IMG_1171.HEIC

# Statistical Inference ## Estimation ### What is Estimation? Estimation is the process of determining likely values for a population parameter based on a sample. ### Types of Estimates * **Point Estimate:** A single value that estimates the population parameter. * **Interval Estimate:** A...

# Statistical Inference ## Estimation ### What is Estimation? Estimation is the process of determining likely values for a population parameter based on a sample. ### Types of Estimates * **Point Estimate:** A single value that estimates the population parameter. * **Interval Estimate:** A range of values within which the population parameter is likely to fall. ### Point Estimation * A point estimate is a single number that is used to estimate an unknown population parameter. * For example, the sample mean $(\bar{x})$ is a point estimate of the population mean $(\mu)$. *Example:* The average height of 100 students is 175 cm. So, the point estimate of the average height of all students in the school is 175 cm. ### Interval Estimation * An interval estimate provides a range within which the population parameter is believed to lie. * It is usually associated with a confidence level, which indicates the probability that the interval contains the true parameter. *Example:* A 95% confidence interval for the average height of all students might be (170 cm, 180 cm), indicating that we are 95% confident that the true average height falls within this range. ### Confidence Interval * A confidence interval is an interval estimate of a population parameter. * It is associated with a confidence level, which indicates the probability that the interval contains the true parameter. *Formula:* Confidence Interval = Point Estimate ± (Critical Value × Standard Error) ### Key Components 1. **Point Estimate:** * The sample statistic used to estimate the population parameter. 2. **Critical Value:** * A value from a standard distribution (e.g., Z-distribution, t-distribution) that determines the width of the confidence interval. 3. **Standard Error:** * A measure of the variability of the sample statistic. ### Confidence Level * The confidence level is the probability that the confidence interval contains the true population parameter. * Common confidence levels are 90%, 95%, and 99%. ### Margin of Error * The margin of error is the amount added and subtracted from the point estimate to form the confidence interval. * It is determined by the critical value and the standard error. *Formula:* Margin of Error= Critical Value × Standard Error ### Z-Interval Use Z-Interval if: * Population standard deviation ($\sigma$) is known. * Sample size is large (\> 30). * Data is normally distributed. *Formula:* $\bar{x} \pm Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ Where: * $\bar{x}$ = Sample mean * $Z_{\alpha/2}$ = Z-score for the desired confidence level * $\sigma$ = Population standard deviation * n = Sample size ### T-Interval Use T-Interval if: * Population standard deviation ($\sigma$) is unknown. * Sample size is small (\< 30). * Data is normally distributed. *Formula:* $\bar{x} \pm t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}$ Where: * $\bar{x}$ = Sample mean * $t_{\alpha/2, n-1}$ = t-score for the desired confidence level and degrees of freedom * s = Sample standard deviation * n = Sample size ### Example 1: Z-Interval A survey of 500 adults found that 60% prefer coffee over tea. Assuming a population standard deviation of 5%, construct a 95% confidence interval for the proportion of adults who prefer coffee. *Solution:* * $\bar{p} = 0.60$ * $n = 500$ * $\sigma = 0.05$ * $Z_{\alpha/2} = 1.96$ (for 95% confidence) $0.60 \pm 1.96 \frac{0.05}{\sqrt{500}} = (0.556, 0.644)$ We are 95% confident that the true proportion of adults who prefer coffee is between 55.6% and 64.4%. ### Example 2: T-Interval A sample of 25 students has a mean test score of 75 with a sample standard deviation of 8. Construct a 90% confidence interval for the population mean test score. *Solution:* * $\bar{x} = 75$ * $n = 25$ * $s = 8$ * $t_{\alpha/2, n-1} = 1.711$ (for 90% confidence and 24 degrees of freedom) $75 \pm 1.711 \frac{8}{\sqrt{25}} = (72.26, 77.74)$ We are 90% confident that the true mean test score for all students is between 72.26 and 77.74. ### Key Points * Estimation involves using sample data to infer population parameters. * Point estimates provide single values, while interval estimates offer a range. * Confidence intervals quantify the uncertainty associated with estimating population parameters. * The choice between Z-Interval and T-Interval depends on whether the population standard deviation is known and the sample size.