Inferences on 'v'ariances Homework; Use (120.05 unless otherwise specied 1. The following summary data on bending strength {lb-in/in} ofjoints is taken from the article Bending Strength of Corner Joints Constructed with Injection Molded Splines.\" A Scientist would like to know if there are any differences in the variability of bending strength based on coding. Assume that each sample came from a normal distribution. The results from this study are given below: Type Sample Sample Sample Std Size Mean Dev 3035 With 10 53.23 5.96 Coding a. List the Null and Alternative Hypotheses b. Test this with a hypothesis test at oi=lllJS c. Give the real-world answer d. Check this hypothesis with a Cl. Does the answer change? Explain 2. An engineer is doing a study in a manufacturing setting on the lengths of paperclips. The sample he collected is in the dataset paperclips.xls. The engineer is concerned that the variance is too large. The variance should no bigger than 5. a. Check to see if the data comes from the normal distribution b. List the null and alternative hypotheses c. Test this with a hypothesis test at ot=l105 d. Give the real-world answer e. Check this hypothesis with a confidence interval. Does the answer change? Explain For the Shapiro-Wilk normality test, we assume that: H g : The data comes from a normal distribution HA : The data does not comes from a normal distribution The p-value for this test is 0.7257 Since 0.7257> 0.05 we fail to reject the null hypothesis. This means that we should assume that the given data comes from a normal distribution. This is reected in the normal probability plot below The plot has points that closely follow the line. p-value=0.7257 9 C) 5' Ln ['3' U1 Ordered Response Data Values 0'1 C) 53 C) 52 -1 o 1' Q Normal Quaniiles max (X) = 10.331 min (X) = 0.7993 Range = max (X) - min (X) = (10.331 - 0.7993) = 9.532 n _X? = X? +X3 + ... + X2 = 30.254 + 106.73 + ...+ 32.395 = 1654.09 i= 1 (n _MIX? ) -(EL, Xi)? (50 * 1654.09) - 72038.45 10665.94 n (n - 1) = 4.353 50 * 49 2450 (n In Ci-1 X?) - (EXi)' S = V4.353 = 2.086 n (n - 1) Lower Fence = Q1 - (1.5 * IQR) = 3.732 - (1.5 * 2.843) = 3.732 - 4.264 = -0.5322 Upper Fence = Q3 + (1.5 x IQR) = 6.575 + (1.5 * 2.843) = 6.575 + 4.264 = 10.839\fStatistic Value Mean 5.368 Median 5.656 Sample Size 50 Q1 3.732 Q3 6.575 Min 0.7993 Max 10.331 IQR 2.843 Range 9.532 Variance 4.353 Standard Deviation 2.086 Lower Fence -0.5322 Upper Fence 10.839