Question

An information technology (IT) consulting firm specializing in health care solutions wants to study communication deficiencies in the health care industry. A random sample of 70 health

care clinicians reveals the following:

• Time wasted in a day due to outdated communication technologies:

X = 45 minutes, S = 10 minutes.

• Thirty-six health care clinicians cite inefficiency of pagers as the reason for the wasted time.

a. Construct a 99% confidence interval estimate for the population mean time wasted in a day due to outdated communication technologies.

b. Construct a 95% confidence interval estimate for the population proportion of health care clinicians who cite inefficiency of pagers as the reason for the wasted time.

Answer 1

Solution:

Given:=45, n=70,s=10

a)Calculation of 99% confidence interval estimate for the population mean time wasted in a day due to outdated communication technologies.

Formula to find confidence interval is  given by Z *s/ where Z =Z0.01/2=Z0.005=2.58(from the distribution table)

So substituting all the values in the above formula, We have 452.58*10/70

=453.084

=[48.084,41.916]

So 99% confidence interval estimate for the population mean time wasted in a day due to outdated communication technologies is in between 48.084 and 41.916.

b.Calculation of 95% confidence interval estimate for the population proportion of health care clinicians who cite inefficiency of pagers as the reason for the wasted time.

Given:Thirty-six health care clinicians cite inefficiency of pagers as the reason for the wasted time.

Proportion,p=X/n=36/70=0.514 and with 95% level of confidence Z=Z0.05/2=Z0.025=1.96

Formula is given by pZ *p(1-p)/n

=0.514(1.96)0.514(1-0.514)/70

0.5140.117

[0.631,0.397]

So 95% confidence interval estimate for the population proportion of health care clinicians who cite inefficiency of pagers as the reason for the wasted time is in between 0.631 and 0.397.