Calculating Confidence Intervals & Testing Hypotheses with Normal Distribution - Prof. Dav, Exams of Statistics

Download Calculating Confidence Intervals & Testing Hypotheses with Normal Distribution - Prof. Dav and more Exams Statistics in PDF only on Docsity! MATH 203 Sample Problem Unit 3 Bottles of liquid soap are normally distributed in weight with a mean of 64 ounces and a standard deviation 0.5 ounces. (a) Let various random samples of size 21 be collected. Compute P(0.47 ≤ S ≤ 0.55) . With n = 21 and = 0.5: P(0.47 ≤ S ≤ 0.55) = P 20 × 0.47 2 0.52 ≤ (n − 1)S 2 2 ≤ 20 × 0.552 0.52         = P 17.672 ≤ 2 (20) ≤ 24.2( ) ≈ 0.37524 . Use 2cdf(17.672, 24.2, 20) (b) Suppose is unknown, but a random sampling of 26 bottles yields S = 0.60. Find a 95% confidence interval for the true standard deviation. Use the 95% chi-square scores from the 2 (n −1) = 2 (25) curve which are L = 13.12 and R = 40.65. (n − 1) × S2 R ≤ ≤ (n − 1) × S2 L 25 × 0.60 2 40.65 ≤ ≤ 25 × 0.60 2 13.12 0.47 ≤ ≤ 0.828 (c) Using S = 0.60 from a sample of 26 bottles, is there evidence, at the 10% level of significance, to reject the claim that = 0.5? State null and alternative hypotheses, give the test-statistic and P -value, and use the P -value to explain your conclusion in detail. With S = 0.6 and n = 26 : We test H0 : = 0.50 vs. Ha > 0.50 . The test stat is (n −1)S2 2 = 25 × 0.62 0.52 = 36 . The P -value is 2cdf(36, 1E99, 25) ≈ 0.0716 . (Use the right tail for Ha > 0.50 .) If = 0.50 were true, then there is a 7.16% chance of obtaining an S of 0.60 or larger with a sample of size 26. There is enough evidence to reject H0 with a 10% level of significance