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Deposits into a bank account are represented by positive numbers. Withdrawals from the

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Paul [167]3 years ago

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A study of 35 golfers showed that their average score on a particular course was 92. The sample standard deviation was 5. We wan

svlad2 [7]

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.