(a) Do the data provide sufficient evidence to indicate that assessor 1 tends to give higher assessments than assessor 2? (Use a = 0.05) State the null and alternative hypotheses. O H,: 4 = 0 versus H: 4, < 0 a O Ho: H4 = 0 versus H: + 0 O Ho: Hd < 0 versus H,: u > 0 * 0 versus H: O Ho: Hd* = 0 versus H: >0 State the test statistic. t = State the p-value. State the conclusion. O H, is rejected. There is sufficient evidence to indicate assessor A gives higher assessments than assessor B. O H, is not rejected. There is sufficient evidence to indicate assessor A gives higher assessments than assessor B. O H, is rejected. There is insufficient evidence to indicate assessor A gives higher assessments than assessor B. O H, is not rejected. There is insufficient evidence to indicate assessor A gives higher assessments than assessor B. (b) Estimate a 95% lower one-sided confidence bound. (Use u, - u,.)

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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17 question has part a,b, and c

In response to a complaint that a particular tax assessor (1) was biased, an experiment was conducted to compare the assessor named in the complaint with another tax assessor (2) from the same office. Eight properties were selected, and each was assessed by both assessors. The assessments (in thousands of dollars) are shown in the table.

| Property | Assessor 1 | Assessor 2 |
|----------|------------|------------|
| 1        | 276.2      | 274.8      |
| 2        | 287.5      | 287.5      |
| 3        | 280.1      | 276.6      |
| 4        | 295.6      | 299.7      |
| 5        | 268.4      | 268.0      |
| 6        | 282.4      | 280.0      |
| 7        | 276.3      | 275.4      |
| 8        | 279.8      | 279.0      |

Use the MINITAB printout to answer the questions that follow. (Use the exact values found in the MINITAB output.)

**Paired T-Test and CI: Assessor 1, Assessor 2**

**Descriptive Statistics**

| Sample     | N | Mean  | StDev | SE Mean |
|------------|---|-------|-------|---------|
| Assessor 1 | 8 | 280.7 | 6.13  | 2.08    |
| Assessor 2 | 8 | 278.96| 6.87  | 2.43    |

**Estimation for Paired Differences**

|         | Mean | StDev | SE Mean | 95% Lower Bound for μ Difference |
|---------|------|-------|---------|-----------------------------------|
| 1.725   | 2.060| 0.728 | 0.345   |

- μ_difference: mean of (Assessor 1 - Assessor 2)

**Test**

- Null hypothesis H₀: μ_difference = 0
- Alternative hypothesis Hₐ: μ_difference > 0

|         |         |
|---------|---------|
| T-Value | 2.37    |
| P-Value | 0.025   |
Transcribed Image Text:In response to a complaint that a particular tax assessor (1) was biased, an experiment was conducted to compare the assessor named in the complaint with another tax assessor (2) from the same office. Eight properties were selected, and each was assessed by both assessors. The assessments (in thousands of dollars) are shown in the table. | Property | Assessor 1 | Assessor 2 | |----------|------------|------------| | 1 | 276.2 | 274.8 | | 2 | 287.5 | 287.5 | | 3 | 280.1 | 276.6 | | 4 | 295.6 | 299.7 | | 5 | 268.4 | 268.0 | | 6 | 282.4 | 280.0 | | 7 | 276.3 | 275.4 | | 8 | 279.8 | 279.0 | Use the MINITAB printout to answer the questions that follow. (Use the exact values found in the MINITAB output.) **Paired T-Test and CI: Assessor 1, Assessor 2** **Descriptive Statistics** | Sample | N | Mean | StDev | SE Mean | |------------|---|-------|-------|---------| | Assessor 1 | 8 | 280.7 | 6.13 | 2.08 | | Assessor 2 | 8 | 278.96| 6.87 | 2.43 | **Estimation for Paired Differences** | | Mean | StDev | SE Mean | 95% Lower Bound for μ Difference | |---------|------|-------|---------|-----------------------------------| | 1.725 | 2.060| 0.728 | 0.345 | - μ_difference: mean of (Assessor 1 - Assessor 2) **Test** - Null hypothesis H₀: μ_difference = 0 - Alternative hypothesis Hₐ: μ_difference > 0 | | | |---------|---------| | T-Value | 2.37 | | P-Value | 0.025 |
**Educational Content for Hypothesis Testing and Confidence Bounds**

(a) **Assessing Evidence for Higher Assessments**
- **Objective:** Evaluate if Assessor 1 tends to give higher assessments than Assessor 2 with a significance level of 0.05.

- **Null and Alternative Hypotheses:**
  - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d < 0 \)
  - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d \neq 0 \)
  - \( H_0: \mu_d < 0 \) versus \( H_a: \mu_d > 0 \)
  - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d = 0 \)
  - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d > 0 \)

- **Test Statistic:**
  - \( t = \) [Input required]

- **P-value:**
  - [Input required]

- **Conclusions:**
  - \( H_0 \) is rejected. Sufficient evidence shows Assessor A gives higher assessments than Assessor B.
  - \( H_0 \) is not rejected. Sufficient evidence indicates Assessor A gives higher assessments than Assessor B.
  - \( H_0 \) is rejected. Insufficient evidence shows Assessor A gives higher assessments than Assessor B.
  - \( H_0 \) is not rejected. Insufficient evidence indicates Assessor A gives higher assessments than Assessor B.

(b) **Estimating a Confidence Bound**
- **95% Lower One-Sided Confidence Bound:**
  - \( \mu_1 - \mu_2 = \) [Input required]

(c) **Assumptions for Valid Inference**
- Select the necessary assumptions for parts (a) and (b):

  - [ ] The assessments must be normally distributed.
  - [ ] The properties must be independently selected.
  - [ ] The sample size must be greater than 5 for each assessor.
  - [ ] The properties must be randomly selected.
  - [ ] The variance of the data sets for both assessors must be equal.

This exercise teaches you to construct hypotheses, calculate statistics, and determine the validity of your results based on assumptions. Understanding
Transcribed Image Text:**Educational Content for Hypothesis Testing and Confidence Bounds** (a) **Assessing Evidence for Higher Assessments** - **Objective:** Evaluate if Assessor 1 tends to give higher assessments than Assessor 2 with a significance level of 0.05. - **Null and Alternative Hypotheses:** - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d < 0 \) - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d \neq 0 \) - \( H_0: \mu_d < 0 \) versus \( H_a: \mu_d > 0 \) - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d = 0 \) - \( H_0: \mu_d = 0 \) versus \( H_a: \mu_d > 0 \) - **Test Statistic:** - \( t = \) [Input required] - **P-value:** - [Input required] - **Conclusions:** - \( H_0 \) is rejected. Sufficient evidence shows Assessor A gives higher assessments than Assessor B. - \( H_0 \) is not rejected. Sufficient evidence indicates Assessor A gives higher assessments than Assessor B. - \( H_0 \) is rejected. Insufficient evidence shows Assessor A gives higher assessments than Assessor B. - \( H_0 \) is not rejected. Insufficient evidence indicates Assessor A gives higher assessments than Assessor B. (b) **Estimating a Confidence Bound** - **95% Lower One-Sided Confidence Bound:** - \( \mu_1 - \mu_2 = \) [Input required] (c) **Assumptions for Valid Inference** - Select the necessary assumptions for parts (a) and (b): - [ ] The assessments must be normally distributed. - [ ] The properties must be independently selected. - [ ] The sample size must be greater than 5 for each assessor. - [ ] The properties must be randomly selected. - [ ] The variance of the data sets for both assessors must be equal. This exercise teaches you to construct hypotheses, calculate statistics, and determine the validity of your results based on assumptions. Understanding
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