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IAI Actuarial Core Principles · Actuarial Statistics · Hypothesis testing and goodness of fit

Paired differences (after minus before) in monthly premium collection efficiency for 9 branches have mean 1.5 and sample standard deviation 3.0. Assuming normal differences, a two-sided 5% test of H0: mean difference = 0 is carried out. The critical value from t with 8 degrees of freedom is 2.306. Which conclusion is correct?

The statistic is 1.5 and H0 is not rejected. The standard error is 3 divided by root 9, which equals 1, so t is 1.5, below the two-sided 5% critical value of 2.306 on 8 degrees of freedom.

  1. At = 0.50, do not reject H0
  2. Bt = 1.50, do not reject H0Correct
  3. Ct = 1.50, reject H0
  4. Dt = 4.50, reject H0
  5. t = 0.50, reject H0

Explanation

Standard error = 3/sqrt(9) = 1. So t = 1.5/1 = 1.5, which is less than 2.306 in absolute value, so H0 is not rejected. Option 0.50 divides by s alone... 1.5/3 = 0.5 is the wrong standard error, and 4.5 multiplies by 3.

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