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 ≠ 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 = 0.106 (round to three decimal places as needed.)

Explanation:

Step1: Identify Hypotheses

The claim is \( \mu
eq 24 \), so the null hypothesis \( H_0 \) is the opposite of the alternative (or the claim's complement for two - tailed). So \( H_0:\mu = 24 \) and \( H_a:\mu
eq24 \), which matches option B.

Step2: Calculate Test Statistic (t - test)

The formula for the t - test statistic is \( t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \)
Given \( \bar{x} = 21.9 \), \( \mu = 24 \), \( s = 4.5 \), \( n = 14 \)
First, calculate the denominator: \( \frac{s}{\sqrt{n}}=\frac{4.5}{\sqrt{14}}\approx\frac{4.5}{3.7417}\approx1.2027 \)
Then, the numerator: \( \bar{x}-\mu=21.9 - 24=- 2.1 \)
Now, \( t=\frac{-2.1}{1.2027}\approx - 1.75 \) (rounded to two decimal places)

Step3: Calculate P - value

For a two - tailed test with \( t=-1.75 \) and degrees of freedom \( df=n - 1=14 - 1 = 13 \)
Using a t - distribution table or calculator, the P - value is the probability that \( T < - 1.75 \) or \( T>1.75 \)
From t - distribution, \( P(T < - 1.75)=0.053 \), so two - tailed P - value is \( 2\times0.053 = 0.106 \) (rounded to three decimal places)

Answer:

• Hypothesis: B. \( H_0:\mu = 24 \), \( H_a:\mu

eq24 \)

• Standardized test statistic: \(-1.75\)
• P - value: \(0.106\)