Question

Unit 6    [Hypothesis Testing/Decision Making]       

For all Hypothesis\Decision Making Problems, please do the following:

            1.         Answer: Is the data discrete or continuous?

            2.        List the assumptions.

3.        State the null and alternate hypothesis.

            4          Write down the proper test statistic, Show all calculations, i.e. impute the numbers.                       .               Set up Decision rules. Show the critical value.    Reject or Fail to reject.  

            5.         State your conclusions.  

                      

6.2.3

A random sample of 16 emergency reports was selected from the files of an ambulance

service. The mean time (computed from the sample data) required for ambulances to

reach their destinations was 13 minutes, with a sample standard variance calculated

to be equal to 9. Assume that the population of times is approximately normally distributed. Can we conclude at the .05 level of significance that the population mean

is different than 10 minutes?

Answer 1

Since data is continuous because time (computed from the sample data) required for ambulances to reach their destinations is measurable.

Here we have -13, s -3, n-16 Hypotheses are Level of significance a-0.05 Test is two tailed Degree of freedom: df -n-1-15 Test Statistics: = (13-10)/(3/sqrt(16)) Critical value: 2.131 If It| >2.131, Reject HO Rejection Region: Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value P-value-0.0012 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: TINV(0.05,15) =TD IST(4,15,2)