Questions: Refer to the results of the accompanying study. Complete parts (a) through (e) below. B. The mean is very close to the sample mean, 306/659, not the mean found in part (b). The standard deviation though is very close the theoretical standard deviation from part (b). C. These values are very close to the theoretical values from part (b). D. The mean is very close to the theoretical mean from part (b), but the standard deviation does not seem to follow the theoretical standard deviation from part (b). (d) Describe the shape of the distributions of outcomes from the simulation in part (c). Compute n p(1-p) where p=0.5 (the proportion stated in the null hypothesis). Based on this result is the distribution of p̂ approximately normal? Since n p(1-p)= 164.75 10 and the sample size is less than 5% of the population size (assuming the TV personality has made at least the necessary number of predictions, which is a reasonable assumption), the distribution of the sample proportion, p̂, is approximately normal. (e) Use the normal model to find P(p̂ ≤ 0.464), where 0.464=306/659, the sample proportion of correct predictions made by the TV personality. P(p̂ ≤ 0.464)=

Refer to the results of the accompanying study. Complete parts (a) through (e) below. B. The mean is very close to the sample mean, 306/659, not the mean found in part (b). The standard deviation though is very close the theoretical standard deviation from part (b). C. These values are very close to the theoretical values from part (b). D. The mean is very close to the theoretical mean from part (b), but the standard deviation does not seem to follow the theoretical standard deviation from part (b). (d) Describe the shape of the distributions of outcomes from the simulation in part (c). Compute n p(1-p) where p=0.5 (the proportion stated in the null hypothesis). Based on this result is the distribution of p̂ approximately normal? Since n p(1-p)= 164.75 10 and the sample size is less than 5% of the population size (assuming the TV personality has made at least the necessary number of predictions, which is a reasonable assumption), the distribution of the sample proportion, p̂, is approximately normal. (e) Use the normal model to find P(p̂ ≤ 0.464), where 0.464=306/659, the sample proportion of correct predictions made by the TV personality. P(p̂ ≤ 0.464)=
Transcript text: Refer to the results of the accompanying study. Complete parts (a) through (e) below. B. The mean is very close to the sample mean, $\frac{306}{659}$, not the mean found in part (b). The standard deviation though is very close the theoretical standard deviation from part (b). C. These values are very close to the theoretical values from part (b). D. The mean is very close to the theoretical mean from part (b), but the standard deviation does not seem to follow the theoretical standard deviation from part (b). (d) Describe the shape of the distributions of outcomes from the simulation in part ( $c$ ). Compute $n p(1-p)$ where $p=0.5$ (the proportion stated in the null hypothesis). Based on this result is the distribution of $\hat{p}$ approximately normal? Since $n p(1-p)=$ 164.75 $\square$ 10 and the sample size is $\square$ less than $5 \%$ of the population size (assuming the TV personality has made at least the necessary number of predictions, which is a reasonable assumption), the distribution of the sample proportion, $\hat{p}$, is approximately normal. (e) Use the normal model to find $\mathrm{P}(\hat{p} \leq 0.464)$, where $0.464=\frac{306}{659}$, the sample proportion of correct predictions made by the TV personality. $P(\hat{p} \leq 0.464)=$ $\square$
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Solution

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Solution Steps

Step 1: Calculate \( n p(1-p) \)

To determine if the distribution of the sample proportion \( \hat{p} \) is approximately normal, we calculate \( n p(1-p) \):

\[ n p(1-p) = 659 \cdot 0.5 \cdot (1 - 0.5) = 164.75 \]

Since \( n p(1-p) = 164.75 \) is greater than 10 and the sample size is less than 5% of the population size, the distribution of the sample proportion \( \hat{p} \) is approximately normal.

Step 2: Calculate \( P(\hat{p} \leq 0.464) \)

Next, we find the probability \( P(\hat{p} \leq 0.464) \) where \( \hat{p} = \frac{306}{659} \).

The mean \( \mu \) of the sampling distribution of \( \hat{p} \) is:

\[ \mu = p = 0.5 \]

The standard deviation \( \sigma \) of the sampling distribution of \( \hat{p} \) is calculated as follows:

\[ \sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5 \cdot (1 - 0.5)}{659}} = \sqrt{\frac{0.25}{659}} \approx 0.0194 \]

Using the Z-score formula, we find:

\[ Z_{end} = \frac{0.464 - 0.5}{0.0194} \approx -1.85 \]

The probability \( P(\hat{p} \leq 0.464) \) is calculated as:

\[ P(\hat{p} \leq 0.464) = \Phi(Z_{end}) - \Phi(-\infty) = \Phi(-1.85) - 0 = 0.0 \]

Final Answer

The results are summarized as follows:

  • \( n p(1-p) = 164.75 \)
  • \( P(\hat{p} \leq 0.464) = 0.0 \)

Thus, the final answers are: \[ \boxed{n p(1-p) = 164.75} \] \[ \boxed{P(\hat{p} \leq 0.464) = 0.0} \]

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