A study of 35 golfers showed that their average score on a particular course was 92. The sample standard deviation was 5. We want to construct a 95% confidence interval to estimate the value of the true population mean. What is the t-value that should be used to compute the confidence interval (to the third decimal place)?

Answer:[tex]t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322[/tex]Step-by-step explanation:We have a sample of size n = 35, [tex]\bar{x} = 92[/tex] and [tex]s = 5[/tex]. As we want to construct a 95% confidence interval to estimate the value of the true population mean [tex]\mu[/tex], we should use the pivotal quantity given by [tex]T = \frac{\bar{X}-\mu}{S/\sqrt{n}}[/tex] which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value [tex]t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322[/tex], i.e., regarding the t-distribution with 34 degrees of freedom, we have an area of 0.025 above 2.0322 and below the probability density function. The rationale behind this is that [tex]P(-2.0322\leq T\leq 2.0322) = 0.95[/tex] because of the simmetry of the t distribution.