# Sampling Distributions of Proportions

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Sampling Distributions of Proportions. 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, it’s the distribution of all possible sample proportions (p). We will use:
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Sampling Distributions of ProportionsSampling 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, it’s the distribution of all possible sample proportions (p)
• We will use:p for the population proportionand p-hat for the sample proportionSuppose we have a population of six people: Alice, Ben, Charles, Denise, Edward, & FrankWhat is the proportion of females?What is the parameter of interest in this population?Draw samples of two from this population.How many different samples are possible?1/3Proportion of females6C2 =15Alice & Ben .5 Alice & Charles .5 Alice & Denise 1 Alice & Edward .5 Alice & Frank .5 Ben & Charles 0 Ben & Denise .5Ben & Edward 0Ben & Frank 0Charles & Denise .5Charles & Edward 0Charles & Frank 0Denise & Edward .5Denise & Frank .5Edward & Frank 0Find the 15 different samples that are possible & find the sample proportion of the number of females in each sample.How does the mean of the sampling distribution (mp-hat) compare to the population parameter (p)?mp-hat = pFind the mean & standard deviation of all p-hats.Formulas:These are found on the formula chart!Correction factor – multiply byDoes the standard deviation of the sampling distribution equal the equation?NO -So – in order to calculator the standard deviation of the sampling distribution, we MUST be sure that our sample size is less than 10% of the population!WHY?We are sampling more than 10% of our population! If we use the correction factor, we will see that we are correct.Assumptions (Rules of Thumb)
• Sample size must be less than 10% of the population (independence)
• Sample size must be large enough to ensure a normal approximation can be used.
• np > 10 & n (1 – p) > 10Why does the second assumption insure an approximate normal distribution?Remember back to binomial distributionsSuppose n = 10 & p = 0.1(probability of a success), a histogram of this distribution is strongly skewed right!np > 10 & n(1-p) > 10 insures that the sample size is large enough to have a normal approximation!Now use n = 100 & p = 0.1 (Now np > 10!) While the histogram is 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 will not make payments on time. The bank recently approved 200 loans.What are the mean and standard deviation of the proportion of clients in this group who may not make payments on time?Are assumptions met?What is the probability that over 10% of these clients will not make payments on time?Yes – np = 200(.07) = 14n(1 - p) = 200(.93) = 186Ncdf(.10, 1E99, .07, .01804) = .0482Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen?np = 200(.5) = 100 & n(1-p) = 200(.5) = 100Since both > 10, I can use a normal curve!Find m & s using the formulas.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 BHS wear contacts. In a sample of 100 students, what is the probability that more than 35% of them wear contacts?Check assumptions!mp-hat = .3 & sp-hat = .045826np = 100(.3) = 30 & n(1-p) =100(.7) = 70Ncdf(.35, 1E99, .3, .045826) = .1376
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