Find the value of the standard score, z, and determine whether to reject the null hypothesis at a 0.01 significance level. Is the alternative hypothesis supported? Ho: μ = 18.1 meters, H₂: μ# 18.1 meters, n = 36, x= 18.5 meters, o = 1.4 meters The value of the standard score is (Round to two decimal places as needed.)

MATLAB: An Introduction with Applications
6th Edition
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Author:Amos Gilat
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Chapter1: Starting With Matlab
Section: Chapter Questions
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The value of the standard score is (round to two decimal places as needed)

The critical value(s) is/are (use a comma to separate the answers as needed. Round to two decimal places as needed.)

Determine whether the alternative hypothesis is supported at a 0.01 significance level.

**Hypothesis Testing and Standard Score Calculation**

In this example, we aim to find the value of the standard score \( z \) and determine whether to reject the null hypothesis at a 0.01 significance level. We will also evaluate if the alternative hypothesis is supported based on the value of \( z \).

**Given:**
- Null Hypothesis (\( H_0 \)): \( \mu = 18.1 \) meters
- Alternative Hypothesis (\( H_1 \)): \( \mu \ne 18.1 \) meters
- Sample size (\( n \)): 36
- Sample mean (\( \overline{x} \)): 18.5 meters
- Population standard deviation (\( \sigma \)): 1.4 meters

To find the standard score (\( z \)):

1. **Formula to calculate \( z \)-score:**
\[ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

2. **Substitute the given values into the formula:**
\[ \overline{x} = 18.5 \]
\[ \mu = 18.1 \]
\[ \sigma = 1.4 \]
\[ n = 36 \]

\[ z = \frac{18.5 - 18.1}{\frac{1.4}{\sqrt{36}}} \]

3. **Calculate the denominator:**
\[ \frac{1.4}{\sqrt{36}} = \frac{1.4}{6} = 0.2333 \]

4. **Calculate the numerator:**
\[ 18.5 - 18.1 = 0.4 \]

5. **Combine the values:**
\[ z = \frac{0.4}{0.2333} \approx 1.71 \]

**Conclusion:**
The value of the standard score is **1.71** (rounded to two decimal places).

To determine whether to reject the null hypothesis, compare the \( z \)-score with the critical value for a 0.01 significance level in a two-tailed test. Since \( \left| z \right| = 1.71 \) does not exceed the critical value of approximately 2.58 for a 0.01 significance level, we do not reject the null hypothesis.

**Interpretation:**
Given the \( z
Transcribed Image Text:**Hypothesis Testing and Standard Score Calculation** In this example, we aim to find the value of the standard score \( z \) and determine whether to reject the null hypothesis at a 0.01 significance level. We will also evaluate if the alternative hypothesis is supported based on the value of \( z \). **Given:** - Null Hypothesis (\( H_0 \)): \( \mu = 18.1 \) meters - Alternative Hypothesis (\( H_1 \)): \( \mu \ne 18.1 \) meters - Sample size (\( n \)): 36 - Sample mean (\( \overline{x} \)): 18.5 meters - Population standard deviation (\( \sigma \)): 1.4 meters To find the standard score (\( z \)): 1. **Formula to calculate \( z \)-score:** \[ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] 2. **Substitute the given values into the formula:** \[ \overline{x} = 18.5 \] \[ \mu = 18.1 \] \[ \sigma = 1.4 \] \[ n = 36 \] \[ z = \frac{18.5 - 18.1}{\frac{1.4}{\sqrt{36}}} \] 3. **Calculate the denominator:** \[ \frac{1.4}{\sqrt{36}} = \frac{1.4}{6} = 0.2333 \] 4. **Calculate the numerator:** \[ 18.5 - 18.1 = 0.4 \] 5. **Combine the values:** \[ z = \frac{0.4}{0.2333} \approx 1.71 \] **Conclusion:** The value of the standard score is **1.71** (rounded to two decimal places). To determine whether to reject the null hypothesis, compare the \( z \)-score with the critical value for a 0.01 significance level in a two-tailed test. Since \( \left| z \right| = 1.71 \) does not exceed the critical value of approximately 2.58 for a 0.01 significance level, we do not reject the null hypothesis. **Interpretation:** Given the \( z
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