Question

Annual salaries for graduates are expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual sling salary is desired. How large a sample should be taken if the desired margin of error is $250.

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Sigrid

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63 Answers

**Step 1:**

Given parameters:

Desired Margin of Error (ME): $250

Confidence Level: 95%

**Step 2:**

Calculate the estimated population standard deviation ( \sigma ):

Given range: $30,000 to $45,000 which covers approximately 6 standard deviations

\sigma = \frac{Range}{6} = \frac{45000 - 30000}{6} = \frac{15000}{6} = 2500

**Step 3:**

Calculate the required sample size ( n ):

Substitute \sigma and z into the formula for n :

n = \left(\frac{1.96 \times 2500}{250}\right)^2

Calculating step by step:

1. \frac{1.96 \times 2500}{250} = 19.6

2. 19.6^2 = 384.16

Therefore, approximately \boxed{384} participants would be required for the sample.

Given parameters:

Desired Margin of Error (ME): $250

Confidence Level: 95%

**Step 2:**

Calculate the estimated population standard deviation (

Given range: $30,000 to $45,000 which covers approximately 6 standard deviations

**Step 3:**

Calculate the required sample size (

Substitute

Calculating step by step:

1.

2.

Therefore, approximately

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