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
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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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
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
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

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