Pregunta
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Withdrawal symptoms may occur when a person using a painkiller stops using it. For a certain, widely taken painkiller, withdrawal symptoms occur in \( 12 \% \) of people who have stopped using the painkiller. Suppose that we will take a random sample of 8 people who have stopped using the painkiller. Let \( \hat{p} \) represent the proportion of people from the sample who experienced withdrawal symptoms. Consider the sampling distribution of the sample proportion \( \hat{p} \). Complete the following. Carry your intermediate computations to four or more decimal places. Write your answers with two decimal places, rounding if needed. (a) Find \( \mu_{\hat{p}} \) (the mean of the sampling distribution of the sample proportion). (b) Find \( \sigma_{\hat{p}} \) (the standard deviation of the sampling distribution of the sample proportion).

Ask by Fowler Smith. in the United States
Jan 27,2025

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- (a) \( \mu_{\hat{p}} = 0.12 \) - (b) \( \sigma_{\hat{p}} = 0.11 \)

Solución

To solve this problem, we need to find the mean and standard deviation of the sampling distribution of the sample proportion \( \hat{p} \). ### Given: - The population proportion \( p = 0.12 \) (12% of people experience withdrawal symptoms). - Sample size \( n = 8 \). ### (a) Finding \( \mu_{\hat{p}} \) The mean of the sampling distribution of the sample proportion \( \hat{p} \) is given by the formula: \[ \mu_{\hat{p}} = p \] Substituting the known value: \[ \mu_{\hat{p}} = 0.12 \] ### (b) Finding \( \sigma_{\hat{p}} \) The standard deviation of the sampling distribution of the sample proportion \( \hat{p} \) is given by the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \] Substituting the known values: \[ \sigma_{\hat{p}} = \sqrt{\frac{0.12(1 - 0.12)}{8}} = \sqrt{\frac{0.12 \times 0.88}{8}} \] Now, let's calculate this step by step. 1. Calculate \( 0.12 \times 0.88 \): \[ 0.12 \times 0.88 = 0.1056 \] 2. Divide by \( n = 8 \): \[ \frac{0.1056}{8} = 0.0132 \] 3. Take the square root: \[ \sigma_{\hat{p}} = \sqrt{0.0132} \approx 0.1149 \] ### Final Answers Now, rounding the answers to two decimal places: - (a) \( \mu_{\hat{p}} = 0.12 \) - (b) \( \sigma_{\hat{p}} \approx 0.11 \) Thus, the final answers are: - (a) \( \mu_{\hat{p}} = 0.12 \) - (b) \( \sigma_{\hat{p}} = 0.11 \)

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To solve the given problem, we need to calculate the mean (\( \mu_{\hat{p}} \)) and standard deviation (\( \sigma_{\hat{p}} \)) of the sampling distribution of the sample proportion \( \hat{p} \). **(a) Finding \( \mu_{\hat{p}} \)**: The mean of the sampling distribution of the sample proportion is given by the formula: \[ \mu_{\hat{p}} = p \] where \( p \) is the population proportion. Given that \( p = 0.12 \) (or 12%) for the painkiller withdrawal symptoms, we find: \[ \mu_{\hat{p}} = 0.12 \] **Expressing this with two decimal places**: \[ \mu_{\hat{p}} = 0.12 \] **(b) Finding \( \sigma_{\hat{p}} \)**: The standard deviation of the sampling distribution of the sample proportion is given by the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \] where \( n \) is the sample size. In this case, \( n = 8 \). First, we calculate \( 1 - p \): \[ 1 - p = 1 - 0.12 = 0.88 \] Now we can substitute into the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{0.12 \cdot 0.88}{8}} \] Calculating the numerator: \[ 0.12 \cdot 0.88 = 0.1056 \] Now dividing by \( n \): \[ \frac{0.1056}{8} = 0.0132 \] Finally, taking the square root: \[ \sigma_{\hat{p}} = \sqrt{0.0132} \approx 0.1149 \] **Expressing this with two decimal places**: \[ \sigma_{\hat{p}} \approx 0.11 \] In summary: - \( \mu_{\hat{p}} = 0.12 \) - \( \sigma_{\hat{p}} \approx 0.11 \)

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