The Random Sample Shown Below Was Selected From Normal Distribution10 The random sample shown below was selected from normal distribution. 10, 3, 4, 10, 3, 6 a. Construct a 99% confidence interval for the population mean. Round to two decimal places as needed. b. Assume that the sample mean and the sample standard deviation remain exactly the same as those just calculated, but are based on a sample of n=25 observations. Repeat part (a). What is the effect of increasing the sample size on the width of the confidence interval? The confidence interval is ( -,-). What is the effect of the sample size on the width of the confidence interval? a. As the sample size increases, the width decreases.
Paper For Above instruction Confidence intervals are vital tools in statistics, providing a range of plausible values for a population parameter based on sample data. They offer an estimate of the uncertainty associated with a sample statistic, such as the mean. When dealing with normal distributions, especially with small sample sizes, the t-distribution is typically used to construct these intervals due to its accommodation of additional variability introduced by estimating the population standard deviation from the sample. This paper explores the construction of confidence intervals for the mean, particularly focusing on how sample size influence the width of these intervals, using a specific sample and extending the analysis to a larger sample size scenario. In the initial scenario, a small sample of six observations—10, 3, 4, 10, 3, and 6—was collected, assumed to be from a population with a normal distribution. The task was to compute a 99% confidence interval for the population mean. From this dataset, the sample mean (x■) and sample standard deviation (s) are calculated as follows: Sample data: 10, 3, 4, 10, 3, 6 Sample mean (x■) = (10 + 3 + 4 + 10 + 3 + 6)/6 = 36/6 = 6.00 Sample standard deviation (s) = √[Σ(xi - x■)² / (n-1)] Calculating deviations: (10-6)²=16, (3-6)²=9, (4-6)²=4, (10-6)²=16, (3-6)²=9, (6-6)²=0 Sum of squared deviations = 16 + 9 + 4 + 16 + 9 + 0 = 54