The Following Random Sample Was Selected From A Normal Distr

The Following Random Sample Was Selected From A Normal Distribution The following random sample was selected from a normal distribution: 11, 13, 5, 18, 19, 4, 15, 5, 5, 8.

Construct a 90% confidence interval for the population mean µ. ≤µ≤ Construct a 95% confidence interval for the population mean µ. ≤µ≤

Paper For Above instruction Introduction Understanding the population mean µ based on sample data is a fundamental aspect of statistical inference.

Confidence intervals provide a range of plausible values for the population mean, reflecting the uncertainty inherent in sampling. In this paper, we will analyze a given sample drawn from a normal distribution, compute the confidence intervals for the mean at 90% and 95% confidence levels, and interpret these results.

Sample Data and Descriptive Statistics The sample data consist of the following values: 11, 13, 5, 18, 19, 4, 15, 5, 5, 8. To proceed with confidence interval calculations, it is necessary to determine the sample mean and sample standard deviation.

Calculating the sample mean (\(\bar{x}\)): \[ \bar{x} = \frac{11 + 13 + 5 + 18 + 19 + 4 + 15 + 5 + 5 + 8}{10} = \frac{103}{10} = 10.3 \] Calculating the sample standard deviation (\(s\)): \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \] Where \(x_i\) are individual data points, and \(n=10\).

Computing the squared deviations:

\[ (11 - 10.3)^2 = 0.49 \\ (13 - 10.3)^2 = 7.29 \\ (5 - 10.3)^2 = 28.09 \\ (18 - 10.3)^2 = 59.29 \\ (19 - 10.3)^2 = 75.69 \\ (4 - 10.3)^2 = 39.69 \\ (15 - 10.3)^2 = 22.09 \\ (5 - 10.3)^2 = 28.09 \\ (5 - 10.3)^2 = 28.09 \\ (8 - 10.3)^2 = 5.29 \] Sum of squared deviations: \[ 0.49 + 7.29 + 28.09 + 59.29 + 75.69 + 39.69 + 22.09 + 28.09 + 28.09 + 5.29 = 293.10 \] Sample variance: \[ s^2 = \frac{293.10}{9} \approx 32.57 \] Sample standard deviation: \[ s = \sqrt{32.57} \approx 5.71

\] Constructing Confidence Intervals Since the population standard deviation is unknown, and the sample size is small (\(n=10\)), we will use the t-distribution to determine the confidence intervals. The formula for the confidence interval is: \[ \bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}} \] Where: - \(\bar{x}\) = 10.3 - \(s\) = 5.71 - \(n\) = 10 - \(t_{\alpha/2, n-1}\) = critical t-value for the specified confidence level and degrees of freedom (\(n-1=9\)) **For 90% Confidence Level:** - \(\alpha = 0.10\) - \(t_{0.05, 9}\) ≈ 1.833 (from t-distribution table) **For 95% Confidence Level:** - \(\alpha = 0.05\) - \(t_{0.025, 9}\) ≈ 2.262 Calculating the standard error (\(SE\)): \[ SE = \frac{s}{\sqrt{n}} = \frac{5.71}{\sqrt{10}} \approx 1.805 \] **90% Confidence Interval:**

\[ 10.3 \pm 1.833 \times 1.805 \approx 10.3 \pm 3.310 \] \[ \text{Lower bound} = 10.3 - 3.310 = 6.99 \] \[ \text{Upper bound} = 10.3 + 3.310 = 13.61 \] **Result:** (6.99, 13.61) **95% Confidence Interval:** \[ 10.3 \pm 2.262 \times 1.805 \approx 10.3 \pm 4.084 \] \[ \text{Lower bound} = 10.3 - 4.084 = 6.216 \] \[ \text{Upper bound} = 10.3 + 4.084 = 14.384 \] **Result:** (6.216, 14.384) Discussion and Interpretation The confidence intervals provide plausible ranges for the true population mean µ based on the sample data.

The 90% confidence interval (6.99, 13.61) implies that, with 90% confidence, the true mean falls within this range. Similarly, the 95% confidence interval (6.216, 14.384) offers a slightly wider range, reflecting increased certainty about the estimate.

These results have practical implications, suggesting that the true average likely lies between approximately 6.2 and 14.4, depending on the confidence level chosen. The wider interval at 95% indicates greater assurance but also reflects more uncertainty due to variability in the sample data.

The use of t-distribution is appropriate given the small sample size and unknown population standard deviation. It compensates for this uncertainty, ensuring more accurate interval estimation compared to normal distribution assumptions.

Limitations and Assumptions Key assumptions include the normality of the data distribution, which is reasonable here given the sample size and the data pattern. However, deviations from normality could affect the validity of the confidence intervals. Additionally, the sample size of 10 is relatively small, which inherently leads to wider confidence intervals and more uncertainty. Larger samples would provide more precise estimates.

Conclusion This analysis demonstrates how confidence intervals can be constructed to estimate the population mean µ from a small sample drawn from a normal distribution. The derived intervals at 90% and 95% confidence levels offer useful bounds, with the wider interval providing greater certainty. Statistical methods such as the t-distribution are crucial for accurate inference when population parameters are unknown, particularly with limited data.

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