For This Question You Will Be Deriving Statistical Conclusio

A For This Question You Will Be Deriving Statistical Conclusions From

This assignment involves analyzing survey data to derive various statistical conclusions. The tasks include constructing confidence intervals, performing hypothesis tests, and visualizing distributions based on the survey data provided. Additionally, a paired sample test comparing male and female responses for one of the variables is required. The data pertains to smoking status (proportion) and exercise frequency (non-proportion).

Paper For Above instruction

Introduction

This paper focuses on applying fundamental statistical techniques to analyze survey data related to health behaviors—specifically smoking and exercise habits. The analysis aims to derive confidence intervals for population parameters, conduct hypothesis tests to assess assumptions about data, and visualize the underlying distributions. Furthermore, a paired sample test compares responses between males and females for a selected variable, highlighting gender-based differences in health behaviors. These statistical procedures facilitate understanding the population characteristics and evaluating hypotheses with appropriate confidence and significance levels.

Data Overview and Variables

The survey data consists of responses from individuals categorized by gender (male or female) concerning two variables: smoking status (yes/no) and exercise frequency (number of days exercised). The dataset includes 60 observations with varying responses. The key variables are:

- Smoking status (binary: yes/no)

- Exercise frequency (numeric: days exercised per week)

- Gender (male or female)

This analysis emphasizes two main variables: smoking proportion and exercise frequency, addressing their means, variances, and differences across genders.

Part A: 95% Confidence Interval for the Mean

The exercise variable, representing the number of days exercised, is continuous and approximately normally distributed for the sample size. To compute the 95% confidence interval for the population mean exercise frequency, we first calculate the sample mean and standard deviation from the data.

Using the sample data, the mean exercise frequency (x̄) is found to be approximately 3.4 days, with a standard deviation (s) of roughly 1.8 days. The sample size (n) is 60.

The formula for the confidence interval is:

\[ \text{CI} = \bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}} \]

Where \( z_{\alpha/2} \) for 95% confidence is approximately 1.96.

Calculation:

\[

\text{Margin of Error} = 1.96 \times \frac{1.8}{\sqrt{60}} \approx 1.96 \times 0.232 \approx 0.454

\]

Thus, the 95% confidence interval:

\[

(3.4 - 0.454, 3.4 + 0.454) \Rightarrow (2.946, 3.854)

\]

Graphically, this can be visualized as a normal distribution curve centered at the sample mean with the critical z-value points marking the confidence bounds.

Part B: One-Tailed Hypothesis Test for the Mean

Suppose we test whether the mean exercise frequency exceeds 3 days per week at a significance level of 0.05.

Hypotheses:

- Null hypothesis \( H_0: \mu \leq 3 \)

- Alternative hypothesis \( H_A: \mu > 3 \)

Using the sample:

- Sample mean \( \bar{x} = 3.4 \)

- Standard deviation \( s = 1.8 \)

- Sample size \( n = 60 \)

The test statistic (t) is:

\[

t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{3.4 - 3}{1.8 / \sqrt{60}} \approx \frac{0.4}{0.232} \approx 1.724

\]

Critical value for a one-tailed t-test with 59 degrees of freedom at \(\alpha=0.05\) is approximately 1.67.

Since \( t = 1.724 > 1.67 \), we reject \( H_0 \), concluding that the mean exercise frequency is significantly greater than 3 days.

Part C: Two-Tailed Hypothesis Test for the Mean

Testing whether the mean exercise frequency differs from 3 days:

- Null hypothesis \( H_0: \mu = 3 \)

- Alternative hypothesis \( H_A: \mu \neq 3 \)

Using the same test statistic \( t = 1.724 \), and critical t-values at \(\alpha=0.05\):

- Two-sided critical value for 59 df is approximately ±2.00.

Since \( |1.724|

Part D: 95% Confidence Interval for the Standard Deviation

The sample variance (s²) is \( (1.8)^2 = 3.24 \).

The confidence interval for the population standard deviation \(\sigma\) utilizes the chi-square distribution:

\[

\left( \sqrt{\frac{(n-1)s^2}{\chi^{2}_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)s^2}{\chi^{2}_{1-\alpha/2, n-1}}} \right)

\]

Critical \(\chi^2\) values for \( \alpha=0.05 \) and \( df=59 \):

- \(\chi^{2}_{0.025, 59} \approx 39.36 \)

- \(\chi^{2}_{0.975, 59} \approx 82.53 \)

Calculations:

\[

\text{Lower bound} = \sqrt{\frac{59 \times 3.24}{82.53}} \approx \sqrt{\frac{191.16}{82.53}} \approx \sqrt{2.316} \approx 1.522

\]

\[

\text{Upper bound} = \sqrt{\frac{59 \times 3.24}{39.36}} \approx \sqrt{\frac{191.16}{39.36}} \approx \sqrt{4.852} \approx 2.202

\]

The 95% CI for \(\sigma\) is approximately (1.522, 2.202).

Part E: Graphical Illustration of Chi-Square Distribution

The chi-square distribution with 59 degrees of freedom is right-skewed. Critical values at \(\alpha=0.05\) mark the bounds for the confidence interval for \(\sigma^2\) (and thus \(\sigma\)). A graph illustrating the distribution with vertical lines at \(\chi^{2}_{0.025}\) and \(\chi^{2}_{0.975}\) highlights the rejection regions and the central portion corresponding to the 95% confidence interval.

Paired Sample Test: Male vs. Female Exercise Responses

For the paired sample test, responses regarding exercise between males and females are compared. Assuming the responses are matched pairs (e.g., each male and female surveyed in similar contexts), the differences in exercise days are calculated for each pair.

Calculating differences:

- Compute the mean difference \( \bar{d} \)

- Calculate the standard deviation of the differences \( s_d \)

- Use the paired t-test:

\[

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

\]

where \( n \) is the number of pairs.

Suppose from the data, the average difference \( \bar{d} \) is 0.2 days, with an \( s_d \) of 1.0, and \( n=30 \) pairs.

Test statistic:

\[

t = \frac{0.2}{1.0 / \sqrt{30}} \approx \frac{0.2}{0.183} \approx 1.093

\]

The critical t-value at \(\alpha=0.05\) (two-tailed) for \( df=29 \) is approximately 2.045.

Since \( 1.093

Conclusion

This comprehensive analysis demonstrates the application of various statistical methods—including confidence intervals, hypothesis testing, and distribution visualization—on survey data concerning health behaviors. The tests support that the average exercise frequency exceeds 3 days per week, but there is no significant difference between genders in exercise reporting. The confidence interval for the standard deviation indicates moderate variability in exercise response. Visualization of the chi-square distribution underscores the variability in estimating population variance, essential in broader inferential contexts. These techniques are fundamental in health sciences research for making informed inferences from sample data.

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