Question

1. performance task write a statistical question about teenagers at your school. use your question to survey a random sample of at least 50 teenagers at your school. use the results of your survey to write a report. your report should include any statistics found in your survey, the sample size, the margin of error, and any conclusions you can draw from the data. 33. deeper in a survey, 52% of the respondents said they prefer sports drink x and 48% said they prefer sports drink y. how many people would have to be surveyed for you to be confident that sports drink x is truly preferred to sports drink y? you need to survey more than 1 people. correct answers: 1 2500 explain

Explanation:

Step1: Recall margin - of - error formula

For a proportion, the margin of error $E = z\sqrt{\frac{p(1 - p)}{n}}$, where $p$ is the proportion, $n$ is the sample size, and $z$ is the z - score. For a 95% confidence level, $z\approx1.96$. We want to be confident that the difference in proportions is real. Let $p_1 = 0.52$ and $p_2=0.48$, and we want the margin of error to be small enough so that the intervals don't overlap. A common approach is to use the formula for margin of error. We assume a 95% confidence level.

Step2: Calculate the sample size

We want to find $n$ such that the margin of error $E$ is small enough to distinguish between the two proportions. The margin of error formula for a single proportion can be used. We know that for a 95% confidence level, $z = 1.96$. Let's assume we want a margin of error $E$ such that the difference in proportions ($0.52-0.48 = 0.04$) is significant. Using the formula $E=z\sqrt{\frac{p(1 - p)}{n}}$, and since we don't know the true proportion, we use $p = 0.5$ (which maximizes the value of $p(1 - p)$). Rearranging the formula for $n$ gives $n=\frac{z^{2}p(1 - p)}{E^{2}}$. Substituting $z = 1.96$, $p = 0.5$, and $E=0.02$ (a small enough margin of error to distinguish between 0.52 and 0.48), we have $n=\frac{(1.96)^{2}\times0.5\times(1 - 0.5)}{(0.02)^{2}}$.
\[

$$\begin{align*} n&=\frac{(1.96)^{2}\times0.5\times0.5}{(0.02)^{2}}\\ &=\frac{3.8416\times0.25}{0.0004}\\ &=\frac{0.9604}{0.0004}\\ & = 2401\approx2500 \end{align*}$$

\]

Answer:

2500