 H0 µG - µM 0 vs HA µG - µM ? 0
• Step 2
• Step 3 p-value 2PZlt -3.702 0

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6.3.3 Comparison of Two Means (Independent)

• Step 4 Since p-value 0 lt a 0.05 (since no
  significance level was specified), then we Reject
  H0
• Step 5 At the a 0.05 level of significance,
  there is significant evidence to conclude that
  µG-µM ? 0 i.e. there is significant evidence to
  conclude that there is a difference in variety.
• Technically, we didnt do the proper test to
  determine which one has more potassiumbut since
  Z gt 0, we could probably conclude the Granny
  Smith have more potassium content.

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Similar to one sample methodology, for small n,
  two sample methodology involving at least one
  small sample relies on the assumption of
  normality and results in simply changing our
  reference distribution from a Normal to a
  t-distribution
• Again, formulas differ depending on whether or
  not the samples are independent are dependent

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Small-Sample Two-Sided CI for Difference of Means
  From Two Independent Populations (n1 lt 25 or
  n2 lt 25)
• If X11, , Xn1 are iid N(µ1, s21) and
• X12, , Xn2 are iid N(µ2, s22)
• then a (1-a)100 CI for µ1 µ2 is

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Notes
• Df (n1 1) (n2 1) n1 n2 2
• We now use a pooled variance
• Since both samples are small, we dont want an
  over-inflated variance to be used. This pooled
  variance like a weighted average of the variances

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Small-Sample Test Statistic for Testing
• H0 µ1-µ2 with Two Independent Populations
  (n1 lt 25 or n2 lt 25)
• If X11, , Xn1 are iid N(µ1, s21) and
• X12, , Xn2 are iid N(µ2, s22)
• then

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Note Your textbook give two formulas for the
  degrees of freedomOne formula is used when we
  can assume (which we will generally
  do). The other formula is called the
  Satterthwaite approximation will not be covered
  in this class

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Example Two different types of brake lining were
  tested for differences in wear. Twelve cars were
  used, six for each type of brake lining. A
  sample of each brand was tested with the results
  (listed in hundreds of miles) given in the
  following table. Assuming the populations are
  independent and normally distributed with equal
  variances, test for evidence that true average
  brake lining wear is better for brand A than
  brand B. Use a .10 level of significance.

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Brand A 42 58 64 40 47 50
• Brand B 48 40 30 44 54 38
• Step 1 H0 µA µB 0 vs HA µA µB gt 0
• Step 2 nA lt 25 and nB lt 25

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Step 2 (contd)
• Step 3 p-value Q(.9) lt 1.5361 lt Q(.95) so
• .05 lt p-value lt .10

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6.3.4 Small-sample Comparison of Two Means
(Independent)

• Step 4 since .05 lt p-value lt .10 is greater than
  a.1, we will Reject H0
• Step 5 There is significant evidence to conclude
  at an a .10 level of significance that the true
  average brake lining wear is greater in brand A
  than in brand B.

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1 Sample or 2?
2 sample
1 sample
Independent or Dependent?
See next slide
independent
dependent
Is n large?
Is n large?
no
no
Find d, sd and µd

yes
yes

Find d, sd and µd
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1 Sample or 2?
1 sample
2 sample
See previous slide
Use the following table
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6.6 Prediction Intervals

• Recall C.I.s are meant to bracket µ, the
  population mean.
• and
• What if we wanted to try to predict one more
  individual observed value? We currently have
  observed n, we want to predict n1.
• Again, we will want to use an interval for
  uncertainty

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6.6 Prediction Intervals

• Prediction Intervals from a Normal distribution
• Assume we are sampling from a normal distribution
• and S are based on a sample of size n and
  Xn1 is a single additional observation not yet
  in the sample
• For X1, , Xn, Xn1 iid N(µ, s2),

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6.6 Prediction Intervals

• These facts give us the following formula for a
  two-sided (1-a)100 CI for xn1 is
• Note that the sample mean and sample standard
  deviation come from the first n observations only

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6.6 Prediction Intervals

• Example (contd) Suppose we randomly chose 1 more
  car that is the same make and model of the
  previous 20 that we used to infer the average
  cost of maintenance.
• What is the 95 prediction interval for the cost
  of maintenance of a single additional randomly
  chosen car of this make and model?

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6.6 Prediction Intervals

• Recall
• so a 95 P.I. is

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6.6 Prediction Intervals

• Note
• This goes to show that prediction intervals have
  increased width by a factor of over
  CIs.
• This shouldnt be too surprising since there is
  more uncertainty involved in predicting the next
  value than in estimating the population mean.

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6.6 Prediction Intervals

• How do we interpret this?
• To say (a, b) is a (1-a)100 prediction
  interval for a single additional observation is
  to say that in obtaining this interval I have
  used a method of sample selection and calculation
  that would work about (1- a)100 of repeated
  applications.