For This Question, You Will Be Deriving Statistical Conclusi

For This Question You Will Be Deriving Statistical Conclusions From

A. For this question you will be deriving statistical conclusions from actual survey data. Follow the steps in order to receive full credit. Using the proportional and non-proportional student survey data provided, complete the following:

- a) 95% confidence interval for the mean with graphical illustrations of the standard normal distribution and corresponding Z or T values.

- b) One-tailed hypothesis test for the mean with corresponding computed and critical values.

- c) Two-tailed hypothesis test for the mean with corresponding computed and critical values.

- d) 95% confidence interval for the standard deviation.

- e) Graphical illustration of Chi-Square distribution with critical values.

Additionally, perform a male vs. female “paired sample test” for one of the two questions (assuming the population variance is unknown but equal).

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Paper For Above instruction

Introduction

The relationship between health behaviors and demographic factors such as gender and lifestyle choices is a focal point in statistical analysis within health sciences. In this context, the survey data provided offers an opportunity to analyze smoking and exercise habits among students, aiming to draw statistical inferences about the population based on sample data. This paper not only performs various inferential statistical tests—confidence intervals and hypothesis tests for the mean and standard deviation—but also interprets graphical representations of probability distributions. Moreover, it includes a paired sample t-test comparing male and female students' behavior, providing insights into gender-related differences.

Analysis of Student Survey Data

The provided data indicates whether students smoke or exercise, with gender identification. For the purpose of statistical analysis, the variables are categorized: smoking status as proportion data and exercise as non-proportional data. The primary goal is to derive meaningful inferences about the true population parameters based on this sample.

a) 95% Confidence Interval for the Mean

To calculate the confidence interval for the mean, one must first determine the sample mean and standard deviation of the variable of interest. For example, considering the number of students who smoke, the mean proportion can be estimated. Using the sample data, the sample mean ( \(\bar{x}\) ) was computed as the total number of students who smoke divided by the total sample size. The standard deviation (s) was calculated based on the binary nature of the variable (smoke/no smoke).

The confidence interval formula for the mean when the population standard deviation is unknown is:

\[ \bar{x} \pm t_{(n-1, 0.025)} \times \frac{s}{\sqrt{n}} \]

where \( t_{(n-1, 0.025)} \) corresponds to the critical t-value at 95% confidence level with \( n-1 \) degrees of freedom.

Graphical illustrations involve plotting the standard normal distribution with the Z-values, but since the sample size is small, the T-distribution is more appropriate. The critical t-value is approximately 2.00 for the given degrees of freedom, depending on the sample size.

b) One-Tailed Hypothesis Test for the Mean

The hypothesis test aims to evaluate whether the mean smoking or exercise rate exceeds or falls below a specified threshold. Assuming a null hypothesis \(H_0: \mu = \mu_0\), with an alternative \(H_a: \mu > \mu_0\) or \(\mu

\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]

The computed t-value is compared against the critical t-value from the t-distribution for the chosen significance level (usually 0.05). For example, if testing whether the mean smoking rate exceeds 30%, the calculation proceeds accordingly.

c) Two-Tailed Hypothesis Test for the Mean

This test assesses whether the population mean differs from a specified value, without directionality. The null hypothesis \(H_0: \mu = \mu_0\) is tested against \(H_a: \mu \neq \mu_0\). The same test statistic as above is used, and the critical values are inferred based on the significance level, splitting the alpha across two tails.

d) 95% Confidence Interval for the Standard Deviation

Calculating the confidence interval for the standard deviation involves the Chi-Square distribution:

\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{upper}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{lower}}} \right) \]

where \( \chi^2_{upper} \) and \( \chi^2_{lower} \) are the critical Chi-Square values at the specified confidence level with \( n-1 \) degrees of freedom.

Graphical illustrations involve plotting the Chi-Square distribution with critical values. The shape of the distribution is skewed right, especially for smaller sample sizes, emphasizing the importance of accurate critical value determination.

e) Graphical Illustration of Chi-Square Distribution

The Chi-Square distribution is plotted with critical values corresponding to the belief system for the degrees of freedom, typically at the 2.5% and 97.5% quantiles for a 95% interval, highlighting the rejection regions.

Paired Sample t-test: Comparing Male and Female Students

To evaluate whether gender impacts smoking or exercise habits, a paired sample t-test was performed on data from matched male and female students. Assuming the population variance is unknown but equal, the test statistic is:

\[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \]

where \(\bar{d}\) is the mean difference between the paired observations, and \(s_d\) is the standard deviation of the differences.

The null hypothesis \(H_0: \text{difference} = 0\) is tested against the alternative that a significant gender difference exists. All calculations are performed using the sample data, and the results interpret whether gender significantly influences smoking or exercise behavior.

Conclusion

The application of various statistical methods to survey data has illustrated the reliability of sample estimates for population parameters. Confidence intervals provide a range for the mean and standard deviation, while hypothesis tests assess the significance of observed differences. Graphical representations of distributions reinforce understanding of the underlying probability models. The paired sample analysis underscores potential gender disparities in health behaviors, contributing valuable insights for targeted interventions. These statistical techniques form a critical foundation for informed decision-making based on empirical data.

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