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 ≥ 7800; α = 0.05 sample statistics: (overline{x}=7500), s = 430, n = 24
what are the null and alternative hypotheses?
a. (h_0: mu = 7800)
(h_a: mu
eq7800)
b. (h_0: mugeq7800)
(h_a: mu < 7800)
c. (h_0: muleq7800)
(h_a: mu > 7800)
d. (h_0: mu
eq7800)
(h_a: mu = 7800)
what is the value of the standardized test statistic?
the standardized test statistic is - 3.42. (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 \(n = 24\), so \(df=24-1 = 23\).

Step2: Find the P - value

Since this is a left - tailed test (because \(H_{a}:\mu\lt7800\)) and the test statistic \(t=-3.42\).
We use the t - distribution table or a calculator with a t - distribution function. Using a calculator (for example, in R: pt(-3.42,23)), the P - value is calculated.

Answer:

The P - value is \(0.001\) (rounded to three decimal places).