Question

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(99 \%\) confidence, \(n=24\) c. \(90 \%\) confidence, \(n=13\)

Short answer

The t critical value for 95% confidence level and sample size 17 is \(2.12\), for 99% confidence level and sample size 24 is \(2.807\), and for 90% confidence level and sample size 13 is \(1.782\).

Step-by-step solution

Confidence Level 95%, Sample Size 17

Find the t critical value for 95% confidence level and sample size 17. The degrees of freedom is \(n-1 = 17 - 1 = 16\). If you check the t-distribution chart or use t-distribution calculator, the t critical value is roughly \(2.12\).

Confidence Level 99%, Sample Size 24

Find the t critical value for 99% confidence level and sample size 24. The degrees of freedom is \(n-1 = 24 - 1 = 23\). If you check the t-distribution chart or use t-distribution calculator, the t critical value is roughly \(2.807\).

Confidence Level 90%, Sample Size 13

Find the t critical value for 90% confidence level and sample size 13. The degrees of freedom is \(n-1 = 13 - 1 = 12\). If you check the t-distribution chart or use t-distribution calculator, the t critical value is roughly \(1.782\).