Answer:
The minimum sample size required is 4610.
Step-by-step explanation:
The (1 - α)% confidence interval for population proportion p is:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The margin of error for this interval is:
[tex]MOE= z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
It is provided that a 99% confidence interval is computed to estimate the percentage of homeowners who own at least two television sets.
Assume that 50% of the homeowners own at least two television sets.
So, the sample proportion is, [tex]\hat p=0.50[/tex].
The critical value of z for 99% confidence interval is:
[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58[/tex]
*Use a z-table for the critical value.
The margin of error is given as, MOE = 0.019.
Compute the value of n as follows:
[tex]MOE= z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)} }{MOE}]^{2}[/tex]
[tex]=[\frac{2.58\times \sqrt{0.50(1-0.50)}}{0.019}]^{2}[/tex]
[tex]=(67.8947)^{2}\\=4609.6903\\\approx4610[/tex]
Thus, the minimum sample size required is 4610.
