#5 The following text appears to be for educational purposes involving statistical analysis and hypothesis testing:---1. **Testing Iron Concentration in Water**: - Does the sample mean concentration exceed the company’s claim of 2.5? Use a 5% level of significance to test this hypothesis based on samples taken from an industrial park. - Calculate a 95% confidence interval for the sample data.2. **Studying Chemical Z in Blood**: - Assess if the average concentration of Chemical Z is higher than 20%. This involves testing at a 5% significance level and calculating a 99% confidence interval from the given data. Table: Concentration Data Across Months\[\begin{array}{|c|c|c|}\hline\text{Month} & \text{Sample 1} & \text{Sample 2} \\\hline\text{Oct} & 3.1 & 1.5 \\\text{Nov} & 2.0 & 1.4 \\\text{Dec} & 1.6 & 1.6 \\\text{Jan} & 3.3 & 2.2 \\\text{Feb} & 3.9 & 3.3 \\\text{Mar} & 3.6 & 3.0 \\\text{Apr} & 3.8 & 3.8 \\\hline\end{array}\]3. **Long Term Graduation Rate**: - Is there sufficient evidence to show that the long term graduation rate is different from 67%? Evaluate using the given sample data from the university. 4. **Fishing Hours Mean Comparison**: - Utilize a 5% level of significance to assess whether there is a significant difference in the mean hours spent fishing from shore versus from a boat.5. **Analysis of Lay Fever Rates**: - Examine two random samples to determine if the rate of hay fever is significantly different between people over 50 and those under 25, using a significance level of 5%.**Note**: Several of these problems have been adapted from textbook examples.**End of Text**.---This text is meant for exercises in statistical hypothesis testing, confidence interval estimation, and comparison studies.

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Description: #5 The following text appears to be for educational purposes involving statistical analysis and hypothesis testing:---1. **Testing Iron Concentration in Water**: - Does the sample mean concentration exceed the company’s claim of 2.5? Use a 5% leve

Given Information : To determine that whether it is better to fish from the shore or a boat . Let , x be a random variable representing the number of hours taken to catch a fish . The table below contains the data taken over several months for B from the shore , and A from a boat . First calculate Mean & standard deviation for the data provided . For shore (B) The sample size is n = 7n=7. The provided sample data along with the data required to compute the sample mean X¯ and sample variance s2 are shown in the table below: X X2 1.6 2.56 1.8 3.24 2 4 3.2 10.24 3.9 15.21 3.6 12.96 3.3 10.89 Sum = 19.4 59.1 The sample mean X¯ is computed as follows: Also, the sample variance s2 is , Therefore, the sample standard deviation s is , For boat (A) : The sample size is n=7. The provided sample data along with the data required to compute the sample mean X¯ and sample variance s2 are shown in the table below: X X2 1.5 2.25 1.4 1.96 1.6 2.56 2.2 4.84 3.3 10.89 3 9 3.8 14.44 Sum = 16.8 45.94 The sample mean X¯ is computed as follows: Also, the sample variance s2 is Therefore, the sample standard deviation s is : The calculated sample means are shown below: \bar X_1 = 2.771X¯1 =2.771 , X¯2=2.4 Also, the calculated sample standard deviations are: s_1 = 0.943s1=0.943s_2 = 0.968 , s2=0.968 and the sample sizes are n1=7 and n2=7. Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Population mean hour per fish using boat and shore is equal . i.e. μ1 = \mu_2μ2 Ha: Population mean hour per fish using boat is different than fishing from shore . i.e. μ1 ≠ \mu_2μ2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. A t-test for two means with unknown population variances and two independent samples is a hypothesis test that attempts to make a claim about the population means (\mu_1μ1 and \mu_2μ2). Rejection Region Based on the information provided, the significance level is \alpha = 0.01α=0.01, and the degrees of freedom are df = 12df=12. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this two-tailed test is tc=3.055, for α=0.01 and df=12. The rejection region for this two-tailed test is R = \{t: |t| > 3.055\}R={t:∣t∣>3.055}. Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: Decision about the null hypothesis Since it is observed that |t| = 0.726 \le t_c = 3.055∣t∣=0.726≤tc=3.055, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.4816, and since p=0.4816≥0.01, it is concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that there is significant difference in the population mean hour per fish using boat compared with fishing from shore , at the 0.01 significance level.

Math

Statistics

5. Is it better to fish from the shore or a boat? Let x be a random variable representing the number of hours taken to catch a fish. The table below contains the data taken over several months for B from the shore, and A from a boat. Oct Dec Jan Feb Mar udy 3.3 AON B (shore) 1.8 3.2 6'E 3.3 1.6 3.6 1.5 1.4 2.2 3.8 9'T Use a 1% level of significance to determine if there is a difference in the population mean hours per fish using a boat compared with fishing from shore. (This is a paired difference problem.)

5. Is it better to fish from the shore or a boat? Let x be a random variable representing the number of hours taken to catch a fish. The table below contains the data taken over several months for B from the shore, and A from a boat. Oct Dec Jan Feb Mar udy 3.3 AON B (shore) 1.8 3.2 6'E 3.3 1.6 3.6 1.5 1.4 2.2 3.8 9'T Use a 1% level of significance to determine if there is a difference in the population mean hours per fish using a boat compared with fishing from shore. (This is a paired difference problem.)

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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The following text appears to be for educational purposes involving statistical analysis and hypothesis testing:

---

1. **Testing Iron Concentration in Water**:
- Does the sample mean concentration exceed the company’s claim of 2.5? Use a 5% level of significance to test this hypothesis based on samples taken from an industrial park.
- Calculate a 95% confidence interval for the sample data.

2. **Studying Chemical Z in Blood**:
- Assess if the average concentration of Chemical Z is higher than 20%. This involves testing at a 5% significance level and calculating a 99% confidence interval from the given data.

Table: Concentration Data Across Months

\[
\begin{array}{|c|c|c|}
\hline
\text{Month} & \text{Sample 1} & \text{Sample 2} \\
\hline
\text{Oct} & 3.1 & 1.5 \\
\text{Nov} & 2.0 & 1.4 \\
\text{Dec} & 1.6 & 1.6 \\
\text{Jan} & 3.3 & 2.2 \\
\text{Feb} & 3.9 & 3.3 \\
\text{Mar} & 3.6 & 3.0 \\
\text{Apr} & 3.8 & 3.8 \\
\hline
\end{array}
\]

3. **Long Term Graduation Rate**:
- Is there sufficient evidence to show that the long term graduation rate is different from 67%? Evaluate using the given sample data from the university.

4. **Fishing Hours Mean Comparison**:
- Utilize a 5% level of significance to assess whether there is a significant difference in the mean hours spent fishing from shore versus from a boat.

5. **Analysis of Lay Fever Rates**:
- Examine two random samples to determine if the rate of hay fever is significantly different between people over 50 and those under 25, using a significance level of 5%.

**Note**: Several of these problems have been adapted from textbook examples.

**End of Text**.

---

This text is meant for exercises in statistical hypothesis testing, confidence interval estimation, and comparison studies.

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