Transcript proportions
Sampling Distributions of Proportions • Toss a penny 20 times and record Thenumber dotplot isof a partial graph of the the heads. sampling distribution of all sample • Calculate the of heads proportions of proportion sample size 20. If I founditall possible sample & mark onthethe dot plot on the proportions – this would be board. approximately normal! 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: Melissa, Jake, Charles, Kelly, Mike, & Brian What is the proportion of females? 1/3 What is the parameter of interest in this population? Proportion of females 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: The mean of the sampling distribution. X pˆ n m pˆ p The standard deviation of the sampling distribution. p1 p pˆ n 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) • Use this formula for standard deviation when the population is sufficiently large, at least 10 times as large as the sample. • Sample size must be large enough to insure a normal approximation can be used. We can use the normal approximation when np > 10 & n (1 – p) > 10 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 npstrongly > 10 & skewed n(1-p) > 10 is right! insures that the sample Now useisnlarge = 100enough & p = 0.1 size to (Now np > 10!) While the histogram is have a normal still strongly skewed right – look approximation! 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? P( pˆ that .1) over .0482 What is the probability 10% of these clients will not make payments on 1- (Normcdf(-∞, 0.1, 0.07, 0.01804) OR: Ncdf(.10, 1E99, .07, .01804) = time? .0482 Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen? = 100 & n(1-p) 200(.5) = 100 . 5 (. 5 ) np = 200(.5) = .0118 ncdf ,.42,.5, Since both a normal curve! > 10, I can use 200 m & using the formulas. Find No – since there is approximately a 1% chance of this happening, I do not believe the student did this. Example Example #1 #3 Assume that 30% of the students at HH wear contacts. In a sample of 100 students, what is the probability that more than 35% of mp-hat = .3 & p-hat = .045826 them wear contacts? np = 100(.3) = 30 & n(1-p) =100(.7) = 70 P(p-hat > .35)= Ncdf(.35, 1E99, .3, .045826)assumptions! = .1376 Check Example #1 STATE: We want to know the probability that a random sample yields a result within 2 percentage points of the true proportion. We want to determine P (.33 pˆ .37) Example #1 PLAN: We have drawn an SRS of size 1500 from the population of interest. The mean of the sampling distribution of p-hat is 0.35: mpˆ 0.35 Example #1 PLAN: We can assume that the population of first-year college students is over 15,000, and are safe to use the standard deviation formula: p(1 p ) (0.35)(0.65) pˆ 0.0123 n 1500 In order to use a normal approximation for the sampling distribution, the expected number of successes and failures must be sufficiently large: np 10 and n(1 p) 10 1500(.35) 10 and 1500(.65) 10 Therefore, pˆ N (0.35,0.0123) Example #1 DO: Perform a normal distribution calculation to find the desired probability: P (.33 pˆ .37) .8961 Example #1 CONCLUDE: About 90% of all SRS’s of size 1500 will give a result within 2 percentage points of true proportion.