How to calculate sample mean and variance given the confidence interval for the normal?











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A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).



What are the values of the sample mean and the variance? (X bar and S)



I got X bar to be 47.3.



I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4










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    up vote
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    A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).



    What are the values of the sample mean and the variance? (X bar and S)



    I got X bar to be 47.3.



    I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).



      What are the values of the sample mean and the variance? (X bar and S)



      I got X bar to be 47.3.



      I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4










      share|cite|improve this question















      A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).



      What are the values of the sample mean and the variance? (X bar and S)



      I got X bar to be 47.3.



      I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4







      statistics normal-distribution standard-deviation variance confidence-interval






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      edited Nov 19 at 1:49

























      asked Nov 19 at 1:15









      kmediate

      115




      115






















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          Yes, you have gotten the mean right, it is the average of the end points.



          $$bar{x} - k S = 44.7$$



          $$bar{x} + k S = 44.7$$



          where $k$ is known to you.



          Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.






          share|cite|improve this answer























          • I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
            – kmediate
            Nov 19 at 1:37











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          1 Answer
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          1 Answer
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          active

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          up vote
          0
          down vote













          Yes, you have gotten the mean right, it is the average of the end points.



          $$bar{x} - k S = 44.7$$



          $$bar{x} + k S = 44.7$$



          where $k$ is known to you.



          Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.






          share|cite|improve this answer























          • I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
            – kmediate
            Nov 19 at 1:37















          up vote
          0
          down vote













          Yes, you have gotten the mean right, it is the average of the end points.



          $$bar{x} - k S = 44.7$$



          $$bar{x} + k S = 44.7$$



          where $k$ is known to you.



          Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.






          share|cite|improve this answer























          • I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
            – kmediate
            Nov 19 at 1:37













          up vote
          0
          down vote










          up vote
          0
          down vote









          Yes, you have gotten the mean right, it is the average of the end points.



          $$bar{x} - k S = 44.7$$



          $$bar{x} + k S = 44.7$$



          where $k$ is known to you.



          Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.






          share|cite|improve this answer














          Yes, you have gotten the mean right, it is the average of the end points.



          $$bar{x} - k S = 44.7$$



          $$bar{x} + k S = 44.7$$



          where $k$ is known to you.



          Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 19 at 1:38

























          answered Nov 19 at 1:20









          Siong Thye Goh

          95.7k1462116




          95.7k1462116












          • I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
            – kmediate
            Nov 19 at 1:37


















          • I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
            – kmediate
            Nov 19 at 1:37
















          I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
          – kmediate
          Nov 19 at 1:37




          I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
          – kmediate
          Nov 19 at 1:37


















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