Question

a researcher studying public opinion of proposed social security changes obtains a simple random sample of 30 adult americans and asks them whether or not they support the proposed changes. to say that the distribution of \\( \hat { p } \\), the sample proportion of adults who respond yes, is approximately normal, how many more adult americans does the researcher need to sample in the following cases? (a) 10% of all adult americans support the changes (b) 15% of all adult americans support the changes (a) the researcher must ask \\( \square \\) more american adults. (round up to the nearest integer.)

Explanation:

Step1: Check the condition for normal approximation

For the sampling distribution of \(\hat{p}\) to be approximately normal, we need \(np\geq5\) and \(n(1 - p)\geq5\).
Let the current sample size be \(n_0 = 30\). Let the new sample size be \(n\).

Case (a): \(p = 0.1\)

We know that \(np\geq5\) and \(n(1 - p)\geq5\). Substituting \(p = 0.1\), from \(np\geq5\), we have \(n\times0.1\geq5\Rightarrow n\geq\frac{5}{0.1}=50\).
Since the current sample size \(n_0 = 30\), the number of additional samples \(n - n_0\).
\(n - n_0=50 - 30=20\)

Step2: Verify \(n(1 - p)\) for case (a)

When \(n = 50\) and \(p = 0.1\), \(n(1 - p)=50\times(1 - 0.1)=50\times0.9 = 45\geq5\)

Answer:

20