A study is to be conducted of the percentage of homeowners who own at least two television sets. At least how large a sample is required if we wish to be 99% confident that the error in estimating this quantity is less than 0.019

The minimum sample size required is 4610.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion p is:

The margin of error for this interval is:

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, .

The critical value of z for 99% confidence interval is:

*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:

Thus, the minimum sample size required is 4610.

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x