Question

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 1

Answer:

$t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$

Step-by-step explanation:

We have a sample of size n = 35, $\bar{x} = 92$ and $s = 5$. As we want to construct a 95% confidence interval to estimate the value of the true population mean $\mu$, we should use the pivotal quantity given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value $t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$, 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 $P(-2.0322\leq T\leq 2.0322) = 0.95$ because of the simmetry of the t distribution.