Questions: Determine the mean of the sampling distribution of p̂. w = 0?0 (Round to two decimal places as needed) Determine the standard deviation of the Sampling distribution of p̂ σ = (Round to three decimal places as needed) (b) What is the probability that in a random sample of 900 adults, more than 32% do not own a credit card? The probability is 0.0038 (Round to four decimal places as needed) Interpret this probability If 100 different random samples of 900 adults were obtained, one would expect to result in more than 32% not owning a credit card

Solution Steps

To solve the given problems, we need to follow these steps:

1. Mean of the Sampling Distribution of \(\hat{p}\): The mean of the sampling distribution of \(\hat{p}\) is equal to the population proportion \(p\).

2. Standard Deviation of the Sampling Distribution of \(\hat{p}\): The standard deviation (also known as the standard error) of the sampling distribution of \(\hat{p}\) is calculated using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the sample size.

3. Probability Calculation: To find the probability that more than 32% of a sample of 900 adults do not own a credit card, we use the normal approximation to the binomial distribution. We calculate the z-score and then use the standard normal distribution to find the probability.

The mean of the sampling distribution of \(\hat{p}\) is given by the population proportion \(p\). Thus, we have: \[ \text{Mean} = \hat{p} = 0.32 \]

The standard deviation (standard error) of the sampling distribution of \(\hat{p}\) is calculated using the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.32(1-0.32)}{900}} \approx 0.0155 \]

To find the probability that more than \(32\%\) of a sample of \(900\) adults do not own a credit card, we first calculate the z-score: \[ z = \frac{0.32 - \hat{p}}{\sigma_{\hat{p}}} = \frac{0.32 - 0.32}{0.0155} = 0.0 \] Using the z-score, we find the probability: \[ P(X > 0.32) = 1 - P(Z \leq 0) = 1 - 0.5 = 0.5 \]

If \(100\) different random samples of \(900\) adults were obtained, we would expect approximately: \[ 50 \text{ samples} \text{ to result in more than } 32\% \text{ not owning a credit card.} \]

Final Answer