Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, [tex]$P$[/tex]-value, critical value(s), and state the final conclusion.

Test the claim that the mean age of the prison population in one city is less than 26 years. Sample data are summarized as [tex]$n=25, \bar{x}=24.4$[/tex] years, and [tex]$s=9.2$[/tex] years. Use a significance level of [tex]$\alpha=0.05$[/tex].

Select the corresponding [tex]$P$[/tex]-value and final conclusion:

A. P-value: [tex]$p=0.1966$[/tex]. There is not sufficient evidence to support the claim that the mean age is less than 26 years.
B. P-value: [tex]$p=0.01966$[/tex]. There is not sufficient evidence to support the claim that the mean age is greater than 26 years.
C. P-value: [tex]$p=0.01966$[/tex]. There is sufficient evidence to support the claim that the mean age is less than 26 years.
D. P-value: [tex]$p=0.1966$[/tex]. There is sufficient evidence to support the claim that the mean age is less than 26 years.



Answer :

To test the claim that the mean age of the prison population in one city is less than 26 years, we will use a one-sample t-test. We'll follow these steps:

### Given Data:
- Sample size ([tex]\( n \)[/tex]) = 25
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 24.4 years
- Sample standard deviation ([tex]\( s \)[/tex]) = 9.2 years
- Population mean ([tex]\( \mu_0 \)[/tex]) = 26 years
- Significance level ([tex]\( \alpha \)[/tex]) = 0.05

### 1. State the Hypotheses
- Null Hypothesis (H₀): [tex]\(\mu \geq 26\)[/tex] (The mean age of the prison population is 26 years or more)
- Alternative Hypothesis (H₁): [tex]\(\mu < 26\)[/tex] (The mean age of the prison population is less than 26 years)

### 2. Calculate the Test Statistic ([tex]\( t \)[/tex]-value)
The test statistic for a one-sample t-test is calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]

Substituting the given values:
[tex]\[ t = \frac{24.4 - 26}{\frac{9.2}{\sqrt{25}}} \][/tex]
[tex]\[ t = \frac{-1.6}{\frac{9.2}{5}} \][/tex]
[tex]\[ t \approx -0.8696 \][/tex]

### 3. Determine Degrees of Freedom
Degrees of freedom ([tex]\( df \)[/tex]) for the t-test is given by:
[tex]\[ df = n - 1 = 25 - 1 = 24 \][/tex]

### 4. Find the Critical Value
For a one-tailed test at [tex]\( \alpha = 0.05 \)[/tex] and [tex]\( df = 24 \)[/tex], the critical value is obtained from t-distribution tables or computational tools:
[tex]\[ t_{critical} \approx -1.7109 \][/tex]

### 5. Calculate the P-value
The P-value is the probability that the test statistic will be less than or equal to the observed test statistic value under the null hypothesis. Using computational tools or tables, the P-value corresponding to [tex]\( t \approx -0.8696 \)[/tex] and [tex]\( df = 24 \)[/tex] is:
[tex]\[ P \approx 0.1966 \][/tex]

### 6. Make the Decision
We compare the P-value with the significance level:
- If [tex]\( P \leq \alpha \)[/tex], reject [tex]\( H₀ \)[/tex].
- If [tex]\( P > \alpha \)[/tex], do not reject [tex]\( H₀ \)[/tex].

Here, [tex]\( P = 0.1966 \)[/tex] and [tex]\( \alpha = 0.05 \)[/tex]. Since [tex]\( 0.1966 > 0.05 \)[/tex], we do not reject the null hypothesis.

### 7. State the Conclusion
Based on the comparison, we conclude:
There is not sufficient evidence to support the claim that the mean age of the prison population is less than 26 years.

### Summary:
- Test Statistic (t-value): [tex]\(\approx -0.8696\)[/tex]
- P-value: [tex]\(\approx 0.1966\)[/tex]
- Critical Value: [tex]\(\approx -1.7109\)[/tex]
- Conclusion: There is not sufficient evidence to support the claim that the mean age is less than 26 years.