Question

An information technology (IT) consulting firm specializing in healthcare 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• Time wasted in a day due to outdated communication technologies: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a. Construct a 99% confidence interval estimate for the population mean time wasted in a day due to outdated communication technologies.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

Answer)

As the population s.d is unknown here we will use t distribution table to estimate the intervals

Given mean = 45

S.d = 10

N = 70

A)

Degrees of freedom is = n-1 = 69

For 69 dof and 99% confidence level, critical value t from t distribution is = 2.65

Margin of error (MOE) = t*s.d/√n = 2.65*10/√70 = 3.17

Interval is given by

(Mean - MOE, Mean + MOE)

[41.83, 48.17].

You can be 99% confident that the population mean (μ) falls between 41.83 and 48.17.

B)

Critical value t from.t distribution for 95% confidence level and 69 dof = 1.99

MOE = t*s.d/√n = 2.38

[42.62, 47.38].

You can be 95% confident that the population mean (μ) falls between 42.62 and 47.38.