Question

in a survey, 21 people were asked how much they spent on their childs last birthday gift. the results were roughly bell - shaped with a mean of $47 and standard deviation of $2. find the margin of error at a 98% confidence level.
give your answer to two decimal places, but for critical value round to 4 decimal places.
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Explanation:

Step1: Find the critical value \( z_{\alpha/2} \)

The confidence level is \( C = 0.98 \). Then \( \alpha=1 - C=1 - 0.98 = 0.02 \), and \( \alpha/2=0.01 \).
Using a standard normal table or a calculator, \( z_{\alpha/2}=z_{0.01}\approx2.3263 \) (rounded to 4 decimal places).

Step2: Calculate the margin of error \( E \)

The formula for the margin of error for a population mean (when the population standard deviation \( \sigma \) is known) is \( E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \).
We are given that \( n = 21 \), \( \sigma=2 \), and \( z_{\alpha/2}=2.3263 \).
Substitute the values into the formula: \( E=2.3263\times\frac{2}{\sqrt{21}} \).
First, calculate \( \sqrt{21}\approx4.5837 \).
Then \( \frac{2}{\sqrt{21}}\approx\frac{2}{4.5837}\approx0.4363 \).
Finally, \( E = 2.3263\times0.4363\approx1.01 \).

Answer:

\( 1.01 \)