AGR EDUC 387 Data Analysis in Social Sciences

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AGR EDUC 387Data Analysis in Social Sciences

• Dr. Jamie Cano, Instructor
• Ms. Kristy Brewer, TA
• January 20, 2005

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Examples

• Above a z score of 1.80
• Between the mean and a z score of 0.43
• Below a z score of 3.00
• Between the mean and a z score of 1.65
• Above a z score of 0.60
• Below a z score of 2.65
• Between z score of 0 and 1.96
• Answers on page 498 of textbook

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Problem

• What is the relative standing of a student with a
  score of 640 on the GRE given a normal curve and
  a mean of 500 and a standard deviation of 100?

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Objective

• Determine the proportion of area and score on a
  normal curve.

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Finding Proportions

• Finding proportion BELOW a score (to left of
  mean)
• Sketch a normal curve and shade in the target
  area (score of 640, mean 500, s 100)
• Plan your solution according to the normal table
• Convert from x to z by expressing 640 as

x - ?
640 - 500
Z

1.40
?
100
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Finding Proportions

• Find the target area
• Area below the mean is .5000
• 640 500 1.40
• Find 1.40 from Table A .4192 (Between)
• .5000 .4192 .9192
• Answer represents that 92 of scores on the GRE
  are 640 or less, or the student ranks at the 92nd
  percentile.
• Or, 8 of the scores are above the score of 640

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Finding Proportions

• Find the target area (Option B)
• Find GRE scores greater than 640 (unshaded area)
• Subtract that proportion from 1.0000 which
  represents all GRE scores
• Column A to z score of 1.40, Column C indicates
  .0808
• 1.0000 - .0808 .9192

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Problem

• The gestation period for humans approximates a
  normal curve with a mean of 270 days and a
  standard deviation of 15 days. What proportion
  of gestation periods will be between 245 and 255
  days?

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Finding Proportions

• Finding proportions BETWEEN two scores
• Sketch a normal curve (gestation period between
  245 and 255 days, mean 270, s 15)
• Plan your solution according to the normal table
• Convert x to z by expressing 255 as

255 - 270
Z

-1.00
15
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Finding Proportions

• And by expressing 245 as

245 - 270
Z
-1.67
15
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Finding Proportions

• Find the target area
• Column A for z score of 1.00 and Column C
  .1587
• Column A for z score of 1.67 and Column C
  .0475
• Subtract the smaller proportion from the larger
  proportion to obtain answer
• .1587 - .0475 .1112
• Thus about 11 of all gestations are between 245
  and 255 days

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Problem

• School officials believe that their students
  intellectual aptitudes approximate the normal
  curve with a mean of 105, and a standard
  deviation of 15. What proportion of students
  IQs should be more than 30 points either above or
  below the mean?

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Finding Proportions

• Finding proportions BEYOND PAIRS of scores
• Sketch a normal curve and shade in the two target
  areas (IQ should be more than 30 points above or
  below the mean, mean 105, s 15)
• Plan your solution according to the normal table
• Convert x to z

135 - 105
75 - 105
Z
2.00
-2.00
15
15
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Finding Proportions

• Find the target area
• Column A for z score of 2.00 and Column C .0228
• Column A for z score of 2.00 and Column C
  .0228
• .0228 .0228 .0456
• About 5 of the IQs are 30 points above or below
  the mean

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Problem

• Exam scores for a biology class approximate a
  normal curve with a mean of 230 and a standard
  deviation of 50. Graded on the curve with
  upper 20 being awarded A. What is the lowest
  score on the exam to earn an A?

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Finding Scores

• Finding a score (to the right of the mean)
• Sketch a normal curve, and on the correct side of
  the mean, draw a line representing the target
  score (lowest score to earn an A, mean 230,
  standard deviation 50)
• Plan your solution according to the normal table
• Remember, .5000 is to the left of the mean and
  .3000 is to the right of the mean
• Also, .2000 is to the right of the score being
  sought

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Finding Scores

• Find the z score
• Need to find z score for .3000 only in Column B,
  or .2000 in Column C
• .2000 (.2005) has a z score of .84
• .3000 (.2995) has a z score of .84
• Convert z score to target score using the
  following formula

X ? (z)(?)
X 230 (.84)(50)
X 272
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Problem

• Annual rainfall in San Francisco approximates a
  normal curve with mean of 22 and a standard
  deviation of 4. What are the rainfalls for the
  more atypical years, defined as the driest 2.5
  and the wettest 2.5 of the years?

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Finding Scores

• Finding PAIRS of scores (on both sides of the
  mean)
• Sketch a normal curve. On either side of the
  mean, draw two lines representing the two target
  scores (2.5 driest, 2.5 wettest, mean 22,
  standard deviation 4)

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Finding Scores

• Plan your solution according to the normal table
• 2.5 on both ends indicates that .0250 on left
  tail, and .9750 on the right tail
• Columm B, because it includes all of the left
  side must subtract the .5000
• Target z score can be found by scanning Column B
  or C
• Column B for .4750 or Column C for .0250
• z 1.96

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Finding Scores

• Convert z score to target score

x ? (z)(?)
x 22 (-1.96)(4)
x 22 (1.96)(4)
x 22 7.84
x 22 7.84
x 14.16
x 29.84