Question

use a t - test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. assume the population is normally distributed. claim: μ ≥ 7900; α = 0.10 sample statistics: (overline{x}=7600), s = 440, n = 20 what are the null and alternative hypotheses? a (h_{0}:mu
eq7900) (h_{a}:mu = 7900) b (h_{0}:muleq7900) (h_{a}:mu>7900) c (h_{0}:mu = 7900) (h_{a}:mu
eq7900) d (h_{0}:mugeq7900) (h_{a}:mu<7900) what is the value of the standardized test statistic? the standardized test statistic is - 3.05. (round to two decimal places as needed.) what is the p - value? p = (round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 20\), so \(df=20-1 = 19\).

Step2: Use the t - distribution table or calculator for P - value

We have a one - tailed test (since \(H_{a}:\mu<7900\)) and \(t=-3.05\).
Using a t - distribution table or a calculator (e.g., in R: pt(-3.05, 19)), we find the P - value.
The P - value is the probability that \(T

Answer:

\(P\approx0.003\)