SOLUTION: In​ 1960, census results indicated that the age at which men in a certain region first married had a mean of 24.3 years. It is widely suspected that young people today are waitin
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Question 1197976: In​ 1960, census results indicated that the age at which men in a certain region first married had a mean of 24.3 years. It is widely suspected that young people today are waiting longer to get married. We want to find out if the mean age of first marriage has increased since then. Complete parts a through f below.
​a) Write appropriate hypotheses.
​ The men in our sample married at an average age of 25.2 ​years, with a standard deviation of 5.4 years. That results in a​ t-statistic of 1.054 . What is the​ P-value for​ this?
The​ P-value is ?
Answer by proyaop(69) (Show Source): You can put this solution on YOUR website!
To address the problem, let's go through the parts step by step:
---
### **a) Write Appropriate Hypotheses**
The null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)) are formulated as follows:
- \( H_0 \): The mean age at which men first marry today is equal to 24.3 years (\( \mu = 24.3 \)).
- \( H_a \): The mean age at which men first marry today is greater than 24.3 years (\( \mu > 24.3 \)).
This is a **one-tailed test** since we are testing if the mean has **increased**.
---
### **b) Compute the \( P \)-value**
#### **Step 1: Recall the Given Data**
- Sample mean (\( \bar{x} \)) = 25.2 years
- Population mean under \( H_0 \) (\( \mu_0 \)) = 24.3 years
- Sample standard deviation (\( s \)) = 5.4 years
- Sample size (\( n \)) = Not directly given but can be inferred using the \( t \)-statistic formula and other values.
#### **Step 2: Use the Formula for the \( t \)-Statistic**
The \( t \)-statistic is calculated as:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
From the problem, we know \( t = 1.054 \). Rearrange the formula to find \( n \):
\[
1.054 = \frac{25.2 - 24.3}{5.4 / \sqrt{n}}
\]
\[
1.054 = \frac{0.9}{5.4 / \sqrt{n}}
\]
\[
5.4 / \sqrt{n} = 0.9 / 1.054 \approx 0.854
\]
\[
\sqrt{n} = 5.4 / 0.854 \approx 6.32
\]
\[
n \approx 6.32^2 \approx 39.95 \approx 40
\]
Thus, the sample size is \( n = 40 \).
---
#### **Step 3: Determine Degrees of Freedom**
Degrees of freedom (\( df \)) for a \( t \)-test:
\[
df = n - 1 = 40 - 1 = 39
\]
---
#### **Step 4: Find the \( P \)-value**
Using a \( t \)-distribution table or software, find the \( P \)-value for \( t = 1.054 \) with \( df = 39 \) for a one-tailed test.
1. From the table or calculator, the \( P \)-value for \( t = 1.054 \) with \( df = 39 \) is approximately \( 0.149 \).
---
### **Final Answer**
The \( P \)-value is:
\[
\boxed{0.149}
\]
This indicates there is no strong evidence to reject the null hypothesis at common significance levels (e.g., \( \alpha = 0.05 \) or \( \alpha = 0.10 \)).
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