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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

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Answer:

The minimum sample size required is 4610.

Explanation:

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


CI=\hat p\pm z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}

The margin of error for this interval is:


MOE= z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}

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,
\hat p=0.50.

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


z_(\alpha/2)=z_(0.01/2)=z_(0.005)=2.58

*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:


MOE= z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}


n=[(z_(\alpha/2)* √(\hat p(1-\hat p)) )/(MOE)]^(2)


=[(2.58* √(0.50(1-0.50)))/(0.019)]^(2)


=(67.8947)^(2)\\=4609.6903\\\approx4610

Thus, the minimum sample size required is 4610.

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