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Chapter Seven Introduction to Sampling Distributions Section 3

Chapter Seven Introduction to Sampling Distributions Section 3 Sampling Distributions for Proportions. Key Points 7.3. Compute the mean and standard deviation for the proportion p hat = r/n Use the normal approximation to compute probabilities for proportions p hat = r/n

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Chapter Seven Introduction to Sampling Distributions Section 3

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  1. Chapter Seven Introduction to Sampling Distributions Section 3 Sampling Distributions for Proportions

  2. Key Points 7.3 • Compute the mean and standard deviation for the proportion p hat = r/n • Use the normal approximation to compute probabilities for proportions p hat = r/n • Construct P-charts and interpret what they tell you

  3. Sampling Distributions for Proportions Allow us to work with the proportion of successes rather than the actual number of successes in binomial experiments.

  4. Sampling Distribution of the Proportion • n= number of binomial trials • r = number of successes • p = probability of success on each trial • q = 1 - p = probability of failure on each trial

  5. Sampling Distribution of the Proportion If np > 5 and nq > 5 then p-hat = r/n can be approximated by a normal random variable (x) with:

  6. The Standard Error for

  7. Continuity Correction • When using the normal distribution (which is continuous) to approximate p-hat, a discrete distribution, always use the continuity correction. • Add or subtract 0.5/n to the endpoints of a (discrete) p-hat interval to convert it to a (continuous) normal interval.

  8. If n = 20, convert a p-hat interval from 5/8 to 6/8 to a normal interval. Note: 5/8 = 0.625 6/8 = 0.75 So p-hat interval is 0.625 to 0.75. Since n = 20, .5/n = 0.025 5/8 - 0.025 = 0.6 6/8 + 0.025 = 0.775 Required x interval is 0.6 to 0.775 Continuity Correction

  9. Suppose 12% of the population is in favor of a new park. • Two hundred citizen are surveyed. • What is the probability that between 10 % and 15% of them will be in favor of the new park?

  10. Is it appropriate to the normal distribution? • 12% of the population is in favor of a new park. p = 0.12, q= 0.88 • Two hundred citizen are surveyed. n = 200 • Both np and nq are greater than five.

  11. Find the mean and the standard deviation

  12. What is the probability that between 10 % and 15%of them will be in favor of the new park? • Use the continuity correction • Since n = 200, .5/n = .0025 • The interval for p-hat (0.10 to 0.15) converts to 0.0975 to 0.1525.

  13. Calculate z-score for x = 0.0975

  14. Calculate z-score for x = 0.1525

  15. P(-0.98 < z < 1.41) 0.9207 -- 0.1635 = 0.7572 There is about a 75.7% chance that between 10% and 15% of the citizens surveyed will be in favor of the park.

  16. Control Chart for Proportions P-Chart

  17. Constructing a P-Chart • Select samples of fixed size n at regular intervals. • Count the number of successes r from the n trials. • Use the normal approximation for r/n to plot control limits. • Interpret results.

  18. Determining Control Limits for a P-Chart • Suppose employee absences are to be plotted. • In a daily sample of 50 employees, the number of employees absent is recorded. • p/n for each day = number absent/50.For the random variable p-hat = p/n, we can find the mean and the standard deviation.

  19. Finding the mean and the standard deviation

  20. Is it appropriate to use the normal distribution? • The mean of p-hat = p = 0.12 • The value of n = 50. • The value of q = 1 - p = 0.88. • Both np and nq are greater than five. • The normal distribution will be a good approximation of the p-hat distribution.

  21. Control Limits Control limits are placed at two and three standard deviations above and below the mean.

  22. Control Limits The center line is at 0.12. Control limits are placed at -0.018, 0.028, 0.212, and 0.258.

  23. Control Chart for Proportions Employee Absences 0.3 +3s = 0.258 0.2 +2s = 0.212 0.1 mean = 0.12 0.0 -2s = 0.028 -0.1 -3s = -0.018

  24. Daily absences can now be plotted and evaluated. Employee Absences 0.3 +3s = 0.258 0.2 +2s = 0.212 0.1 mean = 0.12 0.0 -2s = 0.028 -0.1 -3s = -0.018

  25. Calculator – Chapter 7 • In this chapter use the TI-83 or TI-84 Plus graphing calculator to do any computations with formulas from the chapter. For example, computing the z score corresponding to a raw score from an x bar distribution.

  26. Calculator – Chapter 7 • Example: If a random sample of size 40 is taken from a distribution with mean = 10 and standard deviation = 2, find the z score corresponding to x=9 • We use the z formula: • A Calculator is used to compute • The result rounds to z= -3.16

  27. Statistics are like bikinis.  What they reveal is suggestive, but what they conceal is vital.  ~Aaron Levenstein

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