Question

use a t - test to test the claim about the population mean μ at the given level of significance α using the given statistics. assume the population is normally distributed
claim μ ≠ 24, α = 0.05 sample statistics x̄ = 21.9, s = 4.5, n = 14
a. h₀: μ ≠ 24
hₐ: μ < 24
b. h₀: μ = 24
hₐ: μ ≠ 24
c. h₀: μ ≤ 24
hₐ: μ > 24
d. h₀: μ ≠ 24
hₐ: μ = 24
what is the value of the standardized test statistic?
the standardized test statistic is - 1.75 (round to two decimal places as needed.)
what is the p - value of the test statistic?
p - value = \square (round to three decimal places as needed.)

Explanation:

Step1: Identify test type and parameters

We have a two - tailed t - test (since the alternative hypothesis \(H_a:\mu
eq24\)) with sample size \(n = 14\), sample mean \(\bar{x}=21.9\), sample standard deviation \(s = 4.5\), and hypothesized population mean \(\mu_0 = 24\). The degrees of freedom \(df=n - 1=14 - 1 = 13\). The test statistic \(t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}\) was calculated as \(t=- 1.75\) (from the given information).

Step2: Calculate P - value for two - tailed test

For a two - tailed t - test, the P - value is the probability that \(T_{df}\) is less than \(-|t|\) or greater than \(|t|\), where \(T_{df}\) follows a t - distribution with \(df\) degrees of freedom. We can use a t - distribution table or a statistical software. Using a t - distribution calculator or software with \(df = 13\) and \(t=-1.75\) (the absolute value \(|t| = 1.75\)):

The cumulative probability for \(t=-1.75\) with \(df = 13\) (left - tailed) can be found. The left - tailed probability \(P(T_{13}<-1.75)\) can be calculated. Using a t - table or a calculator (for example, in R: 2*pt(-1.75,13)), we get:

First, find the left - tailed probability for \(t=-1.75\) with \(df = 13\). The value of \(P(T_{13}<-1.75)\approx0.053\). Since it is a two - tailed test, the P - value is \(2\times0.053 = 0.106\) (rounded to three decimal places).

Answer:

\(0.106\)