Sampling Distributions of Proportions • • The dotplot is a partial graph of the Toss a penny 20 times and record sampling distribution of all sample theproportions number ofof heads. sample size 20. If I found all possible sample Calculate thetheproportion of heads proportions this plot wouldon bethe & mark it on the– dot approximately normal! board. What shape do you think the dot plot will have? Sampling Distribution • Is the distribution of possible values of a statistic from all possible samples of the same size from the same population • In the case of the pennies, the We it’s will use: p for the population distribution of all possible sample proportion proportions (p) and p-hat for the sample proportion Suppose we have a population of six people: Alice, Ben, Charles, Denise, Edward, & Frank What is the proportion of females? 1/3 What is the parameter of interest in this population? gender Draw samples of two from this population. How many different samples are possible? 6C2 =15 Find the 15 different samples that are possible & find the sample proportion of the number of females in each sample. Ben & Frank Alice & Ben .5 Charles & Denise Alice & Charles .5 Alice & Denise 1 Charles & Edward Alice & Edward .5 Charles & Frank the mean of the Alice & Frank How does .5 Denise & Edward (mp-hat) Ben & Charlessampling 0 distribution Denise & Frank Ben & Denise compare .5 to the population Edward & Frank parameter (p)? m = p Ben & Edward 0p-hat 0 .5 0 0 .5 .5 0 Find the mean & standard deviation of all p-hats. μpˆ 1 3 & σ pˆ 0.29814 Formulas: μpˆ p σ pˆ These are found on the formula chart! p 1 p n Does the standard deviation of the sampling distribution equal the equation? NO - σ pˆ 1 2 1 3 So3– in order to 0calculate .29814 the 2 standard 3 deviation of the sampling distribution, we WHY? MUST be sure that our sample Correction factor – multiply size is less than 10% of the by We are sampling more than 10% of our population! If we N n population! use the correction factor, we will see that weNare 1 correct. σ pˆ 1 2 3 3 6 2 0.29814 2 6 1 Assumptions (Rules of Thumb) • Sample size must be less than 10% of the population (independence) • Sample size must be large enough to insure a normal approximation can be used. np > 10 & n (1 – p) > 10 Why does the second assumption insure an approximate normal distribution? Remember back to binomial distributions Suppose n = 10 & p = 0.1 (probability of a success), a histogram of this distribution > 10 &skewed n(1-p) >right! 10 isnp strongly insures that the sample size Now use n = 100 & p = 0.1 (Now is large enough to have a np normal > 10!) While the histogram is approximation! still strongly skewed right – look what happens to the tail! Based on past experience, a bank believes that 7% of the people who receive loans μpˆ .07 will not make payments on time. The bank recently approved 200 .93 .07loans. σ pˆ .01804 Yes – 200 What are the mean and standard deviation np = 200(.07) = 14 of the proportion of clients in this group n(1 - p) = 200(.93) = 186 who may not make payments on time? Are assumptions met? Ncdf(.10, 1E99, .07, .01804) = What is the probability that over 10% of .0482 these clients will not make payments on time? Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen? .5(.=5)200(.58) np = 200(.42) = 84 & n(1-p) = 116 ncdf ,. 42 ,. 5 , . 0118 Sinceboth > 10, I can use a normal curve! 200formulas. Find m & s using the No – since there is approximately a 1% chance of this happening, I do not believe the student did this. Assume that 30% of the students at SHS wear contacts. In a sample of 100 students, what is the probability that more than 35% of them wear contacts? mp-hat = .3 & sp-hat = .045826 Check assumptions! np = 100(.3) = 30 & n(1-p) =100(.7) = 70 Ncdf(.35, 1E99, .3, .045826) = .1376