Question

in a survey, 15 people were asked how much they spent on their childs last birthday gift. the results were roughly bell - shaped with a mean of $33 and standard deviation of $5. find the margin of error at a 90% confidence level. give your answer to two decimal places, but for critical value round to 4 decimal places.

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 15\) (\(n<30\)) and the distribution is roughly bell - shaped (assume normal distribution), we use the \(t\) - distribution. The confidence level is \(C = 0.90\), so the significance level \(\alpha=1 - C=1 - 0.90 = 0.10\). The degrees of freedom \(df=n - 1=15 - 1 = 14\).
Using a \(t\) - table or calculator, the critical value \(t_{\alpha/2}\) with \(df = 14\) and \(\alpha/2=0.05\) is \(t_{0.05,14}=1.7613\) (rounded to 4 decimal places).

Step2: Calculate the margin of error formula

The formula for the margin of error \(E\) for a confidence interval for the population mean when the population standard deviation \(\sigma\) is unknown (we use the sample standard deviation \(s\) as an estimate) is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\).
We are given \(s = 5\), \(n = 15\), and \(t_{\alpha/2}=1.7613\).
Substitute the values into the formula: \(E=1.7613\times\frac{5}{\sqrt{15}}\).
First, calculate \(\sqrt{15}\approx3.87298\). Then \(\frac{5}{\sqrt{15}}\approx\frac{5}{3.87298}\approx1.291\).
Then \(E = 1.7613\times1.291\approx2.27\).

Answer:

\(2.27\)