Question

1) Eliminating infections acquired during hospitalization is a health care priority. Researchers aiming to quantify the scale of this problem in the United States reported their survey finding as follows: of 11,282 patients, 452 had 1 or more healthcare-associated infections. Compute a 95% confidence interval of patients having 1 or more health-care associated infections. State the interpretation in context.

Answer 1

Solution :

Given that,

n = 11282

x = 452

Point estimate = sample proportion = = x / n = 452/11282=0.040

1 - = 1- 0.040=0.96

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z _{0.025 = 1.96} ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )   

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.040*0.96) /11282 )

E = 0.0036

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.040 -0.0036 < p <0.040 0.0036

0.0364< p < 0.0436

The 95% confidence interval for the population proportion p is : lower limit =0.0364, upper limit =0.0436