Question

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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 Ho at the 0.015 significance level Reject Ho at the 0.025 significance level, but not at the 0.015 level Reject Ho at the 0.050 significance level, but not at the 0.025 level Reject Ho at the 0.100 significance level, but not at the 0.050 level Do not reject Ho at the 0.100 significance level

Answer 1

Let $\mu$ denotes the true mean time.

To test null hypothesis Ho -2.6 against alternative hypothesis H1 2.6 Let denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, 4.500, o2= 2.0000, o = 1.41, n = 1 The test statistic can be written as Vn (2.6) Z = which under Ho follows a standard normal distribution We reject Ho at 0.10 level of significance if P-value 0.100 or if |ZobsZ0.0500 Now, The value of the test statistic 1.343503 1.34 The critical value =1.644854 P value P(|Z > zobs)= 2 * P(Z> 1.343503) 2 * (1-0.910445) 0.179109 0.1791 Since p-value > 0.100 and zobsZcritical 1.644854, so We fail to reject Ho at 0.10 level of significance Hence, we can conclude that the population mean is not different from 2.6

ans-> Do not reject H₀ at 0.100 significance level.