A sample of 900 members has a mean 3.4 cm and standard deviation 2.61 cm. Test whether the sample is from a large population of mean 3.25 cm and standard deviation 2.61 cm. If the
population is normal and its mean is unknown, find the 95% confidence interval for population
mean.

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2015-10-09T22:27:00+05:30

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N_s = sample size = 900.         μ_s = mean of the sample = 3.4 cm
σ_s = Standard deviation of the sample = 2.61 cm

Population mean μ₀ =3.25
 Population standard deviation = σ₀ = 2.61 cm

student's t = (μ_s - μ₀) / [σ_s / √N_s ] 
            t  = (3.4 - 3.25) * √900 / 2.61 = 1.7241

Find the probability that     -1.7241 <= t <= 1.7241     from the students t distribution table or from a website that calculates these. Here the degrees of freedom (d.f) are 899 = N_s - 1.   Sample size is 900.   

   The probability we get is 0.915.  Hence,  we can say that the sample belongs to the population with a probability 0.915  or  a  confidence level of 91.5%.
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The population is normal.  we don't  know its mean. We know sample size, mean and standard deviation.  We are given confidence level 95% = 0.95

  Normal distribution variable:
          Z = (μ_s - μ₀) / (σ_s/√N_s)
                  Here σ_s = 2.61 cm,  N_s = 900

  We consult the Standard Normal Distribution tables.  We find z such that
          P(0 <=  Z  <=  z ) =  0.95/2 = 0.475,
  or  from the cumulative standard distribution table we find z such that :
             P(-∞ <  Z <=  z ) = 0.5 + 0.95/2 = 0.975

  Either way, we get the value z of variable Z for which the probability
      P( -z <=  Z <=  z) = 95% = 0.95  is  z = 1.96 

            -1.96 <=   Z  <=  1.96
    -1.96 * σ_s <=  √N_s * (μ_s - μ₀)  <= 1.96 * σ_s
        - 0.1705   <= (μ_s - μ₀)  <=  0.1705 cm

As,  μ_s = 3.4 cm :
           3.4 - 0.1705  <= μ₀ <=  3.4+0.1705  cm
         ie.,  range [3.23 cm ,  3.57 ]

That will be the range of the population mean that we will estimate from the available sample data.

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