77.4k views
4 votes
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)?

User Bmacharia
by
6.2k points

1 Answer

4 votes

Answer:


t_(\alpha/2) = t_(0.05/2) = t_(0.025) = 2.0322

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/√(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.

User Arvindh
by
6.0k points