Lab# _______ Name: _____________Go to Sec: _______________ Reg Section: ________STAT303 Sec 508-510 Spring 2006 Exam #11. Why do we use s, the standard deviation, rather than s2, the variance, when describing a sample anddoing calculations?You CANNOT calculate a standard deviation without having first calculated the variance. The standard deviationis merely the positive square root of the variance.A. It s easier to calculate the standard deviation.’NO! see above*B. It s in the same units as the mean, ’ µ.Yes, the variance, since it’s a squared quantity, has squared units.C. You can t determine the variance without knowing the standard deviation.’This is backwords, see above.D. It doesn t matter which one you use since they are the same quantity.’NO! The standard deviation is the positive square root of the variance.E. Exactly two of the above are correct.Obviously, not.2. It is known that the mean price for a carton of milk is $1.00 with a standard deviation,s, of $0.20. Also, the mean price per pound for beef is $2.50 with a standard deviation of $0.60. If a local store, X, sells milk for $0.85 per carton and beef for $2.00 per pound, which of the following is true for the store?You need to use z-scores to compare unlike quantities zmilk = (0.85 1.00)/0.20 = 0.75, zbeef = (2.002.50)/0.60= 0.83. Since beef has a smaller z-score, it is relatively smaller (less than its mean), therefore, cheaper.A. Beef is relatively cheaper since it s $0.50 below the average vs. $0.15 below the average for milk.’We must look at z-scores, how many standard deviations below (or above) the mean, not just how far from the mean.B. Milk is relatively cheaper since it s only $0.85 where beef is $2.00 per pound.’Again, we must look at z-scores. We can’t compare actually dollars.C. You can t compare the two since the average prices are different.’Yes, you can if you look at z-scores.*D. Beef is relatively cheaper since it s 0.83’ s below the average vs. 0.75s below the average for milk.This is just the definition of the z-scores, so it’s correct.E. Milk is relatively cheaper since it s 0.75’ s below the average vs. 0.83s below the average for beef.This doesn’t make sense. Freq Pct Cum Pct65.00 1 1.0 1.0****72.00 2 2.0 3.073.00 3 3.0 6.074.00 6 6.0 12.075.00 3 3.0 15.076.00 5 5.0 20.077.00 7 7.0 27.0****78.00 15 15.0 42.079.00 11 11.0 53.0****80.00 11 11.0 64.081.00 6 6.0 70.082.00 11 11.0 81.0****83.00 3 3.0 84.084.00 7 7.0 91.085.00 2 2.0 93.086.00 3 3.0 96.087.00 2 2.0 98.089.00 2 2.0 100.0****Total 100.0 100.0The Cum Pct column is giving you percentages to the left of the ‘percentiles’ in the first column. Remember that a percentile is a number with that percentage of the observations to the left of it. The 5 Number Summary: the65605550454035Lab# _______ Name: _____________Go to Sec: _______________ Reg Section: ________minimum is the 1st (or 0th) percentile, Q1 is the 25th, x%is the 50th, Q3 is the 75th and the maximum is the 100th. The other thing you need to remember is that you must EXCEED these percentages. ONE of the 81’s is Q3. Not pick the closest one. 3. Which is the correct list of the 5 Number Summary for this table?A. 1, 25, 50, 75, 100B. 65, 75, 79, 84, 89C. 65, 77, 79, 81, 89*D. 65, 77, 79, 82, 89E. 65, 71, 77, 83, 894. Which of the following indicate that the data is skewed left?A. The mode (tallest bin) of the histogram is on the right, and the other bins get continually shorter as you go left.Yes, the tail (short bins) is on the skewed side. B. The boxplot has the median,x%, closer to Q3 and the maximum than to Q1 and the minimum.Yes, the closer the lines (say Q3 to the maximum), the taller the bins.C. The mean, x, is greater than the median, x%. No, this is backwards. The mean is on the skewed side.D. All of the above indicate left skewness.*E. Exactly two of the above (excluding D.)5. Which of the following is/are true about the boxplots(5)?The first and second boxplots are normal, the 3rd is uniform andthe 4th is slightly skewed to the right.A. All of the boxplots are normal.B. Only number 2 is normal since it has outliers on both ends (therefore symmetric).C. Only number 3 is normal.*D. Numbers 1 and 2 are normal.E. You can t determine whether they are normal or not ’with boxplots. We need normal quantile plots instead.Although normal quantile plots are the BEST way to determinewhether a dataset is normal or not, you can decide by looking at a boxplot. | HairMomHair| Black Blond Brown Red |Total-------+-----------------------+----- Black | 1 0 12 0 | 13Blonde | 0 7 10 0 | 17 Brown | 1 13 40 1 | 55 Gray | 0 0 3 0 | 3 Red | 0 0 3 0 | 3 White | 1 0 0 | 1-------+-----------------------+----- Total | 2 21 68 1 | 926. The Two-way table above is showing the relationship between a child s hair color and their mom s. ’ ’Which iof the following is/are true?A. If your mom has gone gray, you must have brown hair.You cannot make conclusions from a sample unless you know if it is a representative, random sample and there are no confounding variables.*B. In this survey, the child most likely has brown hair no matter what color their mom has.Except for the one white-haired mom, brown-haired children is the largest percentage in each row.C. If the child has brown hair, the likelihood of the mom having brown hair is 40/55.No, 40/55 is the proportion of brown-haired children given the mom is brown-haired. The correct proportion would be 40/68.D. All of the above are true.E. Only two of the above are true.Lab# _______ Name: _____________Go to Sec: _______________ Reg Section: ________7. Again using the
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