Question

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A sample of 1081 hospitalized patients in China, with laboratory-confirmed Covid-19, was obtained. Fever was present in 473 of these patients when they were first admitted to the hospital (Source: New England Journal of Medicine, February 28, 2020).

(a) [4 marks] Calculate a 99% confidence interval for the population proportion (p) of Covid-19 patients that have a fever when they are admitted to the hospital.

(b) [1 mark] Provide an interpretation (in context) of your interval estimate in part (a).

Answer 1

Solution :

Given that,

n = 1081

x = 473

Point estimate = sample proportion = = x / n = 473/1081=0.438

1 -   = 1- 0.438 =0.562

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.438*0.562) /1081 )

E = 0.0389

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.438-0.0389 < p < 0.438+0.0389

0.3991< p < 0.4769

The 99% confidence interval for the population proportion p is : 0.3991, 0.4769