Math  /  Data & Statistics

QuestionSAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300 . An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 85%85 \% confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whole number answer.

Studdy Solution

STEP 1

What is this asking? How many first-year students need to be surveyed to be 85% sure that the average SAT score of the sample is within 5 points of the true average SAT score of all first-year students? Watch out! Don't forget to round your answer up to the nearest whole number, since we can't sample a fraction of a student!
Also, remember that the z-score corresponds to the confidence level, not the margin of error.

STEP 2

1. Find the z-score.
2. Calculate the sample size.

STEP 3

We're dealing with an **85% confidence interval**, which means we want to find the z-score that captures 85% of the area under the normal distribution curve.
This means there's 100%85%=15%100\% - 85\% = 15\% of the area left over.

STEP 4

Since the normal distribution is symmetric, this 15% is split equally between the two tails, leaving 15%/2=7.5%15\% / 2 = 7.5\% in each tail.

STEP 5

To use a z-table, we need the area to the *left* of our z-score.
Since 7.5% is in the right tail, we know that 100%7.5%=92.5%100\% - 7.5\% = 92.5\% is to the left.
So we're looking for a z-score corresponding to an area of **0.925**.

STEP 6

Looking up **0.925** in a z-table gives us a z-score of approximately **1.44**.
This is the magic number we'll use in our sample size calculation!

STEP 7

The formula for calculating the sample size nn is: n=(zσE)2 n = \left( \frac{z \cdot \sigma}{E} \right)^2 where zz is the z-score, σ\sigma is the standard deviation, and EE is the desired margin of error.

STEP 8

We know that z=z = \ **1.44**, σ=\sigma = \ **300**, and E=E = \ **5**.
Let's plug these values into our formula: n=(1.443005)2 n = \left( \frac{1.44 \cdot 300}{5} \right)^2

STEP 9

Now, let's **simplify the expression inside the parentheses**: n=(4325)2 n = \left( \frac{432}{5} \right)^2 n=(86.4)2 n = (86.4)^2

STEP 10

**Squaring** the result gives us: n7464.96 n \approx 7464.96

STEP 11

Since we can't have a fraction of a student, we **round up** to the nearest whole number: n=7465 n = 7465

STEP 12

The administrator should sample **7465** students to achieve the desired margin of error.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord