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- Georgia
- University of Georgia
- Statistics
- Statistics 2000
- Oneal
- Review Question's Answers Chapters 1-3.pdf

Jacob F.

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Page 1 of 7 c) 9/16 = .5625 = 56.25% d) 12/16 = .75 = 75% e) Mean = (sum of all 16 pulse rates) / 16 = 2009 / 16 = 125.5625 Median: 16 values so Median is the average of 8 th & 9 th value. (Average of n/2 and n/2 +1 values) Median = (129 + 136)/2 = 132.5. We can check and see that the mean < median so the distribution is skewed left. Page 2 of 7 f) Range = Largest – Smallest = 169 – 72 = 97 Use StatCrunch and get s = 27.522 g) Q 1 = median for the lower half of the values (the first 8 values), so we average the 4 th and 5 th values = (96 + 115)/2 = 105.5 So Q 1 = 105.5. Q 3 = median for the upper half of the values (the last 8 values), so we average the 12 th and 13 th values = (143 + 146)/2 = 144.5 So Q 3 = 144.5. IQR = Q 3 - Q 1 = 144.5 – 105.5 = 39 h) Q 1 – 1.5*IQR = 105.5 – 1.5*39 = 47 Q 3 + 1.5*IQR = 144.5 + 1.5*39 = 203 Since no data values fall below 47 or above 203, there are no outliers. i) Z-score = (value – mean) / standard deviation = (x -⎯x) / s = (72 – 125.5625) / 27.522 = -1.94617 The pulse rate of 72 is 1.94617 standard deviations below the mean pulse rate of 125.5625. Page 3 of 7 2. The median would be the best measure of center because there is an outlier present (which is the 203 value). The mean is sensitive to outliers and would not be a good measure of center. The median is resistant to outliers and would be a good measure of center. 3. r is the correlation coefficient which describes the linear relationship between two variables. From the r value, we can tell if there is a positive or negative linear relationship between x and y, and if there is a weak or strong linear relationship between x and y. r is negative so there is a negative relationship between x and y. r is close to -1 so there is a strong relationship between x and y. An r value of -.88 indicates a strong, negative relationship between x and y. Page 4 of 7 4. a) The slope is -4.2. For each day a bar of soap is used, the predicted weight decreases by 4.2 grams. b) The y-intercept is 78.5. A brand new bar of soap has a predicted weight of 78.5 grams. c) For x = 8, ˆ 78.5 4.2(8)y =− = 78.5 – 33.6 = 44.9 d) Residual = Actual – Predicted = 40 – 44.9 = -4.9 5. a) $1.60 – 2*$0.07 = $1.46 $1.60 + 2*$0.07 = $1.74 So $1.46 to $1.74 represents within 2 standard deviations. By the Empirical Rule, this means that 95% of the gas stations have prices between $1.46 and $1.74. b) 68% of the gas stations means within one standard deviation according to the Empirical Rule. Lower Number = $1.60 – $0.07 = $1.53 Higher Number = $1.60 + $0.07 = $1.67 So 68% of the gas stations have prices between $1.53 and $1.67. Page 5 of 7 6. a) 2.64 standard deviations below the mean corresponds to a Z-score = -2.64. b)Now we use the Z-score formula to find x. -2.64 = (x – 68.2) / 3 -7.92 = (x – 68.2) � x = -7.92 + 68.2 = 60.28 So a male that is 60.28 inches tall has a height that is 2.64 standard deviations below the mean height for all males. Page 6 of 7 7. Low Blood Pressure High Blood Pressure Total Under 50 (Middle-age) 64 51 115 50 and Over (Old) 31 73 104 Total 95 124 219 a)Proportion with high blood pressure = 124 / 219 = .56621 b)Only looking at professors 50 and over = 73 / 104 = .70192 = 70.192% c) Low Blood Pressure High Blood Pressure Under 50 (Middle-age) 64/115 = 0.5565 51/115 = 0.4435 50 and Over (Old) 31/104 = 0.2981 73/104 = 0.7019 Since the proportions in each row seem very different, there does appear to be an association between these two variables. Page 7 of 7 8. a) First we need to find the total number of students in STAT 2000 in Fall 2002: 320 + 650 + 210 + 80 + 30 = 1290 Proportion that scored higher than a 2.0 = (320 + 650)/1290 = 0.75194 b) The mean is just the average, so we need to sum up all the individual GPAs and divide by the total number of students: (320*4.0) (650*3.0) (210*2.0) (80*1.0) (30*0.0) 2.89147 1290 ++++ = c) The median is the GPA right in the middle, the one that is the 50 th percentile. We have a total of 1290 students, so if we do 1290/2 = 645. We need to find the 645 th and 646 th student. The first 320 students got a 4.0, the next 650 students got a 3.0, so somewhere in those next 650 students were the 645 th and 646 th students, so the median = 3.0. d) The mode is the value which has the highest frequency, so the mode for this dataset is the GPA = 3.0 because that group had the most number of students, 650. cricket3 Microsoft Word - Review for Test 1 Solutions

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