assuming the population has an approximate normal distribution, if a sample size $n = 13$ has a sample mean…

assuming the population has an approximate normal distribution, if a sample size $n = 13$ has a sample mean $\bar{x}=42$ with a sample standard deviation $s = 5$, find the margin of error at a 90% confidence level. round the answer to two decimal places.

assuming the population has an approximate normal distribution, if a sample size $n = 13$ has a sample mean $\bar{x}=42$ with a sample standard deviation $s = 5$, find the margin of error at a 90% confidence level. round the answer to two decimal places.

Answer

Explanation:

Step1: Determine degrees of freedom

The degrees of freedom $df=n - 1$. Given $n = 13$, so $df=13 - 1=12$.

Step2: Find the t - value

For a 90% confidence level and $df = 12$, looking up in the t - distribution table, the two - tailed t - value $t_{\alpha/2}=1.782$.

Step3: Calculate the margin of error

The formula for the margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$. Substitute $t_{\alpha/2}=1.782$, $s = 5$, and $n = 13$ into the formula. So $E=1.782\times\frac{5}{\sqrt{13}}\approx1.782\times\frac{5}{3.606}\approx2.47$.

Answer:

$2.47$