Question

1. [7 marks]

   1. (a) Public health authorities want to estimate the proportion of Australians aged 18-65 who caught the flu during winter 2018. If the true proportion is thought to be about 24%, how large a sample will be needed if we want to be 95% sure that our estimate will be within 0.02 of the true value? [3]

   2. (b) Due to a change in the available funding, the sample actually used for estimating the prevalence of flu included 1,712 randomly-selected Australians aged 18-65. Of these, 367 reported having a “flu-like illness” during winter 2018. Use these data to calculate a 90% confidence interval for the proportion of Australians aged 18-65 who had a flu-like illness during winter 2018. [4]

Answer 1

(a)Solution :

Given that,

= 24 % = 0.24

1 - = 1 - 0.24 = 0.76

margin of error = E = 0.02

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = n = (Z_/2 / E)² * * (1 - )

= ( 1.96 /  0.02)² * 0.24 * 0.76

= 1751.76

Sample size = 1752

(b)Solution :

Given that,

n = 1712

x = 367

= x / n = 367/ 1712= 0.214

1 - = 1 - 0.214 = 0.786

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.214 * 0.786) / 1712) = 0.0163

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.214 - 0.0163 < p < 0.214 + 0.0163

0.1977< p < 0.2303

The 90% confidence interval for the population proportion p is : (0.1977 , 0.2303)