A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point estimate for th

Question

A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point estimate for the population standard deviation. Round your answer to three decimal places. 1.48,1.45,1.54,1.52,1.52

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Evelyn 3 weeks 2021-09-22T01:50:05+00:00 2 Answers 0

Answers ( )

    0
    2021-09-22T01:51:25+00:00

    Answer:

    s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132

    And for the deviation we have:

     s= \sqrt{0.00132}=0.0363

    And that value represent the best estimator for the population deviation since:

     E(s) =\sigma

    Step-by-step explanation:

    For this case we have the following data:

    1.48,1.45,1.54,1.52,1.52

    The first step for this cae is find the sample mean with the following formula:

    \bar X =\frac{\sum_{i=1}^n X_}{n}

    And replacing we got:

     \bar X= \frac{1.48+1.45+1.54+1.52+1.52}{5} = 1.502

    And now we can calculate the sample variance with the following formula:

     s^2 =\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

    And replacing we got:

    s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132

    And for the deviation we have:

     s= \sqrt{0.00132}=0.0363

    And that value represent the best estimator for the population deviation since:

     E(s) =\sigma

    0
    2021-09-22T01:51:46+00:00

    Answer:

    Point estimate for the population standard deviation = 0.036

    Step-by-step explanation:

    We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters;

          X                             X – X bar                           (X-Xbar)^{2}    

        1.48                   1.48 – 1.502 = -0.022                 0.000484

        1.45                   1.45 – 1.502 = -0.052                 0.002704

        1.54                   1.54 – 1.502 = 0.038                  0.001444

        1.52                   1.52 – 1.502 = 0.018                  0.000324

        1.52                   1.52 – 1.502 = 0.018                 0.000324      

                                                                               Total = 0.00528    

    Firstly, mean of the data above, X bar = \frac{\sum X}{n}  = \frac{7.51}{5} = 1.502

    Standard deviation of data, S.D. = \sqrt{\frac{\sum (X-Xbar)^{2} }{n-1} }

                                                         = \sqrt{\frac{0.00528}{5-1} } = 0.036

    Therefore, point estimate for the population standard deviation is 0.036 .

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