Reducing the confidence interval with T distribition












1












$begingroup$


My question is the following:



The population: weight of apples (in grams). We do not know anything about the population distribution except that it is a normal distribution.



We want to find a 90% confidence interval for the population mean.



So we take a random sample consists of 9 apples, their weights:



$$199, 198, 200,189, 186, 194, 179, 187, 196$$



Since it is a sample with small size $n=9<30$, and the standard deviation $sigma$ is unknown, we need to estimate the variance, the standard deviation and the mean of the population and use the $t$ (student) distribution to calculate the confidence interval:



Estimator for the population mean is the sample mean:



$$hatmu=overline X=frac{1}{n}sum_{i=1}^{n}x_i=192$$



Estimator for the population variance is the sample variance:



$$hatsigma^2=s^2=frac{1}{n-1}sum_{i=1}^{n}(x_i-overline X)^2=51$$



$$ hatsigma=sqrt {(s^2)}=s=7.141 $$



We want 90% confidence interval:



$$1-alpha=0.9 Rightarrow alpha=0.1$$



Calculation of the confidence interval using T distribution:



$$overline X-frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}leqslant mu leqslant overline X+frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}$$



$$192-frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}leqslant mu leqslant 192+frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}$$



$$ t_{8, frac{0.1}{2}}=1.860 Rightarrow 187.5725 leqslantmuleqslant 196.4274 $$



Now we can say: We are 90% sure that the population mean ($mu$) is in the range of: $[187.5725, 196.4274].$



We can see that the confidence interval has length of $8.8549$.



What is the problem? Confidence interval of $8.8549$ is too large for us, we want to reduce the size of the confidence interval to be at most of size $4$, that is $2$ for each side around the mean, and still keep the confidence level: 90%.



In other words: What will be the sample size if we want to reduce the size of the confidence interval to be at most $4$, with 90% confidence?



In mathematical terms, We should use this formula (to get the new sample size):



$$frac{s}{sqrt{n}}t_{n-1, frac{0.1}{2}}leqslant 2$$



My question is: how can i play with this equation to find the new sample size? i mean: I need to know what $t_{n-1,frac{0.1}{2}}$ is at the time that $n$ is unknown.



It is the first time i work with such problems with $t$ distribution, I solved a lot of problems like that with $z$ distribution (when the sample size $gt 30$ or $sigma$ is known), with $z$ distribution there is no problem to do that because the $z$ value is function of the confidence level we want $(alpha)$ only.



Thanks for reading the question so far, any suggestions?



Thanks!!!






normal-distribution statistical-inference estimation confidence-interval







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    My question is the following:



    The population: weight of apples (in grams). We do not know anything about the population distribution except that it is a normal distribution.



    We want to find a 90% confidence interval for the population mean.



    So we take a random sample consists of 9 apples, their weights:



    $$199, 198, 200,189, 186, 194, 179, 187, 196$$



    Since it is a sample with small size $n=9<30$, and the standard deviation $sigma$ is unknown, we need to estimate the variance, the standard deviation and the mean of the population and use the $t$ (student) distribution to calculate the confidence interval:



    Estimator for the population mean is the sample mean:



    $$hatmu=overline X=frac{1}{n}sum_{i=1}^{n}x_i=192$$



    Estimator for the population variance is the sample variance:



    $$hatsigma^2=s^2=frac{1}{n-1}sum_{i=1}^{n}(x_i-overline X)^2=51$$



    $$ hatsigma=sqrt {(s^2)}=s=7.141 $$



    We want 90% confidence interval:



    $$1-alpha=0.9 Rightarrow alpha=0.1$$



    Calculation of the confidence interval using T distribution:



    $$overline X-frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}leqslant mu leqslant overline X+frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}$$



    $$192-frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}leqslant mu leqslant 192+frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}$$



    $$ t_{8, frac{0.1}{2}}=1.860 Rightarrow 187.5725 leqslantmuleqslant 196.4274 $$



    Now we can say: We are 90% sure that the population mean ($mu$) is in the range of: $[187.5725, 196.4274].$



    We can see that the confidence interval has length of $8.8549$.



    What is the problem? Confidence interval of $8.8549$ is too large for us, we want to reduce the size of the confidence interval to be at most of size $4$, that is $2$ for each side around the mean, and still keep the confidence level: 90%.



    In other words: What will be the sample size if we want to reduce the size of the confidence interval to be at most $4$, with 90% confidence?



    In mathematical terms, We should use this formula (to get the new sample size):



    $$frac{s}{sqrt{n}}t_{n-1, frac{0.1}{2}}leqslant 2$$



    My question is: how can i play with this equation to find the new sample size? i mean: I need to know what $t_{n-1,frac{0.1}{2}}$ is at the time that $n$ is unknown.



    It is the first time i work with such problems with $t$ distribution, I solved a lot of problems like that with $z$ distribution (when the sample size $gt 30$ or $sigma$ is known), with $z$ distribution there is no problem to do that because the $z$ value is function of the confidence level we want $(alpha)$ only.



    Thanks for reading the question so far, any suggestions?



    Thanks!!!






    normal-distribution statistical-inference estimation confidence-interval







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      My question is the following:



      The population: weight of apples (in grams). We do not know anything about the population distribution except that it is a normal distribution.



      We want to find a 90% confidence interval for the population mean.



      So we take a random sample consists of 9 apples, their weights:



      $$199, 198, 200,189, 186, 194, 179, 187, 196$$



      Since it is a sample with small size $n=9<30$, and the standard deviation $sigma$ is unknown, we need to estimate the variance, the standard deviation and the mean of the population and use the $t$ (student) distribution to calculate the confidence interval:



      Estimator for the population mean is the sample mean:



      $$hatmu=overline X=frac{1}{n}sum_{i=1}^{n}x_i=192$$



      Estimator for the population variance is the sample variance:



      $$hatsigma^2=s^2=frac{1}{n-1}sum_{i=1}^{n}(x_i-overline X)^2=51$$



      $$ hatsigma=sqrt {(s^2)}=s=7.141 $$



      We want 90% confidence interval:



      $$1-alpha=0.9 Rightarrow alpha=0.1$$



      Calculation of the confidence interval using T distribution:



      $$overline X-frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}leqslant mu leqslant overline X+frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}$$



      $$192-frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}leqslant mu leqslant 192+frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}$$



      $$ t_{8, frac{0.1}{2}}=1.860 Rightarrow 187.5725 leqslantmuleqslant 196.4274 $$



      Now we can say: We are 90% sure that the population mean ($mu$) is in the range of: $[187.5725, 196.4274].$



      We can see that the confidence interval has length of $8.8549$.



      What is the problem? Confidence interval of $8.8549$ is too large for us, we want to reduce the size of the confidence interval to be at most of size $4$, that is $2$ for each side around the mean, and still keep the confidence level: 90%.



      In other words: What will be the sample size if we want to reduce the size of the confidence interval to be at most $4$, with 90% confidence?



      In mathematical terms, We should use this formula (to get the new sample size):



      $$frac{s}{sqrt{n}}t_{n-1, frac{0.1}{2}}leqslant 2$$



      My question is: how can i play with this equation to find the new sample size? i mean: I need to know what $t_{n-1,frac{0.1}{2}}$ is at the time that $n$ is unknown.



      It is the first time i work with such problems with $t$ distribution, I solved a lot of problems like that with $z$ distribution (when the sample size $gt 30$ or $sigma$ is known), with $z$ distribution there is no problem to do that because the $z$ value is function of the confidence level we want $(alpha)$ only.



      Thanks for reading the question so far, any suggestions?



      Thanks!!!






      normal-distribution statistical-inference estimation confidence-interval







      share|cite|improve this question











      $endgroup$




      My question is the following:



      The population: weight of apples (in grams). We do not know anything about the population distribution except that it is a normal distribution.



      We want to find a 90% confidence interval for the population mean.



      So we take a random sample consists of 9 apples, their weights:



      $$199, 198, 200,189, 186, 194, 179, 187, 196$$



      Since it is a sample with small size $n=9<30$, and the standard deviation $sigma$ is unknown, we need to estimate the variance, the standard deviation and the mean of the population and use the $t$ (student) distribution to calculate the confidence interval:



      Estimator for the population mean is the sample mean:



      $$hatmu=overline X=frac{1}{n}sum_{i=1}^{n}x_i=192$$



      Estimator for the population variance is the sample variance:



      $$hatsigma^2=s^2=frac{1}{n-1}sum_{i=1}^{n}(x_i-overline X)^2=51$$



      $$ hatsigma=sqrt {(s^2)}=s=7.141 $$



      We want 90% confidence interval:



      $$1-alpha=0.9 Rightarrow alpha=0.1$$



      Calculation of the confidence interval using T distribution:



      $$overline X-frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}leqslant mu leqslant overline X+frac{s}{sqrt{n}}t_{n-1, frac{alpha}{2}}$$



      $$192-frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}leqslant mu leqslant 192+frac{7.141}{sqrt{9}}t_{8, frac{0.1}{2}}$$



      $$ t_{8, frac{0.1}{2}}=1.860 Rightarrow 187.5725 leqslantmuleqslant 196.4274 $$



      Now we can say: We are 90% sure that the population mean ($mu$) is in the range of: $[187.5725, 196.4274].$



      We can see that the confidence interval has length of $8.8549$.



      What is the problem? Confidence interval of $8.8549$ is too large for us, we want to reduce the size of the confidence interval to be at most of size $4$, that is $2$ for each side around the mean, and still keep the confidence level: 90%.



      In other words: What will be the sample size if we want to reduce the size of the confidence interval to be at most $4$, with 90% confidence?



      In mathematical terms, We should use this formula (to get the new sample size):



      $$frac{s}{sqrt{n}}t_{n-1, frac{0.1}{2}}leqslant 2$$



      My question is: how can i play with this equation to find the new sample size? i mean: I need to know what $t_{n-1,frac{0.1}{2}}$ is at the time that $n$ is unknown.



      It is the first time i work with such problems with $t$ distribution, I solved a lot of problems like that with $z$ distribution (when the sample size $gt 30$ or $sigma$ is known), with $z$ distribution there is no problem to do that because the $z$ value is function of the confidence level we want $(alpha)$ only.



      Thanks for reading the question so far, any suggestions?



      Thanks!!!






      normal-distribution statistical-inference estimation confidence-interval




      normal-distribution statistical-inference estimation confidence-interval






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 10 '18 at 20:26







      D.Rotnemer

















      asked Dec 8 '18 at 21:37









      D.RotnemerD.Rotnemer

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