Questions: According to a survey in a country, 32% of adults do not own a credit card. Suppose a simple random sample of 400 adults is obtained. Complete parts (a) through (d) below. (a) Describe the sampling distribution of p̂, the sample proportion of adults who do not own a credit card. Choose the phrose that best describes the shape of the sampling distribution of pbelow. A. Not normal because n ≤ 0.05 N and np(1-p)<10 B. Not normal because n ≤ 0.05 N and np(1-p) ≥ 10 C. Approximately normal because n ≤ 0.05 N and np(1-p)<10 D. Approximately normal because n ≤ 0.05 N and np(1-p) ≥ 10 Determine the mean of the sampling distribution of p̂. μp= (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of p̂. σp̂= (Round to three decimal places as needed.) (b) What is the probability that in a random sample of 400 adults, more than 34% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 400 adults were obtained, one would expect to result in more than 34% not owning a credit card. (Round to the nearest integer as needed.)

According to a survey in a country, 32% of adults do not own a credit card. Suppose a simple random sample of 400 adults is obtained. Complete parts (a) through (d) below.
(a) Describe the sampling distribution of p̂, the sample proportion of adults who do not own a credit card. Choose the phrose that best describes the shape of the sampling distribution of pbelow.
A. Not normal because n ≤ 0.05 N and np(1-p)<10
B. Not normal because n ≤ 0.05 N and np(1-p) ≥ 10
C. Approximately normal because n ≤ 0.05 N and np(1-p)<10
D. Approximately normal because n ≤ 0.05 N and np(1-p) ≥ 10

Determine the mean of the sampling distribution of p̂.
μp= (Round to two decimal places as needed.)

Determine the standard deviation of the sampling distribution of p̂.
σp̂= (Round to three decimal places as needed.)
(b) What is the probability that in a random sample of 400 adults, more than 34% do not own a credit card?

The probability is 
(Round to four decimal places as needed.)
Interpret this probability.
If 100 different random samples of 400 adults were obtained, one would expect to result in more than 34% not owning a credit card. (Round to the nearest integer as needed.)
Transcript text: According to a survey in a country, $32 \%$ of adults do not own a credit card. Suppose a simple random sample of 400 adults is obtained. Complete parts (a) through (d) below. (a) Describe the sampling distribution of $\hat{p}$, the sample proportion of adults who do not own a credit card. Choose the phrose that best describes the shape of the sampling distribution of pbelow. A. Not normal because $n \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p})<10$ B. Not normal because $\mathrm{n} \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p}) \geq 10$ C. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p})<10$ D. Approximately normal because $\mathrm{n} \leq 0.05 \mathrm{~N}$ and $\mathrm{np}(1-\mathrm{p}) \geq 10$ Determine the mean of the sampling distribution of $\hat{p}$. $\mu_{\mathrm{p}}=\square$ $\square$ (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of $\hat{p}$. $\sigma_{\hat{p}}=\square$ $\square$ (Round to three decimal places as needed.) (b) What is the probability that in a random sample of 400 adults, more than $34 \%$ do not own a credit card? The probabaity is $\square$ (Round to four decimal places as needed.) Interpet this probability. If 100 different random samples of 400 adults were obtained, one would expect $\square$ to result in more than $34 \%$ not owning a credit card. (Round to the nearest integer as needed.)
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Solution

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Solution Steps

Step 1: Sampling Distribution Description

The sampling distribution of \(\hat{p}\), the sample proportion of adults who do not own a credit card, is approximately normal. This conclusion is based on the condition that \(n \cdot p \cdot (1 - p) \geq 10\). Specifically, we have:

\[ n = 400, \quad p = 0.32 \quad \Rightarrow \quad n \cdot p \cdot (1 - p) = 400 \cdot 0.32 \cdot (1 - 0.32) = 86.4 \geq 10 \]

Thus, the shape of the sampling distribution is described as:

\[ \text{D. Approximately normal because } n \leq 0.05N \text{ and } np(1-p) \geq 10 \]

Step 2: Mean of the Sampling Distribution

The mean of the sampling distribution of \(\hat{p}\) is given by the population proportion \(p\):

\[ \mu_{\hat{p}} = p = 0.32 \]

Step 3: Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of \(\hat{p}\) is calculated using the formula:

\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.32 \cdot (1 - 0.32)}{400}} \approx 0.023 \]

Step 4: Probability Calculation

To find the probability that more than \(34\%\) of adults do not own a credit card, we calculate:

\[ P(\hat{p} > 0.34) \]

Using the Z-score transformation, we find:

\[ Z_{end} = \frac{0.34 - 0.32}{0.023} \approx 0.8696 \]

The probability is calculated as:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(17.1499) = 0.0 \]

Thus, the probability that more than \(34\%\) do not own a credit card is:

\[ P(\hat{p} > 0.34) = 1.0 \]

Step 5: Interpretation of Probability

If \(100\) different random samples of \(400\) adults were obtained, we would expect:

\[ \text{Expected number of samples with } \hat{p} > 0.34 = 100 \cdot P(\hat{p} > 0.34) = 100 \cdot 1.0 = 100 \]

Final Answer

  • Shape of the sampling distribution: D. Approximately normal because \(n \leq 0.05N\) and \(np(1-p) \geq 10\)
  • Mean of the sampling distribution of \(\hat{p}\): \(0.32\)
  • Standard deviation of the sampling distribution of \(\hat{p}\): \(0.023\)
  • Probability that more than \(34\%\) do not own a credit card: \(1.0\)
  • Expected number of samples with more than \(34\%\) not owning a credit card: \(100\)

\[ \boxed{\text{Shape: D, Mean: 0.32, Std Dev: 0.023, Probability: 1.0, Expected Samples: 100}} \]

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