Question

A sample of 49 general practicians (GPs) in Queensland hospitals were surveyed and it was found that the average time they had spent with a patient was 20.2 minutes. It is known the standard deviation, for all GPs in Queensland that spend time with a patient, is 5.6 minutes. The lower limit of a 99% confidence interval for the population mean time (in minutes) spent with a patient is estimated to be _____________ (2 decimal places)

Answer 1

As per the given information given in the question relating to the Queensland hospital

Sample of GPs (n) = 49

Average time spent with patient (x) = 20.2 minutes

Standard deviation of GPs in spending time with patients (S) = 5.6 minutes

Standard Error (SE) of x = S / sqrt of n

Standard Error (SE) of x = 5.6 / sqrt of 49

Standard Error (SE) of x = 5.6 / 7

Standard Error (SE) of x = 0.8

Though the sample size n = 49 and n > 30

So the critical value for 99% confidence limit (Level of significance is 1%) for two tailed test is Ztab=2.58

Margin of error = Critical value x Standard Error (SE) of (x)

Margin of error = 2.58 x 0.8= 2.064

The lower limit of 99% confidence interval for the mean time spent with patient = Average time spent with patient – Margin of error

The lower limit of 99% confidence interval for the mean time spent with patient = 20.2 – 2.064 =18.136     or Approximate 18.14 minutes

The lower limit of 99% confidence interval of the population mean time spent with patient is 18.14 minutes.