Math151 Day29

Math 151 , Day 29, Wednesday, April 14, 2004 hit reload...After class

HW Day29 ReRead sec. 4.3?. (Skip ch. 5) New: Sec. 6.1 Read it: Read it again.
Memorize the tan box on p. 242 (mean and s.d. of sampling dist. of X-bar)

Hand in: the homework from Day28 (section 4.3) Corrected!
Sec.6.1 :Read the problems to see where we're going!
p.303, 6.1 poll of women
6.2 95% confidence?
6.3 density of x-bar, and confidence intervals This problem combines the pictures 6.2 and 6.4 For part d, to draw the confidence interval: just choose any point on the horizontal axis of your graph to be x-bar. Measure off the distance m (half the width of the shaded interval) and extend a bar m wide to the left and the right of your point,below the curve. (Like fig. 6.4, the bars with arrows at the ends. The red dots show what the x-bar is for that confidence interval) Choose another point, and repeat.. If your first x-bar was in the shaded interval, pick your second outside the shaded interval, and vice versa. You should note that if x-bar is in the shaded interval, then the confidence interval bar covers mu (280) and if x-bar isn't, then the bar doesn't.
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Using formula p. 306 for C.I.:
6.6 potassium again.
6.7 comparing CI's for different confidence levels. Also write down the m (margin of error) for each interval.
6.9 comparing CI's for different sample sizes.
6.5 IQ test scores Read pp. 312-13 before doing this one.

Read,
to discuss

Optional

Chapter 6, prep:
SAMPLE from an UNKNOWN population. You should have done this: do it now if you didn't:
Each person take 4 slips from the Birkenstock box,
      find the mean, and your mean + .841.
      Record these for yourself . This is your "Interval Estimate" of the mean of the shoebox population.
          Your "estimate" of the (unknown) population mean µ of the numbers in the shoebox is your sample mean plus or minus the "fudge factor/margin of error" .841.
      Record them also on the sheet going around, and draw the interval on the graph transparency going around.
         If xbar = 8.1       7.259|_____________8.1_____________|8.941
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Review Central Limit Theorem et al (day 27) ,
Homework questions for sec. 4.3
Start here Friday

How big does n have to be for Xbar to have a normal distribution? (about 25 is always good.)

"Fuzzy Central Limit Theorem:"
Data whose variation is due to many small independent random influences will have an approximately normal distribution.
Balls and pins, heights of women, etc.
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #
Chapter 6, Introduction to Inference
Statistical Inference: drawing conclusions about a population from sample data.
Requires: Random sample or Randomized experiment. (Simple Random Sample usually)

First example: Use sample mean xbar to "estimate" (unknown) population mean µ
Mean of 4 grades (HW#4.40) estimates population mean of all 10 ("known"= 69.4) E.g. 69.75, 64.25, 73.5
(Each is a "point estimate")

"xbar IS µ" Never true exactly
"xbar is close to µ" True for most xbars, depending on "close", and sample size n.
"xbar is probably close to µ" ["probably"?? It is or it isn't. We have "confidence" ours is a close one]
"xbars are usually close to µ" True.

If we have only one sample -- one xbar, we have NO IDEA if ours is one of the "close" ones.
Quantify "usually" and "close."

Fall 2002: 33% (16 of 48) xbars recorded were within 1 of µ. (between 68.4 and 70.4).
83% (40 of 48) xbars recorded were within 4 of µ. (between 65.4 and 73.4).
94% (45 of 48) xbars recorded were within 5 of µ. (between 64.4 and 74.4).

Fall 2001: 65% (15 of 23) xbars recorded were within 4 of µ. (between 65.4 and 73.4).

78% (18 of 23) xbars recorded were within 5 of µ. (between 64.4 and 74.4).

Spring 2002: 83% (35 of 42) xbars recorded were within 4 of µ. (between 65.4 and 73.4).

90% (38 of 42) xbars recorded were within 5 of µ. (between 64.4 and 74.4).

(I excluded impossible xbars)

Note tradeoff between "close"(accuracy) and "usually" (confidence)

Interval estimate: xbar + margin of error (fudge factor) estimates population mean µ (69.4)

    69.75 + 1:   "µ is between 65.75 and 73.75" True
    69.75 + 4:   "µ is between 65.75 and 73.75" True
     73.5 + 4:    "µ is between 69.5 and 77.5" False
     73.5 + 5:    "µ is between 68.5 and 78.5" True
      64.25 + 4:   "µ is between 60.25 and 68.25" False
      64.25 + 5:   "µ is between 59.25 and 69.25" False

Confidence interval estimate of a(n unknown) population parameter:

an Interval constructed from the data, +
a Confidence level C: where

by this method

Confidence Interval of the form estimate + margin-of-error for the mean with Confidence level C: (p.306)

the estimate is xbar
margin of error m is : z* times Standard deviation of sample mean

Probability C is between -z* and +z*

(Table A, or Table C, t dist. bottom row)

Standard deviation of sample mean: Sigma /sqrt(n)

know

m = z* (sigma)/ sqrt(n),

CI is xbar + z* (sigma)/ sqrt(n)

Example: Sample of size 9 from a Normal population with unknown mean and pop. s.d. sigma = 6, xbar = 12.
Find a 90% CI estimate for the unknown mean µ: z* = 1.645, (sigma)/ sqrt(n) = 6/3=2, so m = 3.290;
CI is 12 + 3.290, or 8.710 to 15.290.

The Birkenstock box contains numbers from a normally distributed population, with population standard deviation 2.
You each constructed a 60% confidence interval for the unknown mean:
    n = 4.
    Standard deviation of sample mean = 2/sqrt(4) = 2/2 = 1
    z* for C = 60% is .841, so margin of error m is .841 times 1= .841.
How many people captured the true mean?
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