Question

if n = 31, $\bar{x}$(x - bar)=45, and s = 20, construct a confidence interval at a 90% confidence level. assume the data came from a normally distributed population. give your answers to one decimal place.

Explanation:

Step1: Determine degrees of freedom

The degrees of freedom $df=n - 1$. Given $n = 31$, so $df=31-1=30$.

Step2: Find the t - value

For a 90% confidence level and $df = 30$, the two - tailed t - value $t_{\alpha/2}$ from the t - distribution table is approximately $t_{0.05,30}=1.697$.

Step3: Calculate the margin of error

The formula for the margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$. Substitute $t_{\alpha/2}=1.697$, $s = 20$, and $n = 31$ into the formula. $E=1.697\times\frac{20}{\sqrt{31}}\approx1.697\times\frac{20}{5.568}\approx1.697\times3.592\approx6.1$.

Step4: Construct the confidence interval

The confidence interval for the population mean $\mu$ is given by $\bar{x}-E<\mu<\bar{x} + E$. Substitute $\bar{x}=45$ and $E\approx6.1$ into the formula. So $45 - 6.1<\mu<45+6.1$.

Answer:

$38.9<\mu<51.1$