Question

13. Extra Credit: (10 points) Listed below are weights (grams) from a simple random sample ot "wheat" pennies. U.S. Mint specifications now requir deviation of 0.0230 g for weights of pennies. Use a 0.01 to test the claim that wheat pennies are manufactured so that weights have a standard deviation equal to 0.0230 g e a standard significance level 2.5024 2.5298 2.4998 2.4823 2.5163 2.5222 2.4900 2.4907 2.5017

Answer 1

Given Data: 2.5024, 2.5298, 2.4998, 2.4823, 2.5163, 2.5222, 2.49, 2.4907, 2.5017 We calculate sample mean and std deviation from given data Sample Mean,2 x^- = sigma (x)/n = 22.5352/9 = 2.503911 Sample Variance, s^2 = sigma (x-x^-)62/n-1 = 0.0002522 Sample std dev, s = squareroot s^2 = squareroot 0.0002522 almostequalto 0.01588 We want to test two tailed hypothesis: H_0: = alpha = 0.023 H_a: sigma notequalto 0.023 (Null Hypothesis) (alternative Hypothesis also called H_1) Given: s = 0.01588 Rightarrow s^2 = 0.000252 , n = 9 Significance level alpha = 0.01 Degrees of freedom df = n -1 = 8 Since this is two tailed test, we need area of alpha = 0.01 on either side of critical value. This means area of 0.005 on right of positive critical value Critical values are given by X_1^2 = alpha/2 = x_0.995^2 and X_1-alpha/2^2 = X_0.995^2 with 8 degree of freedom we can use chi-square table or following excel commands