Confidence intervals #4 - Questions and Answers

Question 1 Explain what is the test statistic used for constructing confidence interval on a proportion. Answer Step 1 The formula for 100(1-α)% confidence interval for population proportion is, CI = p̂ ± ME = p̂ ± zα/2 √(p̂(1-p̂)/n) Here, p̂: the sample proportion ME: Margin of Error n: sample size. z_α/2: critical value. (1-α): Confidence level Step 2 Hence, z test statistic is used for constructing confidence interval on a proportion. For testing population proportion, the test statistic is defined as z_test = (p̂ - p) / √(p(1-p)/n) Here, p̂: sample proportion p: population proportion n: sample size Question 2 Find the critical value z_α/2 that corresponds to the given confidence level. 96% Answer Confidence level = 96% Level of significance (α) = 1 - 0.96 = 0.04 The critical z_α/2 value at 0.04 level of significance from the standard normal table = 2.054. Question 3 Decresing the sample size, while holding the confidence level and the variance the same, will do what to the length of wour confidence interval? Answer Confidence Interval: A confidence interval is an estimate of the population proportion, e.g, population mean, and proportion. It provides lower and upper values where the population parameter is likely to be contained. The sample size should be appropriate for the best estimate of the population parameter. Factors that Affect Confidence Intervals: There are three factors that determine the size of the confidence interval for a given confidence level: Sample size Percentage Population size Sample Size: The larger your sample size, the more sure you can be that their answers truly reflect the population. This indicates that for a given confidence level, the larger your sample size, the smaller your confidence interval. However, the relationship is not linear (i.e., doubling the sample size does not halve the confidence interval). Increasing the sample size decreases the width of confidence intervals, because it decreases the standard error. Answer: "Make it bigger" Explanation: Sample Size: Smaller sample sizes generate wider intervals. There is an inverse square root relationship between confidence intervals and sample sizes. If you want to cut your margin of error in half, you need to approximately quadruple your sample size. Question 4 Let f(x) = 4 - x²/3 + 12/x³. Find the open intervals on which f is increasing (decreasing). Then determine the x- coordinates of all relative maxima (minima). f is increasing on the intervals f is decreasing on the intervals The relative maxima of f occur at x = The relative minima of f occur at x = Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none". In the last two, your answer should be a comma separated list of x values or the word "none". Answer f(x) = 4 - x²/3 + 12/x³ Take derivative and find out critical points to get maxima and minima Apply power rule f'(x) = -2/3 * x^(-2/3) + 12 * (-3x^(-4)) f'(x) = -2/(3x^(2/3)) - 36/x⁴ f'(x) = -2/(3x^(2/3)) + 12/x^5 Set the derivative = 0 and solve for x LCD is x^3, multiply each term by x³ -2x^(4/3) - 36 = 0 -2x^(10/3) - 36x³ = 0 2x - 12 = 0 2x = 12 x= 6 x = 0 makes denominator 0 and it is undefined So x = 0, x = 6 Break the numbers line into three intervals using 0 and 6 Check each interval using derivativeLet x = -1, x = 1, x = 7 f'(x) = -2/(3x^(2/3)) - 36/x⁴ f'(-1) = -2/(3(-1)^(2/3)) - 36/(-1)⁴ = 12 - 10 = -10 f'(1) = -2/(3(1)^(2/3)) - 36/(1)⁴ = -2/3-36 = -10 f'(7) = -2/(3(7)^(2/3)) - 36/(7)⁴ = 2/343 - 12/49 Derivative is from positive in two intervals So increasing intervals are (-∞,0)U(6,∞) Decreasing interval (0,6) There is a break in graph of f(x)atx = 0 The derivative goes from negative to positive at x = 6 So relative minima at x = 6 There is no relative maxima Question 5 Scott wants to find a 95% confidence interval for the average GPA for Allegany College of Maryland students. He gets the records for 89 student. The results were roughly bell- shaped with a mean of 2.3 and standard deviation of 0.7. Construct a 95% confidence interval in the form: Estimate ± margin of error. Answer Step 1 Confidence interval are of utmost importance in statistics. It forms an integral part of hypothesis testing. Step 2 The estimate is the mean = 2.4 The margin of error is z₀.₀₂₅ * σ/√n = (1.96) * 0.7/√89 = 0.4986 n is the sample size, σ is the standard deviation.