Question

in a survey, 25 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 $4. find the margin of error at a 90% confidence level. give your answer to two decimal places.

Explanation:

Step1: Determine the z - score

For a 90% confidence level, the significance level $\alpha=1 - 0.90 = 0.10$, and $\alpha/2=0.05$. Looking up in the standard normal distribution table, the z - score $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step2: Identify sample size, standard deviation

The sample size $n = 25$, and the sample standard deviation $s = 4$.

Step3: Calculate the margin of error formula

The formula for the margin of error $E$ for a confidence interval when the population standard - deviation is unknown (but sample size is reasonably large or population is normally distributed) is $E = z_{\alpha/2}\frac{s}{\sqrt{n}}$.
Substitute $z_{\alpha/2}=1.645$, $s = 4$, and $n = 25$ into the formula:
$E=1.645\times\frac{4}{\sqrt{25}}$.
First, calculate $\sqrt{25}=5$. Then $\frac{4}{\sqrt{25}}=\frac{4}{5}=0.8$.
Finally, $E = 1.645\times0.8=1.316$.

Answer:

$1.32$