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.
2. 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 $E = z\sqrt{\frac{p(1 - p)}{n}}$. To be confident of a difference between two proportions (52% and 48%), we want a small margin of error. A common $z$-value for 95% confidence is $z = 1.96$. Let $p=0.52$ (we can also use $p = 0.48$, the result will be the same). We want to find $n$ such that the difference in proportions is significant. A rule - of - thumb is to make the margin of error small enough so that the intervals don't overlap. For a small difference like 52% - 48% = 4%=0.04, we set $E = 0.02$ (half of the difference).

Step2: Rearrange the margin - of - error formula for $n$

Starting from $E = z\sqrt{\frac{p(1 - p)}{n}}$, we square both sides: $E^{2}=z^{2}\frac{p(1 - p)}{n}$. Then we can solve for $n$: $n=\frac{z^{2}p(1 - p)}{E^{2}}$.

Step3: Substitute values

Substitute $z = 1.96$, $p = 0.52$ (or $p=0.48$), and $E = 0.02$ into the formula. First, calculate $p(1 - p)=0.52\times(1 - 0.52)=0.52\times0.48 = 0.2496$. Then $z^{2}=(1.96)^{2}=3.8416$. So $n=\frac{3.8416\times0.2496}{(0.02)^{2}}=\frac{3.8416\times0.2496}{0.0004}\approx\frac{0.96}{0.0004}=2400$. Rounding up, we need $n = 2500$.

Answer:

2500