Module Knowledge Check
Questiong
Justin:
An existing inventory for a test measuring self-esteem indicates that the scores have a standard deviation of 12 . A psychologist gave the self-esteem test to a random sample of 80 individuals, and their mean score was 59 . Construct a $90 \%$ confidence interval for the true mean of all test scores. Then give its lower fimit and upper timit.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)
\begin{tabular}{l}
\hline Lower limit: $\mathbb{~}$ \\
Upper limit:
\end{tabular}

Step 1 ：Given values are the sample mean (\(x_{bar}\)) as 59, standard deviation (\(\sigma\)) as 12, sample size (\(n\)) as 80, and z-score for 90% confidence interval (\(z\)) as 1.645.

Step 2 ：Calculate the margin of error using the formula \(z \times \frac{\sigma}{\sqrt{n}}\). Substituting the given values, we get the margin of error as approximately 2.207.

Step 3 ：Calculate the lower limit of the confidence interval using the formula \(x_{bar} - \text{margin of error}\). Substituting the given values, we get the lower limit as approximately 56.793.

Step 4 ：Calculate the upper limit of the confidence interval using the formula \(x_{bar} + \text{margin of error}\). Substituting the given values, we get the upper limit as approximately 61.207.

Step 5 ：Round the lower and upper limits to one decimal place. The lower limit is \(\boxed{56.8}\) and the upper limit is \(\boxed{61.2}\).