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    COT No. 4sampling Distribution With Replacement

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    edward louie serrano
    Here are the steps to solve this problem: 1. List all possible samples of size 2: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) 2. Calculate the mean of each sample (2,2) - 2 (2,4) - 3 (2,6) - 4 (4,2) - 3 (4,4) - 4 (4,6) - 5 (6,2) - 4 (6,4) - 5 (6,6) - 6

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    © All Rights Reserved
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    COT No. 4sampling Distribution With Replacement

    Uploaded by

    edward louie serrano
    34 views15 pages
    Here are the steps to solve this problem: 1. List all possible samples of size 2: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) 2. Calculate the mean of each sample (2,2) - 2 (2,4) - 3 (2,6) - 4 (4,2) - 3 (4,4) - 4 (4,6) - 5 (6,2) - 4 (6,4) - 5 (6,6) - 6

    Original Description:

    Presentation on Sampling Distribution

    Original Title

    COT-no.-4sampling-distribution-with-replacement

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    Here are the steps to solve this problem: 1. List all possible samples of size 2: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) 2. Calculate the mean of each sample (2,2) - 2 (2,4) - 3 (2,6) - 4 (4,2) - 3 (4,4) - 4 (4,6) - 5 (6,2) - 4 (6,4) - 5 (6,6) - 6

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    34 views15 pages

    COT No. 4sampling Distribution With Replacement

    Uploaded by

    edward louie serrano
    Here are the steps to solve this problem: 1. List all possible samples of size 2: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) 2. Calculate the mean of each sample (2,2) - 2 (2,4) - 3 (2,6) - 4 (4,2) - 3 (4,4) - 4 (4,6) - 5 (6,2) - 4 (6,4) - 5 (6,6) - 6

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 15

    REVIEW

    PROBLEM
    Find the sample mean from the frequency
    distribution table below.
    Scores Frequency
    40-42 2
    43-45 5
    46-48 8
    49-51 12
    52-54 6
    55-57 4
    58-60 3
    SOLUTION
    Scores f Class Mark(x) fx

    40-42 2 41 82

    43-45 5 44 220

    46-48 8 47 376

    49-51 12 50 600

    52-54 6 53 318

    55-57 4 56 224

    58-60 3 59 177

    ∑f = 40 ∑fx = 1,997
    2 3
    5
    “A population consists of three numbers
    (2, 3, 5). Consider all possible samples of
    size 2 which can be drawn with replacement
    from the population. What are the possible
    samples of size 2?”.

    3
    Answer:
    (2, 2), (2, 3), (2, 5)

    (3, 3), (3, 2), (3, 5)

    (5, 5), (5, 2), (5, 3)


    SAMPLING
    DISTRIBUTION
    What is sampling distribution?

    The probability distribution that


    describes the probability for each
    mean of all samples with the same
    sample size “n”.
    Objective:
    COMPUTE THE MEAN,
    VARIANCE, STANDARD
    DEVIATION OF SAMPLING
    DISTRIBUTION OF MEANS
    WITH
    “REPLACEMENT”
    A population consists of three numbers
    (2, 3, 5). Consider all possible samples of size
    2 which can be drawn with replacement from
    the population.
    Compute the following:
    a. List down all the samples of size 2
    b. Mean of each sample and the mean of the sampling
    distributions.
    c. Variance of the sampling distributions of means
    d. Standard deviation of the sampling distribution of
    means
    STEP 1:
    POSSIBLE SAMPLES OF SIZE 2
    (2, 2), (2, 3), (2, 5)

    (3, 3), (3, 2), (3, 5)

    (5, 5), (5, 2), (5, 3)


    STEP 2:
    Observation Sample Mean(x) (x - µ) (x - µx)²

    1 (2, 2) 2 -1.33 1.7689


    2 -0.83
    (2, 3) 2.5 0.6889
    3 (2, 5) 3.5 0.17 0.0289
    4 (3, 3) 3 -0.33 0.1089
    5 -0.83 0.6889
    (3, 2) 2.5
    6 (3, 5) 4 0.67 0.4489
    7 (5, 5) 5 1.67 2.7889
    8 (5, 2) 3.5 0.17 0.0289
    9 (5, 3) 4 0.67 0.4489
    ∑(x - µx)²
    ∑x =30
    = 7.0001
    PAIR ACTIVITY

    A population consists of three numbers (2,


    4,). Consider all possible samples of size 2
    which can be drawn with replacement from
    the population.
    Compute the following:
    a. Mean of the sampling distribution of means
    b. Variance of the sampling distribution of means
    c. Standard deviation of the sampling distribution
    of means
    ANSWER

    Observation Sample Mean x - µx (x - µx)²


    (x)
    1 (2, 2) 2 -1 1
    2 (2, 4) 3 0 0
    3 (4, 4) 4 1 1
    4 (4, 2) 3 0 0
        ∑x= 12   ∑(x - µx)²
    =2
    GROUP ACTIVITY
    A population consists of three numbers (2, 4, 6).
    Consider all possible samples of size 2 which can
    be drawn with replacement from the population.
    a. Mean of the sampling distribution of means
    b. Variance of the sampling distribution of
    means
    c. Standard deviation of the sampling distribution
    of means

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