Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript

Sampling Distributions for Proportions Allow us to work with the proportion of successes rather than the actual number of successes in binomial experiments. • • • • Sampling Distribution of the r Proportion pˆ 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 Sampling Distribution of the Proportion pˆ r 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 The Standard Error for p̂ The standard deviation of the p̂ sampling distributi on p̂ pq n 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. Continuity Correction 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 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? 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. Find the mean and the standard deviation pˆ p 0.12 pˆ pq .12(.88) 0.023 n 200 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. Calculate z-score for x = 0.0975 0.0975 0.12 z 0.98 0.023 Calculate z-score for x = 0.1525 0.1525 0.12 z 1.41 0.023 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. Control Chart for Proportions P-Chart 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. 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. Finding the mean and the standard deviation Suppose pˆ p 0.12 then pˆ pq .12(.88) 0.046 n 50 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. 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 Control Limits The center line is at 0.12. Control limits are placed at -0.018, 0.028, 0.212, and 0.258. 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 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