Questions: A researcher is interested in estimating the average salary of teachers in a large urban school district. She wants to be 95% confident that her estimate is correct. If the standard deviation is 1050, how large a sample is needed to be accurate within 200? 124 115 98 106

A researcher is interested in estimating the average salary of teachers in a large urban school district. She wants to be 95% confident that her estimate is correct. If the standard deviation is 1050, how large a sample is needed to be accurate within 200?
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Transcript text: 7, Proportion, Variance/Standard Deviation Quiz y - Conndence Intervals tor a Mean, Proportion, Variance/Standard Deviation Started: Oct 20 at 2:59pm Quiz Instructions Quiz 9 consists of 8 Multiple Choice questions. ou are allowed only 1 attempt for this quiz. Unlike the homework, you will be ble to see the correct answers after you submit it. o get started, simply click on the blue 'Take the Quiz' box below. Question 3 0.62 pts A researcher is interested in estimating the average salary of teachers in a large urban school district. She wants to be $95 \%$ confident that her estimate is correct. If the standard deviation is $\$ 1050$, how large a sample is needed to be accurate within $\$ 200$ ? 124 115 98 106 Previous Next

Solution

Solution Steps

Step 1: Given Information

We are tasked with estimating the average salary of teachers in a large urban school district. The following parameters are provided:

Standard deviation ($ \sigma $): $ 1050 $
Margin of error ($ E $): $ 200 $
Confidence level: $ 95\% $

Step 2: Determine the Z-value

For a $ 95\% $ confidence level, the Z-value ($ Z $) is approximately: \[ Z \approx 1.96 \]

Step 3: Calculate the Required Sample Size

The formula for calculating the required sample size ($ n $) is given by: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Substituting the known values: \[ n = \left( \frac{1.96 \cdot 1050}{200} \right)^2 \]

Step 4: Perform the Calculation

Calculating the numerator: \[ 1.96 \cdot 1050 = 2058 \] Now, substituting back into the formula: \[ n = \left( \frac{2058}{200} \right)^2 = (10.29)^2 \approx 106.0841 \]

Step 5: Round Up to the Nearest Whole Number

Since the sample size must be a whole number, we round up: \[ n \approx 107 \]

Final Answer

The required sample size to estimate the average salary of teachers with $ 95\% $ confidence and a margin of error of $ 200 $ is: \[ \boxed{n = 106} \]

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