Question

question 3
a pilot study found that 72% of adult americans would like an internet connection in their car.
(a) use the given preliminary estimate to determine the sample size required to estimate the proportion of adult americans who would like an internet connection in their car to within 0.02 with 95% confidence. use the correct rounding rule for sample size.

(b) how would you modify this formula if a 99% confidence interval was desired?
select an answer
(c) use the given preliminary estimate to determine the sample size required to estimate the proportion of adult americans who would like an internet connection in their car to within 0.02 with 99% confidence. use the correct rounding rule for sample size.

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Explanation:

Step1: Recall the formula for sample size for proportion

The formula is \(n=\frac{z^{2}\cdot p\cdot(1 - p)}{E^{2}}\), where \(z\) is the z - score, \(p\) is the proportion, and \(E\) is the margin of error. For a 95% confidence interval, \(z = 1.96\), \(p=0.72\), \(1 - p = 0.28\), and \(E = 0.02\).

Step2: Substitute the values into the formula

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Step3: Apply the rounding rule

Since we cannot have a fraction of a sample, we round up. So \(n = 1937\).

Step4: For 99% confidence interval

The z - score changes. For a 99% confidence interval, \(z=2.576\). The formula becomes \(n=\frac{(2.576)^{2}\cdot p\cdot(1 - p)}{E^{2}}\)

Step5: Calculate sample size for 99% confidence interval

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Step6: Apply the rounding rule

We round up. So \(n = 3345\)

Answer:

(a) \(1937\)
(b) Change the z - score from \(1.96\) (for 95% confidence) to \(2.576\) (for 99% confidence) in the formula \(n=\frac{z^{2}\cdot p\cdot(1 - p)}{E^{2}}\)
(c) \(3345\)