Question

use a t - test to test the claim about the population mean \\(\mu\\) at the given level of significance \\(\alpha\\) using the given sample statistics. assume the population is normally distributed.\
claim: \\(\mu \geq 7800\\); \\(\alpha = 0.05\\) \quad sample statistics: \\(\bar{x} = 7500\\), \\(s = 420\\), \\(n = 24\\)\
\
a. \\(h_0: \mu \geq 7800\\)\
\\(h_a: \mu < 7800\\) \quad \quad \quad \quad b. \\(h_0: \mu = 7800\\)\
\\(h_a: \mu \
eq 7800\\)\
c. \\(h_0: \mu \leq 7800\\)\
\\(h_a: \mu > 7800\\) \quad \quad \quad \quad d. \\(h_0: \mu \
eq 7800\\)\
\\(h_a: \mu = 7800\\)\
\
what is the value of the standardized test statistic?\
the standardized test statistic is - 3.5. (round to two decimal places as needed.)\
\
what is the p - value?\
\\(p = \square\\) (round to three decimal places as needed.)

Explanation:

Step1: Identify the test type and distribution

This is a t - test for the population mean. The test statistic is \(t=- 3.5\) (from the given information), and the degrees of freedom \(df=n - 1=24 - 1 = 23\). Since the alternative hypothesis is \(H_{a}:\mu<7800\), this is a left - tailed test.

Step2: Calculate the P - value

We need to find the probability that \(t_{df = 23}<-3.5\). Using a t - distribution table or a statistical software (such as a TI - 84 Plus or R), for a t - distribution with \(df = 23\) and \(t=-3.5\), the P - value is the area to the left of \(t=-3.5\) under the t - curve.
Using a t - table or a calculator function (for example, in R, pt(-3.5,23)), we find that the P - value is approximately \(0.001\) (more precisely, using a calculator: the cumulative distribution function for t - distribution with \(df = 23\) at \(t=-3.5\) gives \(P(t_{23}<-3.5)\approx0.0009\), which rounds to \(0.001\) when rounded to three decimal places).

Answer:

\(0.001\)