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IAI Actuarial Core Principles · Actuarial Statistics · Random sampling and sampling distributions

A sample of n = 16 is drawn from a normal population with unknown variance. The sample mean is 105 and the sample standard deviation (divisor n-1) is 12. The test statistic for H0: mu = 100 has what value, and what is its null distribution?

The statistic is 1.67 and it follows a t distribution with 15 degrees of freedom. It is (105-100) divided by 12/4 = 3, and the t distribution is used because the variance is estimated from a sample of 16.

  1. A1.67, standard normal
  2. B0.42, t with 15 degrees of freedom
  3. C1.67, t with 16 degrees of freedom
  4. D1.67, t with 15 degrees of freedomCorrect
  5. 6.67, t with 15 degrees of freedom

Explanation

t = (105-100)/(12/sqrt16) = 5/3 = 1.67. The population variance is unknown and the population is normal, so t has n-1 = 15 degrees of freedom. Using 12 without dividing by sqrt(n) gives 0.42, and the standard normal ignores estimation of sigma.

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