Today

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Todays lesson

• Confidence intervals for the expected value of a
  random variable.
• Determining the sample size needed to have a
  specified probability of a Type II error and
  probability of a Type I error.

2
Confidence Interval for the Mean of a Normally
Distributed Random Variable

• ASS-U-ME that Y is normally distributed with
  unknown mean µ and known standard deviation (say
  100).
• Confidence interval for the mean is the set of
  reasonable values based on the data observed.

3
Example Problem 1

• The random variable Y is normally distributed
  with unknown mean and standard deviation 100. A
  random sample of 25 observations is taken from Y
  and has mean value 515. What is the 99 percent
  confidence interval for the mean of Y?

4
Solution to Problem 1

• Find the standard error of the mean
• standard deviation of Y/square root of sample
  size100/square root of 2520
• Find the factor for the size of the confidence
  interval
• use 1.960 for 95 percent CI
• use 2.576 for 99 percent CI

5
Solution to Problem 1

• Multiply standard error and factor
• 2.5762051.52
• Add and subtract this product to the mean
• Left end point is mean-51.52463.48
• Right end point is mean51.52566.52
• In real life, round off CI to non-obsessive
  numbers.

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Statement of Solution

• The 99 percent confidence interval for the
  expected value of Y (mean of Y) is the interval
  between 463.48 and 566.52.
• The difference between the left and right end
  points of the CI is a measure of how much
  sampling variability is present in the
  experimental results.

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Example Problem 2

• Test the following two situations and select the
  answer that describes your conclusions
• I. Test H0 E(Y)500, a0.01 against H1 E(Y) not
  equal to 500.
• II. Test H0 E(Y)600, a0.01 against H1 E(Y)
  not equal to 600.

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Example Problem 2 Options

• A) Reject null hypothesis I and reject null
  hypothesis II.
• B) Reject null hypothesis I and accept null
  hypothesis II.
• C) Accept null hypothesis I and reject null
  hypothesis II.
• D) Accept null hypothesis I and accept null
  hypothesis II.

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Solution to Problem 2

• Use the confidence interval calculated in problem
  1, the interval between 463.48 and 566.52.
• Null mean in I is 500, which is in 99 percent CI
  hence accept in I.
• Null mean in II is 600, which is not in 99
  percent CI hence reject in II.
• Answer is C.

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Hints and Reminders

• READ YOUR COMPUTER OUTPUT.
• Use the numbers that the computer calculates.
• MAKE SURE THAT THE PARAMETER IN THE QUESTION AND
  THE PARAMETER YOU ARE CALCULATING ARE THE SAME!

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Example Question 3

• The random variable Y is normally distributed
  with unknown mean and unknown standard deviation.
  A random sample of 4 observations is taken from Y
  The mean is 515, and the unbiased estimate of the
  variance is 8100. What is the 99 percent
  confidence interval for the mean of Y?

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Solution to Problem 3

• Recognize that this problem requires the use of
  Students t (the standard deviation is not
  known).
• Find the estimated standard error of the mean
• square root of the unbiased estimate of the
  variance/square root of sample size90/square
  root of 445

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Solution to Problem 3 Continued

• Find the degrees of freedom for the estimate of
  the unknown standard deviation
• size of sample-14-13.
• Find the factor for the size of the confidence
  interval
• stretch 1.960 for 95 percent CI for 3 df, 3.182
• stretch 2.576 for 99 percent CI for 3 df, 5.841

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Solution to Problem 3 Continued

• Multiply standard error and factor
• 5.84145262.8
• Add and subtract this product to the mean
• Left end point is mean-262.8252.2
• Right end point is mean262.8777.8
• In real life, round off CI to non-obsessive
  numbers.

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How to Use Student t Confidence Interval for Mean

• Exactly the same as the use of the normal
  confidence interval for the mean.

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Example Problem 4

• Test the following two situations and select the
  answer that describes your conclusions
• I. Test H0 E(Y)500, a0.01 against H1 E(Y) not
  equal to 500.
• II. Test H0 E(Y)600, a0.01 against H1 E(Y)
  not equal to 600.

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Example Problem 4 Options

• A) Reject null hypothesis I and reject null
  hypothesis II.
• B) Reject null hypothesis I and accept null
  hypothesis II.
• C) Accept null hypothesis I and reject null
  hypothesis II.
• D) Accept null hypothesis I and accept null
  hypothesis II.

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Solution to Example Problem 4

• Use the confidence interval calculated in problem
  3, the interval between 252.2 and 777.8.
• Null mean in I is 500, which is in 99 percent CI
  hence accept in I.
• Null mean in II is 600, which is also in 99
  percent CI hence reject in II.
• Answer is D.

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Determining Sample Size

• Design in a statistical study is crucial for
  success.
• Key issue is how large does the sample size have
  to be.
• There is a key formula for determining the sample
  size.

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One-sample test sample size parameters

• ASS-U-ME sampling for Y, a normally distributed
  random variable.
• Null hypothesis values
• E0, expected value of Y under the null
• s0, standard deviation of Y (a SINGLE value drawn
  from Y) under the null
• a, the level of significance
• za, the percentile from the standard normal
  corresponding to a.

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One-sample test sample size parameters (continued)

• Alternative hypothesis values
• E1, expected value of Y under the alternative
• s1, standard deviation of Y (a SINGLE value drawn
  from Y) under the alternative
• ß, the probability of a Type II error.
• zß, the percentile from the standard normal
  corresponding to ß.

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Sample Size Formula

• Use a sample size n that is as large or larger
  than

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Example Problem Scenario

• A research team will test the null hypothesis
  that E(Y)1000 at the 0.05 level of significance
  against the alternative that E(Y)lt1000. When the
  null hypothesis is true, Y has a normal
  distribution with standard deviation 600.

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Standard Warm-up Problem

• What is the standard deviation of the mean of 900
  observations under the null hypothesis?
• Solution
• The standard error is the standard deviation of Y
  under H0 divided by the square root of the sample
  size.
• 600/square root of 900600/3020.

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A Sometime Warm-up Problem

• What is the critical value of the null hypothesis
  in the scenario when using the average of 900
  observations as the test statistic.
• Solution
• E0 sign zastandard error of test statistic
• left sided test, hence use -
• 1000 - 1.64520967.1

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Example Problem 5

• What is the probability of a Type II error for an
  alternative in which Y is normally distributed,
  E(Y)950, and its standard deviation is 600 using
  the average of a random sample of 900 as the test
  statistic of the null hypothesis in the scenario?

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Solution to Example Problem 5

• The probability of a Type II error ß is equal to
  Pr1Accept H0 Pr1Test statisticgtcritical
  value Pr1Sample meangt967.1.
• Under alternative, sample mean is
• normal
• with mean 950
• with standard error 20

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Solution to Example Problem 5 Continued

• The problem now becomes what is the probability
  that a normally distributed random variable with
  mean 950 and standard deviation 20 is larger than
  967.1?
• Find standard units value of 967.1.
• (967.1-950)/200.855
• What is PrZgt0.855?
• This is about 0.197.

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Example Problem 6

• What is the smallest sample size so that the
  probability of a Type II error is 0.05 when the
  (alternative) distribution of Y is normally,
  E(Y)950, and its standard deviation is 600. The
  test statistic is the average of a random sample
  of n as the test statistic of the null hypothesis
  in the scenario.

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Solution

• Use formula
• For null,
• E01000, s0600, a0.05, za1.645
• For alternative,
• E1950, s1600, ß0.05, zß1.645

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Solution Continued

• Plug and chug
• square root of sample size required is 39.48
• Required sample size is the square of 39.48,which
  is 1558.6.
• Round up to 1559.
• This is optimistic. How do you account for
  nonresponse?

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Review of Todays lecture

• One sample procedures (today confidence
  intervals).
• Determining Sample Size
• When you solve problem, think about the meaning
  of the answer.