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Math  /  Data & Statistics

QuestionWithdrawal symptoms may occur when a person using a painkiller stops using it. For a certain, widely taken painkiller, withdrawal symptoms occur in 16%16 \% of people who have stopped using the painkiller. Suppose that we will take a random sample of 10 people who have stopped using the painkiller. Let pundefined\widehat{p} represent the proportion of people from the sample who experienced withdrawal symptoms. Consider the sampling distribution of the sample proportion p^\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 μp\mu_{p} (the mean of the sampling distribution of the sample proportion). μp^=□\mu_{\hat{p}}=\square (b) Find σp^\sigma_{\hat{p}} (the standard deviation of the sampling distribution of the sample proportion). □\square σp^=⨿\sigma_{\hat{p}}=\amalg

Studdy Solution

STEP 1

1. The sample size is n=10 n = 10 .
2. The probability of experiencing withdrawal symptoms is p=0.16 p = 0.16 .
3. The sampling distribution of the sample proportion pundefined\widehat{p} is approximately normal for large sample sizes, but since n=10 n = 10 is small, we assume the binomial distribution applies.

STEP 2

1. Calculate the mean of the sampling distribution of the sample proportion μp^\mu_{\hat{p}}.
2. Calculate the standard deviation of the sampling distribution of the sample proportion σp^\sigma_{\hat{p}}.

STEP 3

The mean of the sampling distribution of the sample proportion μp^\mu_{\hat{p}} is given by the formula:
μp^=p\mu_{\hat{p}} = p
Substitute the given probability:
μp^=0.16\mu_{\hat{p}} = 0.16

STEP 4

The standard deviation of the sampling distribution of the sample proportion σp^\sigma_{\hat{p}} is given by the formula:
σp^=p(1−p)n\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
Substitute the given values:
σp^=0.16×(1−0.16)10\sigma_{\hat{p}} = \sqrt{\frac{0.16 \times (1 - 0.16)}{10}}
Calculate the expression inside the square root:
=0.16×0.8410= \sqrt{\frac{0.16 \times 0.84}{10}}
=0.134410= \sqrt{\frac{0.1344}{10}}
=0.01344= \sqrt{0.01344}
≈0.1159\approx 0.1159
Round to two decimal places:
σp^≈0.12\sigma_{\hat{p}} \approx 0.12
The answers are: (a) μp^=0.16\mu_{\hat{p}} = 0.16 (b) σp^=0.12\sigma_{\hat{p}} = 0.12

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