Answer:
Sample size n
1269.15
Explanation:
From the information given ,
At 99% of confidence interval,
the level of significance ∝ = 1 - 0.99
the level of significance ∝ = 0.01
the critical value for 99% of confidence interval is:
![\mathtt{(\alpha )/(2) = (0.01)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/ioycmypf9jctql1p0susv3cuwcnbu4kpas.png)
= 0.005
The value for z from the standard normal tables
= 2.58
The Margin of error E= 3% = 0.03
The formula to determine the sample size n used can be expressed as follows:
![\mathtt { n = ((z_(\alpha/2))/(E))^2 \ \hat p (1 - \hat p) }](https://img.qammunity.org/2021/formulas/mathematics/college/jyateofd81mg9cld0c9gabivdkx3cubuzs.png)
where;
= 22% = 0.22
Then:
![\mathtt { n = ((2.58)/(0.03))^2 \ * 0.22 * (1 - 0.22) }](https://img.qammunity.org/2021/formulas/mathematics/college/oxnrvvn5xxmpmton3hrhmud0dsjmpdty08.png)
![\mathtt { n = (86)^2 \ * 0.22 * (0.78) }](https://img.qammunity.org/2021/formulas/mathematics/college/5c7u84c2wrg5h6awxnemavv5kd50pe1v65.png)
![\mathtt { n = 7396 \ * 0.22 * (0.78) }](https://img.qammunity.org/2021/formulas/mathematics/college/6c9025pin3u5paohlx16mytj33sne748n5.png)
n = 1269.1536
Sample size n
1269.15