Question

1. the drama club believes that the proportion of students who bring in holiday treats to share during the last week of school is different from 50%. (they arent sure if more students will bring treats... or if everyone will forget.) hypotheses: $h_0: p = 0.50$ $h_a: p \

eq 0.50$ evidence: a random sample of 160 students is taken, and 92 of them brought holiday treats. $\hat{p} = \square$ (unreduced fraction) unusualness: $z = \square$ (three decimal places) chances: p - value = $\square$ (three decimal places)

Explanation:

Step1: Calculate sample proportion

$\hat{p} = \frac{\text{Number of successes}}{\text{Sample size}} = \frac{92}{160}$

Step2: Simplify the fraction

Divide numerator and denominator by 4: $\frac{92\div4}{160\div4} = \frac{23}{40}$

Step3: Compute test statistic (z-score)

Formula: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$
Substitute values: $p_0=0.50$, $n=160$, $\hat{p}=0.575$
$z = \frac{0.575 - 0.50}{\sqrt{\frac{0.50\times0.50}{160}}} = \frac{0.075}{\sqrt{0.25/160}} = \frac{0.075}{\sqrt{0.0015625}} = \frac{0.075}{0.039528} \approx 1.897$

Step4: Find P-value for two-tailed test

Z=1.897 corresponds to one-tailed area ≈0.0289; two-tailed P-value=2×0.0289≈0.0578

Answer:

$\hat{p} = \frac{23}{40}$, z ≈ 1.897, P-value ≈ 0.058