rice in the southern part of the county and would like to test the claim. A simple random rices is taken from each region. The results are shown below. ean riance Oservations oled Variance pothesized Mean Difference tat T<=t) one-tail Critical one-tail Southern Northern 155.056 168.889 345.938 560.928 18 18 453.433 0 34 1.949 0.030 2.441 P = Ex:
rice in the southern part of the county and would like to test the claim. A simple random rices is taken from each region. The results are shown below. ean riance Oservations oled Variance pothesized Mean Difference tat T<=t) one-tail Critical one-tail Southern Northern 155.056 168.889 345.938 560.928 18 18 453.433 0 34 1.949 0.030 2.441 P = Ex:
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![A real estate agent believes that the mean home price in the northern part of a county is higher than the mean price in the southern part of the county and would like to test the claim. A simple random sample of housing prices is taken from each region. The results are shown below.
### Table of Results:
- **Southern Region:**
- Mean: 155.056
- Variance: 345.938
- Observations: 18
- **Northern Region:**
- Mean: 168.889
- Variance: 560.928
- Observations: 18
- **Pooled Variance:** 453.433
- **Hypothesized Mean Difference:** 0
- **Degrees of Freedom (df):** 34
- **t Statistic:** 1.949
- **P(T<=t) one-tail:** 0.030
- **t Critical one-tail:** 2.441
- **P(T<=t) two-tail:** 0.060
- **t Critical two-tail:** 2.728
- **Confidence Level:** 99%
### Graph Explanation:
The graph displays a normal distribution curve. There is a shaded area under the curve representing the p-value. The t-value is marked on the x-axis at the point where it occurs.
- **p-value (p):** Enter p-value
- **t-value (t):** Enter t-value
### Calculations:
- **n (Sample size):** 9
- **Degrees of freedom:** Input
- **Point estimate for the southern part of the county (x̄ₛ):** Enter value
- **Point estimate for the northern part of the county (x̄ₙ):** Enter value](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafa77d54-ffb2-4772-a246-c955b3ec6a36%2Fa372be1e-64fe-4dd4-8e9c-7d0b95b9986a%2Fnssxayv_processed.png&w=3840&q=75)
Transcribed Image Text:A real estate agent believes that the mean home price in the northern part of a county is higher than the mean price in the southern part of the county and would like to test the claim. A simple random sample of housing prices is taken from each region. The results are shown below.
### Table of Results:
- **Southern Region:**
- Mean: 155.056
- Variance: 345.938
- Observations: 18
- **Northern Region:**
- Mean: 168.889
- Variance: 560.928
- Observations: 18
- **Pooled Variance:** 453.433
- **Hypothesized Mean Difference:** 0
- **Degrees of Freedom (df):** 34
- **t Statistic:** 1.949
- **P(T<=t) one-tail:** 0.030
- **t Critical one-tail:** 2.441
- **P(T<=t) two-tail:** 0.060
- **t Critical two-tail:** 2.728
- **Confidence Level:** 99%
### Graph Explanation:
The graph displays a normal distribution curve. There is a shaded area under the curve representing the p-value. The t-value is marked on the x-axis at the point where it occurs.
- **p-value (p):** Enter p-value
- **t-value (t):** Enter t-value
### Calculations:
- **n (Sample size):** 9
- **Degrees of freedom:** Input
- **Point estimate for the southern part of the county (x̄ₛ):** Enter value
- **Point estimate for the northern part of the county (x̄ₙ):** Enter value
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
From the given output.
There are 18 observations in southern and northern part of the country.
Hence, n=18
From the output, the degrees of freedom is,
df=34
Step by step
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