Question

Q1 Ten individuals are chosen at random, from a normal population and their weights (in kg)
are found to be 63, 63, 66, 67, 68, 69, 70, 70, 71 and 71. In the light of this data set, test the
claim that the mean weight in population is 66 kg at 5% level of significance. (10)

Answer 1

sample size = n = 10
mean of the sample = μ_s = 678/10 = 67.8
standard deviation of the sample : s = 2.856

Mean of the population = μ = 66 kg
Significance level = 5%        =>     Probability = 95%

       Null hypothesis is that the population mean is μ = 66 kg.

student's t test  t = (μ_s - μ) / (s/√n)  =>    t = (67.8 - 66) / (2.856/√10)
             t = 1.993

For a sample of 10 individuals, the number of degrees of freedom is 9.  Consult the table of Student's t distribution.  P(-2.262 < t < 2.262) = 95%  .  This is the two sided probability value, one has to read.
     or,     P( | t | > 2.262) = 5%

If the population mean and sample mean are with in :  2.262 * 2.856/√10 = 2.04 kg, then the sample belongs to the population.  Or, the population mean given is in the acceptable range of the sample mean,  with a confidence level of  95%.

So we will accept under 5% level of significance, a population mean in the range, 67.8 +- 2.04 kg, with a confidence of 95%.  ie.,  65.76 kg  to  69.84 kg.  Given value 66 kg is just inside this range.

  Another way, as the computed t value is | 1.993 | < 2.262, the given population mean is possible.  The claim is accepted.