Question

1.
(a) Public health authorities want to estimate the proportion of Australians aged 18-65 who caught the flu during winter 2019. If the true proportion is thought to be about 20%, how large a sample will we need if we want to be 90% sure that our estimate will be within 0.01 of the true value?
(b) Due to a change in the available funding, the sample actually used for estimating the prevalence of ‘flu included 2,253 randomly-selected Australians aged 18-65. Of these, 528 reported having a ‘flu-like illness during winter 2019. Use these data to calculate a 95% confidence interval for the proportion of Australians aged 18-65 who had a ‘flu-like illness during winter 2019.

Answer 1

Answer:

(a).

Public health authorities want to estimate the proportion of Australians aged 18-65 who caught the flu during winter 2019.

If the true proportion is thought to be about 20%, how large a sample will we need if we want to be 90% sure that our estimate will be within 0.01 of the true value:

The margin of error (E)=0.01

=0.20

1-=1-0.20=0.80

At 90% confidence interval level the Z is,

=1-90%=1-0.90=0.10

We know that,

Therefore,

= 27060.25 0.20 0.80

=4329.64

Therefore, the sample size(n)=4330

(b).

Due to a change in the available funding, the sample actually used for estimating the prevalence of ‘flu included 2,253 randomly-selected Australians aged 18-65. Of these, 528 reported having a ‘flu-like illness during winter 2019.

Use these data to calculate a 95% confidence interval for the proportion of Australians aged 18-65 who had a ‘flu-like illness during winter 2019:

Here,

x=528

n=2253

At 95% confidence interval level the Z is,

=1-95%=1-0.95=0.05

We know that,

=0.01749263

=0.0175

A 95% confidence interval for the proportion of Australians aged 18-65 who had a flu-like illness during winter 2019 is given by,