HARDMAN DSC 210 Take-Home Questions For The Final Exam Tues

2 HARDMAN DSC 210 Take-Home Questions for the Final Exam Tues, Dec 15, points

1. A sample of 120 students yielded a sample mean GPA of 3.35, and a sample standard deviation equal to 0.28. Calculate the 98% confidence interval for the population mean GPA. SHOW YOUR WORK!

2. In a sample of 125 statistics students, 35 said that finding confidence intervals for population proportions was their favorite topic. Give the 99% confidence interval estimate of the population proportion of statistics students whose favorite topic is finding confidence intervals for p. SHOW YOUR WORK!

3. What effect does increasing the sample size have on the length of a CI? EXPLAIN YOUR ANSWER

4. Find the value of the Z-critical number that you would use to construct a 75% CI for . SHOW YOUR WORK! Just the number will earn exactly 0 points, even if it is the correct number!

Paper For Above Instruction

Confidence intervals (CIs) are vital tools in statistical inference, providing a range of plausible values for a population parameter based on sample data. Understanding how to calculate these intervals and interpret their implications is fundamental in statistics. This paper addresses key aspects of constructing confidence intervals for both means and proportions, the impact of sample size on CI length, and critical Z-values associated with specific confidence levels.

Calculating the 98% Confidence Interval for the Population Mean GPA

Given data: sample size (n) = 120, sample mean (x̄) = 3.35, sample standard deviation (s) = 0.28, confidence level = 98%. Since the sample size is sufficiently large, we can safely apply the z-distribution for the CI calculation.

First, determine the standard error (SE):

SE = s / √n = 0.28 / √120 ≈ 0.28 / 10.954 ≈ 0.0255

Next, identify the critical z-value (z) for a 98% confidence level. The z-value corresponds to an alpha (α) of 2%. Since the confidence level is two-sided, α/2 = 0.01. From standard normal tables, z ≈ 2.33.

Calculate the margin of error (ME):

ME = z* × SE = 2.33 × 0.0255 ≈ 0.0593

Construct the confidence interval:

Lower bound = x̄ - ME = 3.35 - 0.0593 ≈ 3.2907

Upper bound = x̄ + ME = 3.35 + 0.0593 ≈ 3.4093

Hence, the 98% confidence interval for the population mean GPA is approximately (3.291, 3.409).

Estimating the Confidence Interval for Proportion of Students Favoring Confidence Intervals

Given data: sample size (n) = 125, number favoring (x) = 35. The sample proportion (p̂) = x / n = 35 / 125 = 0.28. Confidence level = 99%. For large samples, the normal approximation to the binomial is appropriate.

Calculate the standard error (SE) for the proportion:

SE = √[p̂(1 - p̂) / n] = √[0.28 × 0.72 / 125] ≈ √[0.2016 / 125] ≈ √0.0016128 ≈ 0.0402

The z-value for a 99% confidence interval (α/2 = 0.005) is approximately 2.576.

Calculate the margin of error:

ME = z* × SE = 2.576 × 0.0402 ≈ 0.1036

Construct the confidence interval:

Lower bound = p̂ - ME ≈ 0.28 - 0.1036 ≈ 0.1764

Upper bound = p̂ + ME ≈ 0.28 + 0.1036 ≈ 0.3836

Thus, the 99% confidence interval estimate for the proportion of students who favor confidence intervals is approximately (0.176, 0.384).

Impact of Increasing Sample Size on Confidence Interval Length

Increasing the sample size directly affects the length of a confidence interval. Specifically, the length of a CI is proportional to the standard error, which is inversely related to the square root of the sample size (n). As n increases, the standard error decreases because:

SE = s / √n for means, and similarly, for proportions, SE = √[p̂(1 - p̂)/n].

Reducing the standard error results in a narrower margin of error, thus producing a shorter confidence interval. This means that larger samples lead to more precise estimates of the population parameter. Conversely, smaller samples yield wider intervals, indicating greater uncertainty.

In essence, increasing the sample size enhances the accuracy and reliability of the estimate by reducing the width of the confidence interval, thereby narrowing the plausible range for the population parameter.

Finding the Z-critical Number for a 75% Confidence Interval

The Z-critical number (z*) is the value on the standard normal distribution that captures the central (1 - α) proportion of the distribution, where α is the total significance level. For a 75% confidence interval, the significance level is 25%, split symmetrically across both tails of the distribution, so α/2 = 0.125.

To find z* corresponding to a cumulative probability of (1 - α/2) = 0.875, we consult standard normal distribution tables or statistical software.

From standard Z-tables, the z-value that corresponds to a cumulative probability of 0.875 is approximately 1.15.

Therefore, the critical Z-value (z*) for constructing a 75% confidence interval is approximately 1.15.

Conclusion

Constructing confidence intervals involves understanding the relationship between the sample data, the desired confidence level, and the critical values of the standard normal distribution. The calculations for means and proportions demonstrate how sample size and variability influence the precision of the estimates. Increasing the sample size reduces the confidence interval's width, offering more precise insights into the population parameters. Identifying the correct Z-critical value is essential for accurate interval construction, especially at atypical confidence levels such as 75%.

References

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  • Wackerly, D., Mendenhall, W., & Scheaffer, R. (2014). Mathematical Statistics with Applications. Cengage Learning.
  • Newcombe, R. G. (1998). Two-sided confidence intervals for the single proportion: Comparison of seven methods. Statistics in Medicine, 17(8), 873-890.
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