Question

The administrators for a local hospital wish to estimate the average number of days required for in-patient between the ages of 25 and 34. A random sample of 40 patients between these ages produced a sample mean and a sample standard equal to 5.4 and 3.1 days, respectively.

a) What is the best point estimate for the mean length of stay for the population of patients from which the sample was drawn?

b) Calculate the estimated standard error of the sample mean? Interpret this value.

c) For a 95% confidence level, calculate the margin of error. Interpret this value.

d) Construct a 95% confidence interval for the mean length of stay for the population of patients from which the sample was drawn. Interpret this value.

e) Would it be reasonable to conclude that average number of days required for in-patient treatment for this age group in less than 4 days? EXPLAIN why or why not.

Answer 1

a)

best point estimate for mean length of stay = 5.4

b)

sample std dev ,    s =    3.1000
Sample Size ,   n =    40
Standard Error , SE = s/√n =   3.1000   / √    40   =   0.4902
it means sample mean deviates from actual mean of population from 0.4902

c)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   39          
't value='   tα/2=   2.0227   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.1000   / √   40   =   0.4902
margin of error , E=t*SE =   2.0227   *   0.49015   =   0.9914

d)

confidence interval is                   
Interval Lower Limit = x̅ - E =    5.40   -   0.991428   =   4.4086
Interval Upper Limit = x̅ + E =    5.40   -   0.991428   =   6.3914
95%   confidence interval is (   4.41   < µ <   6.39   )

The mean length of stay for the population of patients from which the sample was drawn will lie between (   4.41   < µ <   6.39   )

e)

confidance interval DOES NOT contain 4 so it would not be reasonable to conclude that average number of days required for in-patient treatment for this age group in less than 4 days