There are 3 parts to this questionThe 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 metersTo 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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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.)

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

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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