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    Sampling Distribution-2: Unit-5

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    PAVAN YADAV
    The document discusses estimating the difference between two binomial parameters (p1 and p2) by taking random samples from two binomial populations. It states that the difference between the sample proportions (p1^ - p2^) will be used as the point estimate for p1 - p2. As the sample sizes (n1 and n2) increase, p1^ - p2^ will be approximately normally distributed with mean p1 - p2 and variance that takes into account the sample sizes and population proportions. The document provides an example problem calculating the probability that the average drying time of a paint sample (XA) is more than 1 hour greater than the average drying time of another paint sample (XB).

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    Sampling Distribution-2: Unit-5

    Uploaded by

    PAVAN YADAV
    17 views51 pages
    The document discusses estimating the difference between two binomial parameters (p1 and p2) by taking random samples from two binomial populations. It states that the difference between the sample proportions (p1^ - p2^) will be used as the point estimate for p1 - p2. As the sample sizes (n1 and n2) increase, p1^ - p2^ will be approximately normally distributed with mean p1 - p2 and variance that takes into account the sample sizes and population proportions. The document provides an example problem calculating the probability that the average drying time of a paint sample (XA) is more than 1 hour greater than the average drying time of another paint sample (XB).

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    PROBABILITY AND STATISTICS

    Original Title

    Unit5.2

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    The document discusses estimating the difference between two binomial parameters (p1 and p2) by taking random samples from two binomial populations. It states that the difference between the sample proportions (p1^ - p2^) will be used as the point estimate for p1 - p2. As the sample sizes (n1 and n2) increase, p1^ - p2^ will be approximately normally distributed with mean p1 - p2 and variance that takes into account the sample sizes and population proportions. The document provides an example problem calculating the probability that the average drying time of a paint sample (XA) is more than 1 hour greater than the average drying time of another paint sample (XB).

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    17 views51 pages

    Sampling Distribution-2: Unit-5

    Uploaded by

    PAVAN YADAV
    The document discusses estimating the difference between two binomial parameters (p1 and p2) by taking random samples from two binomial populations. It states that the difference between the sample proportions (p1^ - p2^) will be used as the point estimate for p1 - p2. As the sample sizes (n1 and n2) increase, p1^ - p2^ will be approximately normally distributed with mean p1 - p2 and variance that takes into account the sample sizes and population proportions. The document provides an example problem calculating the probability that the average drying time of a paint sample (XA) is more than 1 hour greater than the average drying time of another paint sample (XB).

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 51

    Sampling Distribution-2

    Unit-5
    Marginal error
    If the unknown proportion p is not expected to be too close to 0 or 1, we can
    establish a confidence interval for p by considering the sampling distribution of
    P^. Designating a failure in each binomial trial by the value 0 and a success by
    the value 1, the number of successes, x, can be interpreted as the sum of n values
    consisting only of 0 and 1s, and ˆp is just the sample mean of these n values. Hence,
    by the Central Limit Theorem, for n sufficiently large, P^ is approximately normally
    distributed with mean
    Marginal error:
    Consider the problem where we wish to estimate the difference between two binomial
    parameters p1 and p2. For example, p1 might be the proportion of smokers with lung
    cancer and p2 the proportion of nonsmokers with lung cancer, and the problem is to
    estimate the difference between these two proportions. First, we select independent
    random samples of sizes n1 and n2 from the two binomial populations with means n1p1
    and n2p2 and variances n1p1q1 and n2p2q2, respectively;
    then we determine the numbers x1 and x2 of people in each sample with lung cancer
    and form the proportions ˆp1 = x1/n and ˆp2 = x2/n. A point estimator of the
    difference between the two proportions, p1 − p2, is given by the statistic ^P1 − ^ P2.
    Therefore, the difference of the sample proportions, ˆp1 − ˆp2, will be used as the
    point estimate of p1 − p2.

    Choosing independent samples from the two populations ensures that the variables
    ^P1 and ^ P2 will be independent, and then by the reproductive property of the
    normal distribution established in Theorem 7.11, we conclude that ^ P1 − ^P2 is
    approximately normally distributed with mean
    Paint Drying Time: Two independent experiments are run in which two different types of
    paint are compared. Eighteen specimens are painted using type A, and the drying time, in
    hours, is recorded for each. The same is done with type B. The population standard
    deviations are both known to be 1.0.
    Assuming that the mean drying time is equal for the two types of paint, find
    P( ￣ XA− ￣ XB > 1.0), where ￣ XA and ￣ XB are average drying times for samples
    of size nA = nB = 18.
    The television picture tubes of manufacturer A have a mean lifetime of 6.5 years and a
    standard deviation of 0.9 year, while those of manufacturer B have a mean lifetime of 6.0 years
    and a standard deviation of 0.8 year. What is the probability that a random sample of 36 tubes
    from manufacturer A will have a mean lifetime that is at least 1 year more than the mean
    lifetime of a sample of 49 tubes from manufacturer B?

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