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Chapter Seven Introduction to Sampling Distributions Section 3 Sampling Distributions for Proportions 1 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 2 Sampling Distributions for Proportions Allow us to work with the proportion of successes rather than the actual number of successes in binomial experiments. 3 Sampling Distribution ofr the ˆ p Proportion n • • • • 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 r ˆ p is read " p - hat" n 4 Sampling Distribution ofr the pˆ Proportion n If np > 5 and nq > 5 then p-hat = r/n can be approximated by a normal random variable (x) with: pˆ p and p̂ pq n 5 The Standard Error for p̂ The standard deviation of the p̂ sampling distributi on p̂ pq n 6 Continuity Correction • When using the normal distribution (which is continuous) to approximate phat, 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. 7 Continuity Correction If n = 20, convert a p- • Since n = 20, hat interval from 5/8 .5/n = 0.025 to 6/8 to a normal interval. • 5/8 - 0.025 = 0.6 • 6/8 + 0.025 = 0.775 Note: 5/8 = 0.625 6/8 = 0.75 • Required x interval is So p-hat interval is 0.625 0.6 to 0.775 to 0.75. 8 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? 9 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. 10 Find the mean and the standard deviation pˆ p 0.12 pˆ pq .12(.88) 0.023 n 200 11 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. 12 Calculate z-score for x = 0.0975 0.0975 0.12 z 0.98 0.023 13 Calculate z-score for x = 0.1525 0.1525 0.12 z 1.41 0.023 14 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. 15 Control Chart for Proportions P-Chart 16 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. 17 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. 18 Finding the mean and the standard deviation Suppose pˆ p 0.12 then pˆ pq .12(.88) 0.046 n 50 19 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. 20 Control Limits Control limits are placed at two and three standard deviations above and below the mean. pq 0.12(0.88) p2 0.12 2 0.12 0.092 n 50 pq 0.12(0.88) p3 0.12 3 0.12 0.138 n 50 21 Control Limits The center line is at 0.12. Control limits are placed at -0.018, 0.028, 0.212, and 0.258. 22 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 23 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 24 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. 25 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 x x z We use the z formula: • A Calculator is used to compute • The result rounds to z= -3.16 x 26 27 28 29 30 31 32 • Statistics are like bikinis. What they reveal is suggestive, but what they conceal is vital. ~Aaron Levenstein 33 34