Help please ### Hypothesis Testing Example 1#### Null Hypothesis: The population average is 40.0 cm#### Alternative Hypothesis:The population average is not 40.0 cm#### Rejection Region:Two-Tailed Test where t < -1.98 and t > 1.98#### Test Statistic:t = 3.75#### p-value:p = 0.0003#### Conclusion:Reject the null hypothesis since the test statistic is in the rejection region and the p-value is less than the significance level.### Hypothesis Testing Example 2#### Null Hypothesis: The population average is 40.6 cm#### Alternative Hypothesis:The population average is not 40.6 cm#### Rejection Region:Two-Tailed Test where t < -1.99 and t > 1.98#### Test Statistic:t = 3.75#### p-value:p = 0.0002#### Conclusion:Reject the null hypothesis since the test statistic is in the rejection region and the p-value is less than the significance level.---In these examples, we see the process of hypothesis testing, where certain conditions or averages of populations are tested statistically to determine if they statistically significantly differ from a hypothesized average. The key steps involve determining the null and alternative hypotheses, defining the rejection region, computing the test statistic, and then comparing the p-value to a significance threshold to make a conclusion. **Hypothesis Testing Example: Average Head Circumference of Two-Month-Old Babies**A random sample of 100 two-month-old babies is obtained, and the mean head circumference is found to be 40.6 cm. Assume that the population standard deviation is known to be 1.6 cm. Using a 5% significance level, determine if the average circumference of all two-month-old babies is not 40.0 cm.### Hypotheses:- **Null Hypothesis (H₀):** The population average is 40.0 cm- **Alternative Hypothesis (H₁):** The population average is not 40.0 cm### Rejection Region:- **Two-Tailed Test:** z less than -1.96 or z greater than 1.96### Test Statistic:- $ z = 3.75 $### p-value:- $ p = 0.0002 $### Conclusion:Reject the null hypothesis since the test statistic is in the rejection region and the p-value is less than the significance level (0.05).---In this case, the test statistic of $ z = 3.75 $ falls outside the range of -1.96 to 1.96, which makes it significant. The p-value of 0.0002 corroborates this finding, indicating that there is strong evidence against the null hypothesis. Thus, we conclude that the average head circumference of two-month-old babies is significantly different from 40.0 cm.

Given, n=100 x¯=40.6 σ=1.6 μ=40

A random sample of 100 two-month old babies is obtained, and the mean head circumference is found to be 40.6 cm. Assume that the population standard deviation is known to be 1.6 cm. Using a 5% significance level, determine if the average circumference of all two-month old babies is not 40.0 cm. Null Hypothesis: The population average is 40.6 cm Alternative Hypothesis: The population average is not 40.6 cm Rejection Region: Two-Tailed Test z less than -1.96 and z greater than 1.96 Test Statistic: z = 3.75 p-value: p = 0.0002 Conclusion: Reject the null hypothesis since the test statistic is in the rejection region and p-value is less than the significance O Null Hypothesis: The population average is 40.0 cm Alternative Hypothesis: The population average is not 40.0 cm Rejection Region: Two-Tailed Test z less than -1.96 and z greater than 1.96 Test Statistic: z = 3.75 p-value: p = 0.0002 Conclusion: Reject the null hypothesis since the test statistic is in the rejection region and p-value is less than the significance

A random sample of 100 two-month old babies is obtained, and the mean head circumference is found to be 40.6 cm. Assume that the population standard deviation is known to be 1.6 cm. Using a 5% significance level, determine if the average circumference of all two-month old babies is not 40.0 cm. Null Hypothesis: The population average is 40.6 cm Alternative Hypothesis: The population average is not 40.6 cm Rejection Region: Two-Tailed Test z less than -1.96 and z greater than 1.96 Test Statistic: z = 3.75 p-value: p = 0.0002 Conclusion: Reject the null hypothesis since the test statistic is in the rejection region and p-value is less than the significance O Null Hypothesis: The population average is 40.0 cm Alternative Hypothesis: The population average is not 40.0 cm Rejection Region: Two-Tailed Test z less than -1.96 and z greater than 1.96 Test Statistic: z = 3.75 p-value: p = 0.0002 Conclusion: Reject the null hypothesis since the test statistic is in the rejection region and p-value is less than the significance

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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Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

Help please

### Hypothesis Testing Example 1

#### Null Hypothesis:
The population average is 40.0 cm

#### Alternative Hypothesis:
The population average is not 40.0 cm

#### Rejection Region:
Two-Tailed Test where t < -1.98 and t > 1.98

#### Test Statistic:
t = 3.75

#### p-value:
p = 0.0003

#### Conclusion:
Reject the null hypothesis since the test statistic is in the rejection region and the p-value is less than the significance level.

### Hypothesis Testing Example 2

#### Null Hypothesis:
The population average is 40.6 cm

#### Alternative Hypothesis:
The population average is not 40.6 cm

#### Rejection Region:
Two-Tailed Test where t < -1.99 and t > 1.98

#### Test Statistic:
t = 3.75

#### p-value:
p = 0.0002

#### Conclusion:
Reject the null hypothesis since the test statistic is in the rejection region and the p-value is less than the significance level.

---

In these examples, we see the process of hypothesis testing, where certain conditions or averages of populations are tested statistically to determine if they statistically significantly differ from a hypothesized average. The key steps involve determining the null and alternative hypotheses, defining the rejection region, computing the test statistic, and then comparing the p-value to a significance threshold to make a conclusion.

**Hypothesis Testing Example: Average Head Circumference of Two-Month-Old Babies**

A random sample of 100 two-month-old babies is obtained, and the mean head circumference is found to be 40.6 cm. Assume that the population standard deviation is known to be 1.6 cm. Using a 5% significance level, determine if the average circumference of all two-month-old babies is not 40.0 cm.

### Hypotheses:

- **Null Hypothesis (H₀):** The population average is 40.0 cm
- **Alternative Hypothesis (H₁):** The population average is not 40.0 cm

### Rejection Region:
- **Two-Tailed Test:** z less than -1.96 or z greater than 1.96

### Test Statistic:
- $ z = 3.75 $

### p-value:
- $ p = 0.0002 $

### Conclusion:
Reject the null hypothesis since the test statistic is in the rejection region and the p-value is less than the significance level (0.05).

---

In this case, the test statistic of $ z = 3.75 $ falls outside the range of -1.96 to 1.96, which makes it significant. The p-value of 0.0002 corroborates this finding, indicating that there is strong evidence against the null hypothesis. Thus, we conclude that the average head circumference of two-month-old babies is significantly different from 40.0 cm.

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