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Statistics 1000 Homework Set 1

by: Ashley Reuben

Statistics 1000 Homework Set 1 BIOSC 0370

Marketplace > University of Pittsburgh > Biological Sciences > BIOSC 0370 > Statistics 1000 Homework Set 1
Ashley Reuben
GPA 3.9
Anthony Bledsoe

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About this Document

Statistics 1000. Odd number problems from Introduction to the Practice of Statistics by Moore: Eighth Edition
Anthony Bledsoe
Class Notes
Statistics, Homework Set One, Odd Number Problems
25 ?




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This 7 page Class Notes was uploaded by Ashley Reuben on Monday January 25, 2016. The Class Notes belongs to BIOSC 0370 at University of Pittsburgh taught by Anthony Bledsoe in Summer 2015. Since its upload, it has received 29 views. For similar materials see Ecology in Biological Sciences at University of Pittsburgh.


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Date Created: 01/25/16
Ashley Reuben Homework 1 118 After looking at the stemplots for vitamin D in Figures 17 and 18 Figure 17 is the more preferable approach in plotting the data In Figure 18 the stem has two sets of the same variable in the stem one representing boys and the other representing girls both that is wrong Since the data is comparing boys and girls the data can just be combined into one plot but separated by putting girls on the left and boys on the right of the stem as shown in Figure 17 134 A Total Weight 2498 Total Percent 998 Total Percent Recycled 239 Comparsion of Percents l PercentOf39lotaI Pe rcentRecycIed B C Pie Chart of Material V Of the two graphs from part B and C I would prefer using the Bar Graph I think it had a better representation of the data considering you can approximately see the numerical values of the percentages whereas in the pie chart it shows no numerical values for the percentages It is also a better comparison of both percentages You are able to see which is greater than the other for each category As for the pie chart it might be hard for someone to decipher which category is larger considering some of the sections of both pie charts almost look equal even though they really aren t 142 The distribution of the dates from the coins would be skewed to the left because most of the dates will be more recent dates were as there will be fewer dates from the 19005 Newer coins are more common than older coins 35 95 quotHE 155 115 quotIE5 144 Hiiogram of IQ Nom39d 153995F39libl Frequency IQ As far as the shape of the histogram of IQ scores the most data lies within the middle The data doesn39t totally skew to one side and there is no symmetry within the graph The distribution is unimodal with one clear peak between 100 and 110 The center of the distribution is located at the median which is 110 The spread of the data has a pretty wide range ranging from 70 to 140 however most of the data lies near the center Outliers of the data would include numbers between 70 90 and numbers greater than 135 A majority of the data is lies close to the median The midpoint for the data is 110 therefore making the midpoint close to 100 Using the graph to determine the midpoint it is clear that the midpoint is clearly above 100 considering most of the IQ scores are larger than 100 168 A Dotplot of Percenl39Aloohol 00000000000000000000 0000000000 PC0000000000000000000 0000000000000 m 00000000000000000000 00000000000 000 00000000000 0000000 0 0 0 0 00 00 00 00 U H O U06 0 u 16 32 64 8 0 96 112 PercerItAlcohol Hidiogram of PercentAlcohol 70 601 8 Frequency 3o 20 1 5 2 a a a I 2 f I choose a mmmho39 Dot Plot to represent the numerical aspect of the data because I thought it was clearer than a stem and leaf plot You can clearly see where the most percent alcohol values lie without regards to the brands of the beer For the graphical aspect of the data I choose a histogram because of the amount of different types of brand named beer made it dif cult to use a bar graph or a pie chart If there were less types of brands then I think a bar graph would have been better B As far as outliers within the data anything less than 32 and above 83 would be classi ed as an outliers These outliers include O39Doul39s 040 Budweiser Select 55 240 Miller Genuine Draft quot64quot 280 Flying Dog Gonzo 920 Sierra Nevada Bigfoot 960 Flying Dog Wild Dog Coffee Stout 990 Flying Dog Horn Dog 1020 and Flying Dog Double Dog 1150 169 A The mean with outliers is 52 and the mean without outliers is 49 The median with outliers is 49 and the median without outliers is 49 The only thing that changed when negating the outliers was the mean These changes occur because since the outliers are signi cantly different than the rest of the data set they affect the mean average of the data when doing the calculation B The standard deviation with outliers is 14 01 is 44 and Q3 is 56 When the outliers are removed the standard deviation is 082 01 is 44 and Q3 is 56 Removing outliers has the same effect as it does when you remove the outliers when calculating the mean 170 A As far as the distributions of calories in the beer increases from left to right The median of the distribution is 150 and the mean is 154 The data has a wide range of the amount of calories in different brands of beer but most of the data is within the same range without regards to the outliers B The outliers to percent alcohol in some brands of beer are the same outliers in regards to the calories The outliers for the calories include Budweiser Select 55 55 Miller Genuine Draft quot64quot 64 O39Doul39s 70 Flying Dog Gonzo 271 Flying Dog Wild Dog Coffee Stout 288 Flying Dog Double Dog 313 Flying Horn Dog 314 and Sierra Nevada Bigfoot 330 178 Number of books read per year 2 6 8 9 15 40 46 47 50 With the smallest amount still in the data set the median would be 15 If the smallest observation were to be removed the median would then be 275 increasing by a signi cant amount 190 A Using Minitab to calculate the mean of the 10 given numbers for the data set it is veri ed that the mean of the data is 15 The stand deviation for the data set was calculated to be 54 B After creating a new data set with the 10 missing values set to 15 the mean was calculated to be 15 The standard deviation was calculated to be 37 C Using this data yet it is clear that adding the 10 addition numbers did not change the mean of the data set However after adding the 10 numbers the standard deviation decreased Eff 192 2 3 11 12 12 The thought process I used to come up with my 5 numbers in my data set is that since I already knew the median was 11 I automatically put 11 in the middle of the set I was also told that the mean of the data was 8 Since there are 5 total possible numbers I can use multiplied 8 mean times 5 possible numbers to get 40 lfl wanted my mean to be 8 the total of the 5 numbers had to equal up to 40 because 40 divided by 5 is 8 1102 228 338 174 228338 402 The range of 997 of the students39 scores is from 174 to 402 1104 Z250 28838 Z 05789 1106 Area between Area to the e Area to the left 340 and 370 of 370 984 of 340 9174 Alf l Z 340 28838 137 Z 370 28838 216 67 of students have scored between 340 and 370 1108 Z 84 x 22838 84 x 2599 260 1110 Distribution Plot Norrrd Mea120 StDev5 Density 9 E 001 000 20 2822 1128 A Z gt 155 00606 B Z lt 155 09394 C Z gt 070 7580 D 070 lt Z lt 155 06974 1130 A 058 B 029 1134 Jessica Z 18251498316103 Ashley Z 2821554 120 1136 Z 20601498316 178 X 17854 215 311 311 on ACT is equivalent to 2060 on SAT 1138 20401498316172 09564 95th percentile 1140 11585 08508 2 104 X 104316 1498 182663 1827 The top 15 for SAT scores will be above 1827


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