Question

1. The mean time between claims for high-risk policyholders is supposedly 2.6. A researcher believes this mean time is incorrect and conducts a hypothesis test. He observed one high-risk policyholder, which resulted in a time of 4.5. Assume these times are normally distributed with variance 2. What is the result of the test? and show the computation process. • Reject H, at the 0.015 significance level • Reject H, at the 0.025 significance level, but not at the 0.015 level • Reject H, at the 0.050 significance level, but not at the 0.025 level • Reject H, at the 0.100 significance level, but not at the 0.050 level . Do not reject H, at the 0.100 significance level

Multiple choice. Choose one of the bullets.

Answer 1

Given that population mean y = 2.6. Population standard deviation = V2 The sample mean is I = 4.5 The null hypothesis is Ho:u= 2.6 The Alternative hypothesis is H:46 2.6(two tailed test) The test statistic: Z-*-u 4.5-2.6 = 1.34 From excel function the value of p is obtained for two tailed tests is as follows: p=2x[1 - NORMSDIST(1.34)] =2 x[1-0.9099) = 2x[0.090123] = 0.180245 Since the value of p is greater than 0.100, the correct option is ‘Do not reject H at the 0.100 significance level.