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Sampling Distributions of Proportions

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. We will use: p for the population proportion and p-hat for the sample proportion.

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Sampling Distributions of Proportions

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  1. Sampling Distributions of Proportions

  2. Sampling Distribution • Is the distribution of possible values of a statistic from all possible samples of the same size from the same population We will use: p for the population proportion and p-hat for the sample proportion

  3. Suppose we have a population of six people: Alice, Ben, Charles, Denise, Edward, & Frank What 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? 1/3 Proportion of females 6C2 =15

  4. Alice & Ben .5 Alice & Charles .5 Alice & Denise 1 Alice & Edward .5 Alice & Frank .5 Ben & Charles 0 Ben & Denise .5 Ben & Edward 0 Ben & Frank 0 Charles & Denise .5 Charles & Edward 0 Charles & Frank 0 Denise & Edward .5 Denise & Frank .5 Edward & Frank 0 Find 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 = p Find the mean & standard deviation of all p-hats.

  5. Formulas: These are found on the formula chart!

  6. 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

  7. 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) = 14 n(1 - p) = 200(.93) = 186 Ncdf(.10, 1E99, .07, .01804) = .0482

  8. Assume that 30% of the students at ESH 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 = .045826 np = 100(.3) = 30 & n(1-p) =100(.7) = 70 Ncdf(.35, 1E99, .3, .045826) = .1376

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