A simple random sample of 56 men results in a standard deviation of 10.09 beats per minute. Use the sample results to test the claim that men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute at the 5% significance level. Assume that the sample is from a normally distributed population.**What are the correct hypotheses? (Select the correct symbols and values.)**- $H_0:$ [Select an answer] beats per minute- Original claim = [Select an answer]- $H_1:$ [Select an answer] beats per minute**Enter the critical values, along with the significance level and degrees of freedom $\chi^2_{(\alpha, df)}$ below the graph. (Graph is for illustration only. No need to shade.)****Graph Description:** - A diagram titled "χ²-Distribution."- The graph shows a right-skewed curve, representing the chi-square distribution, with the x-axis labeled with values from 0 to 210 and the probability on the y-axis ranging from 0 to 0.12.- $χ^2_{\text{left}}$ < \[ \_\_\_\_\ \]- $χ^2_{\text{right}}$ = \[ \_\_\_\_\ \](Round to three decimal places.)**Test Statistic =** \[ \_\_\_\_\ \] (Round to three decimal places.)**Decision: [Select an answer]****Conclusion:** [Select an answer] the claim that men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute.

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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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A simple random sample of 56 men results in a standard deviation of 10.09 beats per minute. Use the sample results to test the claim that men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute at the 5% significance level. Assume that the sample is from a normally distributed population.

**What are the correct hypotheses? (Select the correct symbols and values.)**

- $H_0:$ [Select an answer] beats per minute
- Original claim = [Select an answer]

- $H_1:$ [Select an answer] beats per minute

**Enter the critical values, along with the significance level and degrees of freedom $\chi^2_{(\alpha, df)}$ below the graph. (Graph is for illustration only. No need to shade.)**

**Graph Description:**

- A diagram titled "χ²-Distribution."
- The graph shows a right-skewed curve, representing the chi-square distribution, with the x-axis labeled with values from 0 to 210 and the probability on the y-axis ranging from 0 to 0.12.

- $χ^2_{\text{left}}$ < \[ \_\_\_\_\ \]

- $χ^2_{\text{right}}$ = \[ \_\_\_\_\ \]

(Round to three decimal places.)

**Test Statistic =** \[ \_\_\_\_\ \] (Round to three decimal places.)

**Decision: [Select an answer]**

**Conclusion:** [Select an answer] the claim that men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute.

Expert Solution

Step 1

The claim of the test is that men have pulse rates that have a standard deviation equal to the population standard deviation $(σ)$ of 10.63 beats per minute. The hypothesis is,

Null hypothesis:

$H_{0} : σ = 10.63$ beats per minutes

Alternative hypothesis:

$H_{1} : σ \neq 10.63$ beats per minutes

The original claim is men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute. That is, $H_{0}$ .

Thus, the original claim is $H_{0}$ .

Step 2

The sample size is 56. The significance level is 5%. The degrees of freedom is,

$\begin{array}{rcl} d f & = & n - 1 \\ = & 56 - 1 \\ = & 55 \end{array}$

The degrees of freedom is 55.

Critical value for $χ_{L}^{2}$ :

The right-tail area is,

$\begin{array}{rcl} \frac{1 + α}{2} & = & \frac{1 + 0.95}{2} \\ = & 0.975 \end{array}$

The critical value of at 55 degrees of freedom can be obtained using the excel formula “=CHISQ.INV.RT(0.975,55)”. The critical value is 36.398.

The value of $χ_{L}^{2}$ is 36.398.

Critical value for $χ_{U}^{2}$ :

The right-tail area is,

$\begin{array}{rcl} \frac{1 - α}{2} & = & \frac{1 - 0.95}{2} \\ = & 0.025 \end{array}$

The critical value of at 55 degrees of freedom can be obtained using the excel formula “=CHISQ.INV.RT(0.025,55)”. The critical value is 77.380.

The value of $χ_{U}^{2}$ is 77.380.

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