Question

Question 17 3 pts Small Sample Mean Problem. The maker of potato chips uses an automated packaging machine to pack its 20-ounce bag of chips. At the end of a shift 15 bags are selected at random and tested to see if the equipment needs to be readjusted. After one shift, a sample of 15 bags yielded the following data. mean - 19.45 s - .64 n = 15. Based on a BOE-.354 for a 95% Confidence Interval for the sample, we would be able to reject the Null Hypothesis for that mu = 20 for a two-tailed test. O True O False D Question 18 3 pts Small Sample Mean Problem. Each year the Heritage Foundation creates an Index of Economic Freedom for countries around the word based on assessments of freedom on property rights, corruption, business freedom, and labor freedom, among many others. We have a sample of 31 countries taken from the 2016 data. The following is a summary of this sample. Economic Freedom STEM Leaf 4* 64 5 09 18 22 24 32 42 5 60 65 74 75 76 77 91 93 6 00 02 12 19 22 30 31 659 66 75 87 7126 39 43 754 77 Mean Median Min Max Variance Std Deviaton CV Count 61.174 60.000 46.400 77.700 63.004 7.938 12.975 A value of 4164 is an index of 46.4 The Heritage Foundation feels an index below 60 reflects a country that is mostly unfree. If we were to conduct a test to see if the sample estimate is above 60 at an alpha level -10, what is (are) the critical t-value(s) for this test? -1.310 and 1.310 1310 1.697 -1.697 and 1.697

Answer 1

17. Here BOE=0.354 and mean=19.45

So CI is $mean\pm BOE=19.45\pm 0.354=(19.096,19.804)$

As mean=20 is not in the CI, so we can reject null hypothesis that H0:mu=20

Hence answer is True.