Question

in a survey, 16 people were asked how much they spent on their childs last birthday gift. the results were roughly bell - shaped with a mean of $31 and standard deviation of $4. find the margin of error at a 99% confidence level. give your answer to two decimal places.

Explanation:

Step1: Determine degrees of freedom

The sample size $n = 16$. Degrees of freedom $df=n - 1=16 - 1 = 15$.

Step2: Find the t - value

For a 99% confidence level and $df = 15$, from the t - distribution table, $t_{\alpha/2}=2.947$.

Step3: Calculate margin of error formula

The formula for the margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $s = 4$ (standard deviation) and $n = 16$.

Step4: Substitute values

$E=2.947\times\frac{4}{\sqrt{16}}=2.947\times\frac{4}{4}=2.95$ (rounded to two decimal places)

Answer:

$2.95$