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.
### 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.
