Question 1170686: In a survey, 25 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $43 and standard deviation of $8. Find the margin of error at a 80% confidence level.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's calculate the margin of error for this problem.
**Given:**
* Sample size (n) = 25
* Sample mean (x̄) = $43
* Sample standard deviation (s) = $8
* Confidence level = 80%
**1. Find the Critical t-value (t_c)**
Since the sample size is small (n = 25) and the population standard deviation is unknown, we will use the t-distribution.
* Degrees of freedom (df) = n - 1 = 25 - 1 = 24
* Confidence level = 80%, so α = 1 - 0.80 = 0.20
* Since it's a two-tailed test, α/2 = 0.20 / 2 = 0.10
Now, we need to find the t-value from the t-distribution table or using a calculator with df = 24 and α/2 = 0.10.
* t_c ≈ 1.318
**2. Calculate the Margin of Error (E)**
The formula for the margin of error is:
E = t_c * (s / √n)
Plug in the values:
E = 1.318 * (8 / √25)
E = 1.318 * (8 / 5)
E = 1.318 * 1.6
E = 2.1088
**3. Round to a Reasonable Number of Decimal Places**
Let's round to two decimal places, since we're dealing with money:
E ≈ $2.11
**Answer:**
The margin of error at an 80% confidence level is approximately $2.11.
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