Knewton Alta Lesson 6 Assignment

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7. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column.

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10. Select the top visible segment and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, then click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot and click Format Selection on the tool bar above.

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9. Select the bottom section of the bars and set their Fill to blank. Click the bottom segment of one of the rectangles in the plot, and then click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button.

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A company is hiring people for a data entry position. The faster they can type, the better. In interviewing prospective employees, the hiring manager has acquired a sample of typing speeds (in words per minute, or WPM) from a test she offers as part of the interview process. She decides it is wise to analyze the data to get a sense of how fast people can type. The data is provided below. Use Excel to construct a box and whisker plot for the dataset.

1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1 copy the header for the dataset from A1. 5. In cell C2 write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 55.5, Q1 is 59.65, the median is 68.3, Q3 is 76.425, and the maximum is 119.5. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9 write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot and click Format Selection on the tool bar above. In the Format Error Bars window, click the Vertical Error Bars tab, click Plus button under Direction, and click the Cap button under End Style. Under Error Amount click Custom, and then press Specify Value. In the resulting pop up click on the Positive Error Value field and select the data in cell C12. 15. Create errors bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. For this data set the vertical axis was changed to run from 55.0 to 120.0 with major ticks in increments of 5 and minor ticks marks in increments of 1. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. The completed box plot should look like attached image.

Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time when compared to her team. Compute the z-scores for Angie and Beth. Provide your answer below: Angie: z= Beth: z=

Angie: z= -1.25 Beth: $z= -2 Angie's team has a mean time of μ=27.2 with a standard deviation of σ=0.8. The z-score corresponding to Angie's swim time of x=26.2 seconds is z=x−μσ=26.2−27.2/0.8= −1.25 Beth's team has a mean time of μ=30.1 with a standard deviation of σ=1.4. The z-score corresponding to Beth's swim time of x=27.3 seconds is z=x−μσ=27.3−30.1/1.4= −2

The following data set provides information about the City of Somerville Assessors Valuation for the fiscal year 2016. For commercial buildings, what is the sample standard deviation of the living area? Round your answer to the nearest hundred. Provide your answer below: standard deviation =

standard deviation = 24800 The standard deviation is the square root of the sample variance of the living area of the five commercial buildings. The sample variance is 615,032,637. Take the square root, rounding to the nearest hundred. √615,032,637=24,800

In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount click Custom, and then press Specify Value. In the resulting pop-up click on the Positive Error Value field and select the data from the row labeled Maximum−Q3 in the differences table. In the images above and below this is equivalent to writing "=Sheet1!$E$12:$G$12."

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6. Create a stacked column chart type from the quartile differences. Select all the dataset headers and the five rows beneath them in the differences table. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types.

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Given the following box-and-whisker plot, decide if the data is skewed or symmetrical. Select the correct answer below: The data are skewed to the left. The data are skewed to the right. The data are symmetric.

The data are skewed to the left. Note that the whisker on the left is much longer than the whisker on the right. So there are several much smaller values on the left. Therefore, the data are skewed left. This could represent a researcher studying the age of first trying tobacco. Since there is a legal requirement of a certain age, older people will try tobacco first while few try younger.

9 (continued). Set Fill to blank for the new bottom segment and the top segment as well. It may be a good time to adjust the vertical axis so that its maximum is slightly above the maximum of all data values and the minimum is below the minimum of all data values. The maximum across all data sets in the images above is 5.09 and the minimum is 3.73. For the image below, the vertical axis was changed to run from 3.50 to 5.25 with major tick marks every 0.25 and minor tick marks every 0.05.

The two remaining segments are going to form the box part of the box and whisker plot. To obtain the whiskers, we'll use the differences to appropriately size error bars that will extend from either side of the two segments.

Once the box and whisker plot has been constructed for each dataset, it is time to analyze the dataset with the box and whisker plot. The box and whisker plot is a visualization of the five-number summary, which is the minimum, first quartile, median, third quartile, and the maximum. The quartiles, calculated in steps 2 and 3, separate equal amounts of data. Between each pair of numbers lies 25% of the data points.

This means that between any three adjacent numbers in the five-number summary is 50% of the data. The box part of the box and whisker plot contains all of the data values between Q1 and Q3, which is approximately 50% of the data. The width of the box is called the interquartile range, or IQR for short. If the data is highly concentrated around the median, the box will be thin. If the left whisker and left side of the box is thin while the right side of the box and right whisker are long, then the dataset is right skewed. If the right whisker and right side of the box is thin while the left side of the box and left whisker are long, then the dataset is left skewed. If the two halves of the box are equally wide and the two whiskers are equally wide, the data is symmetric. If the whiskers and the two halves of the box are all the same width, the dataset is uniform. The thinner the box is relative to the whiskers, the more concentrated the data is about the median. The width of the entire box and whisker plot is the range of the dataset.

The following data values represent the daily amount spent by a family each day during a 7 day summer vacation. Find the standard deviation of this data set: $96,$125,$80,$110,$75,$100,$121 Round the final answer to one decimal place. Provide your answer below: standard deviation =

standard deviation = 17.7 The population standard deviation is the square root of the population variance. Since we just found that the population variance to be 314.3, the sample standard deviation is √314.3≈17.7.

A policy think tank focused on teenage obesity was able to get several pediatricians to release anonymous data on the weights in pounds of sixteen-year-olds who had come to their office this year. A sample of 20 of the weights is included below. Construct a box and whisker plot using Excel and the QUARTILE.INC function to select the correct plot below.

1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 110, Q1 is 129.75, the median is 146.5, Q3 is 173, and the maximum is 329. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selection on the toolbar above. In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field and select the data in cell C12. 15. Create errors bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. For this dataset, the vertical axis was changed to run from 100 to 350 with major tick marks in increments of 25 and minor ticks marks in increments of 5. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. The completed box plot should look something like the following image.

A company sells classes on its speed-reading technique, which it advertises to customers through a free, online survey. The results of 20 of these tests are included below. Use Excel and the QUARTILE.INC function to construct a box and whisker plot for the dataset. What is the value of the first quartile?

222 1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 191, Q1 is 222, the median is 239.5, Q3 is 257, and the maximum is 302. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment, and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selection on the tool bar above. In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field and select the data in cell C12. 15. Create error bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. Looking at the box and whisker plot, the first quartile, which is where the bottom whisker intersects with the bottom side of the box, is 222.

The auditors for a health insurance company are reviewing the bills from client's stays at hospitals from last year. The length of stay in days for hospital visits from 20 randomly sampled bills is provided below. Use Excel and the QUARTILE.INC function to construct a box and whisker plot for the dataset. What is the value of the third quartile? Round your answer to two decimal places.

7.25 1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 2, Q1 is 5.75, the median is 6, Q3 is 7.25, and the maximum is 9. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Designtab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Formattab, and below that click the Format Selectionbutton. In the Format Data Serieswindow that pops up, click the Filltab, and then click the No Fillradio button. 14. Select the top visible segment, and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selectionon the toolbar above. In the Format Error Bars window, click the Vertical Error Barstab, click the Plus button under Direction, and click the Capbutton under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field and select the data in cell C12. 15. Create error bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selectionagain. In the Format Error Bars window, click the Vertical Error Barstab, click the Minusbutton under Direction, and click the Capbutton under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. Looking at the box and whisker plot, the third quartile, where the top whisker intersects with the top side of the box, is 7.25.

A veterinary researcher is studying a particular type of dog called the Australian Cattle Dog. The researcher has acquired data on a sample of 30 dogs, including the weight in pounds of each of the dogs. The dog weight dataset is provided below. Using Excel, calculate the mean, median, and mode of the dataset (round your answers to one decimal place as needed). Provide your answer below: Mean = Median = Mode =

Mean = 40.6 Median = 40.6 Mode = 38.5 The mean, median, and mode can be calculated quickly and easily in Excel using the built-in functions for these calculations. Open the accompanying dataset in Excel. The range of the data is A2:A31. In B1 type "Mean", in B2 type "=AVERAGE(", select the data or write their range, and then hit ENTER. In C1 type "Median", in C2 type "=MEDIAN(", select the data or write their range, and then hit ENTER. In D1 type "Mode", highlight cell range D2:D5, type "=MODE.MULT(", select the data or write their range, and then hit CTRL+SHIFT+ENTER. Rounding the results in B2 and C2 to one decimal place should result in a mean of 40.6 and a median of 40.6. All cells given to MULT.MODE, the cell range D2:D5, should show a value of 38.5. This indicates 38.5 is the only mode.

The following data set represents the ages of all 6 of Nancy's grandchildren. 11,8,5,6,3,9 To determine the "spread" of the data, would you employ calculations for the sample standard deviation, or population standard deviation for this data set? Select the correct answer below: Use calculations for sample standard deviation Use calculations for population standard deviation

Use calculations for population standard deviation To determine if sample standard deviation or population standard deviation should be used, determine if the data set represents data values collected from the entire population, or from a subset of the population. If the data values represent data collected from a subset of the population, then the sample standard deviation should be used. If the data values represent data collected from the entire population of interest, then the population standard deviation should be used. In this case, the population standard deviation should be used because the data set represents all of, that is the total population of, Nancy's grandchildren.

5. Under the cell below the first row of data in the new table write "=E3-E2", where E3 is the corresponding cell in the quartiles table. Copy and paste this new cell to the remaining three cells underneath, and then copy and paste the column of five cells to every column corresponding to a dataset in the differences table.

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The members of the marketing team for a bank want to know how saturated the market is before the bank releases a proposed new credit card to the general public. The marketing team polls 20 randomly selected customers by email to ask how many credit cards they already own. The dataset is included below. Use Excel and the QUARTILE.INC function to construct a box and whisker plot for the dataset. What is the value of the minimum for this dataset?

1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 0, Q1 is 1, the median is 2, Q3 is 3, and the maximum is 4. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment, and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selection on the tool bar above. In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field, and select the data in cell C12. 15. Create error bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. Looking at the box and whisker plot, the minimum, which is the bottom side of the bottom whisker, is 0.

A consumer report was released concerning the prices of various food products. The report listed the monthly average price of a pound of beef for 20 different months. Construct a box and whisker plot using Excel and the QUARTILE.INC function. Then, choose the correct answer below.

1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 5.138, Q1 is 5.26275, the median is 5.332, Q3 is 5.46425, and the maximum is 5.687. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment, and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selection on the tool bar above. In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field and select the data in cell C12. 15. Create errors bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. For this dataset, the vertical axis was changed to run from 5.1 to 5.7 with major tick marks in increments of 0.1 and minor ticks marks in increments of 0.025. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. The completed box plot should look something like the following image.

The following set of data represents newly opened credit cards per employee, find the sample standard deviation: 10, 3, 10, 3, 9 Round your answer to ONE decimal place. Provide your answer below: standard deviation = credit cards

standard deviation = 3.7 credit cards Remember that the sample standard deviation is the square root of the sample variance. When we find that the sample variance is 13.5, we can find that the sample standard deviation is √13.5 = 3.7. The larger the standard deviation (further away each value is from the mean), the wider range of newly opened credit cards per employee. This would say the newly opened credit cards per employee is 7±3.7 (units).

Based on the z-scores calculated above for Angie and Beth, which swimmer had the fastest time when compared to her team? Select the correct answer below: Angie Beth Angie and Beth perform equally well. There is not enough information for a comparison.

Beth Angie has a z-score of −1.25 and Beth's z-score is −2. Both Angie and Beth have negative z-scores, meaning they both swim in less time than their team's mean time. In terms of swim times, lower values are faster times, so Beth has the faster swim time when compared to her team.

The following data set provides infomation about the City of Somerville Assessors Valuation for the fiscal year 2016. Which statements are true about the pattern of data for the sample standard deviations of the commercial buildings total assessed land value and total assessed parcel value, and the residential buildings total assessed land value and total assessed parcel value? Select all that apply. Select all that apply: Commercial buildings have a greater standard deviation in both categories than residential. The standard deviation for Commercial Total Assessed Land Value is only two times the standard deviation for Residential Total Assessed Land Value. The largest difference in standard deviation is from Residential Total Assessed Land Value to Commercial Total Assessed Parcel Value. The smallest decrease in standard deviation is from Residential Total Assessed Parcel Value to Residential Total Assessed Land Value.

Commercial buildings have a greater standard deviation in both categories than residential. The largest difference in standard deviation is from Residential Total Assessed Land Value to Commercial Total Assessed Parcel Value. The smallest decrease in standard deviation is from Residential Total Assessed Parcel Value to Residential Total Assessed Land Value. Commercial buildings do have a greater standard deviation in both categories than residential. The standard deviation for Commercial Total Assessed Land Value (4361842) is not more than two times the standard deviation for Residential Total Assessed Land Value (97477). The standard deviation for Commercial Total Assessed Land Value is about 45 times more than the standard deviation for Residential Total Assessed Land Value. The largest difference in standard deviation is from Residential Total Assessed Land Value to Commercial Total Assessed Parcel Value. The smallest decrease in standard deviation is from Residential Total Assessed Parcel Value to Residential Total Assessed Land Value.

The histograms below are of data sets representing four different stocks. Which has the smallest standard deviation?

Remember that the standard deviation is a measure of how spread out the data is. If the values are concentrated around the mean, then a data set has a lower standard deviation. A histogram with fewer values and higher frequencies has a lower standard deviation than a histogram which has a shorter hill and a wider range of values. A smaller standard deviation also suggests less risk in that particular stock.

A student studying statistics wants to look at data for his favorite sport, American football. He collects data on the lengths of 100 field goals from various games over several seasons. Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Sample standard deviation Since the data is from only a sample of games, and not all of the games for all of the seasons, he should calculate a sample standard deviation.

A researcher for an organization that collects and reports on crime has data on the murder rates for only 10 states from a certain year. The murder rate is the number of murders per 100,000 inhabitants. Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Sample standard deviation The researcher has data on murder rates for only 10 states, and not all of the states, so a sample standard deviation would be more appropriate.

A researcher with an organization dedicated to studying trends in crime and the causes underlying crime has been focusing on the effects of unemployment. The unemployment rates for 10 states are provided below. Use a TI-83, TI-83 Plus, or TI-84 to calculate the sample standard deviation and the sample variance. Round your answers to one decimal place. Standard Deviation = Variance =

Standard Deviation = 1.6 Variance = 2.6 To determine the sample standard deviation and sample variance for a data set {x1,x2,...,xn} using a graphing calculator (such as a TI-83, TI-83 Plus, or TI-84), follow these steps: Enter data into List L1. Press STAT and then ENTER to get to the list edit screen. Run 1-Var Stats. Press STAT, then press RIGHT, and then press ENTER to paste 1-Var Stats to the home screen. Press ENTER again. This shows the sample standard deviation Sx. Access and square Sx from the statistics variable screen to get the sample variance. Press VARS, then 5, then 3, then x2, and then ENTER. The sample standard deviation is s≈1.6 and the sample variance is s2≈2.6, rounding eachto one decimal place.

The Bureau of Labor Statistics compiles and makes publicly available data from a range of different sectors of the economy. One number it reports is a weighted average of the costs of certain goods, called the Consumer Price Index (CPI). The CPI is related to the price but is not a dollar amount. The rise in prices is used as a metric of inflation. To make the metric more reliable, a chained CPI was created. The regular CPI does not update the types of goods it averages often enough to reflect the market trends, like a switch from apples to oranges resulting from a rise in apple prices. The chained CPI is a measure of price that takes this into account, and there are several bills that are tied to the chained CPI index in order to determine the payout the bill allots each year. The monthly average chained CPI for urban apparel for all urban consumers for 20 consecutive months is provided below. The data are not seasonally adjusted. Construct a box and whisker plot using Excel and the QUARTILE.INC function to choose the correct plot below.

1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 89.02, Q1 is 90.1975, the median is 91.66, Q3 is 94.825, and the maximum is 99.15. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab and then click the No Fill radio button. 14. Select the top visible segment, and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selection on the tool bar above. In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field and select the data in cell C12. 15. Create errors bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. For this dataset, the vertical axis was changed to run from 88 to 100 with major tick marks in increments of 1 and minor ticks marks in increments of 0.25. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. The completed box plot should look like the attached image.

The following data set provides information of Households by Total Money Income, Race, and Hispanic Origin of Householder. Looking at the data set or chart for household income for all races in 2015, what percent of households are in the category range that contains the mean? Select the correct answer below: 12.7% 16.7% 12.1% 14.1%

12.1% The mean is $79.263, which is in the category of $75,000 to $99,999. This range represents 12.1% of all races.

The following lists of data represent five separate departments' employees used vacation days per year. Which of the following lists of data has the smallest standard deviation? Select the correct answer below: 30, 21, 19, 17, 16, 32, 26, 25, 19, 16 5, 11, 15, 7, 5, 9, 8, 16, 14, 11 25, 24, 28, 18, 32, 34, 34, 22, 28, 19 17, 19, 17, 18, 17, 16, 16, 16, 17, 20 9, 16, 14, 22, 20, 9, 19, 16, 21, 8

17, 19, 17, 18, 17, 16, 16, 16, 17, 20 Remember that standard deviation is a measure of how spread out the values are. The list 17, 19, 17, 18, 17, 16, 16, 16, 17, 20 has the smallest standard deviation because its values are all relatively close together. A smaller standard deviation also suggests less variablity (more consistent) in this department's used vacation days per employee per year. This department could see this as each employee taking equal vacation days each year.

The following lists of data represent five separate departments' employees time spent in meetings for a week. Which of the following lists of data has the largest standard deviation? Select the correct answer below: 9, 11, 13, 10, 13, 13, 13, 9, 9, 10 24, 15, 21, 23, 9, 22, 12, 21, 20, 13 15, 18, 18, 18, 18, 15, 15, 15, 16, 16 23, 24, 23, 24, 26, 25, 26, 23, 26, 24 10, 12, 13, 11, 11, 14, 14, 11, 12, 12

24, 15, 21, 23, 9, 22, 12, 21, 20, 13 Remember that standard deviation is a measure of how spread out the values are. The list 24, 15, 21, 23, 9, 22, 12, 21, 20, 13 has the largest standard deviation because its values are all relatively spread apart. The department could see this as having inconsistent (or variability) weekly meeting times for each employee.

Based on the z-scores calculated above for Stephan's electric bills in IL and FL, in which state is his electric bill higher, when compared to their respective distributions? Select the correct answer below: Illinois Florida Stephan's electric bills in both states are comparable. There is not enough information for a comparison.

Illinois Both z-scores are negative, meaning Stephan's bills are below the average in both states. His Florida bill is 0.75 standard deviations below the Florida state mean, but his Illinois bill is only 0.625 standard deviations below the Illinois state mean. This means his bill was higher in Illinois, when compared to the state distributions of electric bills.

Karl and Fredo are basketball players who want to find out how they compare to their team in points per game. The mean amount of points per game and standard deviations for their team were calculated. Karl's z-score is 0.9. Fredo's z-score is −0.65. Which of the following statements are true about how Karl and Fredo compare to their team in points per game? Select all that apply. Select all that apply: Karl scores 0.9 more points per game on average than his teammates. Fredo's average points per game is 0.65 standard deviations greater than his teammates's average points per game. Karl's average points per game is 0.9 standard deviations greater than his teammates' average points per game. Fredo's average points per game is closer to the team's mean than Karl's.

Karl's average points per game is 0.9 standard deviations greater than his teammates' average points per game. Fredo's average points per game is closer to the team's mean than Karl's. The z-score is the number of standard deviations a data value is from the mean of the data set. Karl's average points per game is 0.9 standard deviations greater than his team mean. Fredo's average points per game is 0.65 standard deviations less than his team mean. But, since Fredo's z-score is −0.65, the distance from this to the mean is less than 0.9. since |−0.65|<0.9. So, Fredo's average points per game is closer to the team's mean than Karl's.

A student of statistics and fan of baseball is looking over the player stats for a list she is compiling of "top thirty players nobody remembers." The data for the batting averages of these 30 players are provided below. Using Excel, calculate the mean, median, and mode of the dataset (round your answers to three decimal places as needed). Provide your answer below: Mean = Median = Mode =

Mean = 0.271 Median = 0.270 Mode = 0.301 The mean, median, and mode can be calculated quickly and easily in Excel using the built-in functions for these calculations. Open the accompanying dataset in Excel. The range of the data is A2:A31. In B1 type "Mean", in B2 type "=AVERAGE(", select the data or write their range, and then hit ENTER. In C1 type "Median", in C2 type "=MEDIAN(", select the data or write their range, and then hit ENTER. In D1 type "Mode", highlight cell range D2:D5, type "=MODE.MULT(", select the data or write their range, and then hit CTRL+SHIFT+ENTER. Rounding the results in B2 and C2 to three decimal places should result in a mean of 0.271 and a median of 0.270. All cells given to MULT.MODE, the cell range D2:D5, should show a value of 0.301. This indicates 0.301 is the only mode

A budding collector is looking over her growing coin collection, recording various details about each coin in a notebook. The collector has 17 rare coins in her collection. She wants to find the standard deviation of the ages of the coins in her collection. Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Population standard deviation She is looking at the distribution of ages of all 17 rare coins in her collection, which is the entire population of her rare coin collection. So a population standard deviation would be more appropriate.

The following data set provides infomation about the City of Somerville Assessors Valuation for the fiscal year 2016. Find the standard deviations for the commercial buildings total assessed land value and total assessed parcel value, and the residential buildings total assessed land value and total assessed parcel value. Which has the smallest standard deviation? Select the correct answer below: Commercial Total Assessed Land Value Residential Total Assessed Land Value Residential Total Assessed Parcel Value Commercial Total Assessed Parcel Value

Residential Total Assessed Land Value Residential Total Assessed Land Value's standard deviation is the smallest for those four, due to the narrowest range of data. Residential Total Assessed Land Value's standard deviation is 97,477.

A manufacturing plant produces custom hardware for specific applications in construction. One particular kind of bolt that is intended to have a length of 44mm is produced by two different machines, A and B. Tristan is an employee in the department that produces this bolt. He randomly selects 12 bolts from each machine and measures the length of each one. The lengths of the bolts, in millimeters, are shown in the table below. Treat the given data sets as samples. Based on the z-scores you found above, which machine is more likely to produce a bolt that is 43.97mm long? Select the correct answer below: Since the absolute value of the z-score for Machine A is less than the absolute value of the z-score for Machine B, Machine A is more likely to produce a bolt that is 43.97mm long. Since the absolute value of the z-score for Machine B is less than the absolute value of the z-score for Machine A, Machine B is more likely to produce a bolt that is 43.97mm long. Since the absolute value of the z-score for Machine A is greater than the absolute value of the z-score for Machine B, Machine A is more likely to produce a bolt that is 43.97mm long. Since the absolute value of the z-score for Machine B is greater than the absolute value of the z-score for Machine A, Machine B is more likely to produce a bolt that is 43.97mm long.

Since the absolute value of the z-score for Machine A is less than the absolute value of the z-score for Machine B, Machine A is more likely to produce a bolt that is 43.97mm long. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for the bolt produced by Machine A, 1.1, is less than the absolute value of the z-score for the bolt produced by Machine B, 1.35, Machine A is more likely to produce a bolt that is 43.97mm long.

A student studying statistics wants to look at data for his favorite sport, American football. He collects data on the lengths of 100 field goals from various games over several seasons. The data are provided below. Use Excel to calculate the sample standard deviation and the sample variance. Round your answers to one decimal place. Provide your answer below: Standard Deviation = Variance =

Standard Deviation = 9.1 Variance = 82.8 To determine the sample standard deviation and sample variance for a data set {x1,x2,...,xn} using Excel, follow these steps: Open the included data with Excel. The data should occupy cell range A2:A101. Select cell B3 and type "=STDEV.S(", select the range A2:A101, and then hit ENTER. This gives the sample standard deviation. Select cell B4 and type "=VAR.S(", select the range A2:A101, and then hit ENTER. This gives the sample variance. The sample standard deviation is s≈9.1 and the sample variance is s2≈82.8, rounding each to one decimal place.

8. Select and delete the legend because it refers to the data from the differences table and some of this data will be made invisible in the following steps. The differences are just telling Excel how far apart the start of each part of the box and whisker plot is from the previous part. The bottom doesn't start at 0, but at the minimum of the data set, which is the top of the bottom section. So our box plot should not show the segment from 0 up to the bottom of the bar.

Steps (7) and (8) produce the image below.

Given the following histogram, decide if the data is skewed or symmetrical. Select the correct answer below: The data are skewed to the left. The data are skewed to the right. The data are symmetric.

The data are skewed to the right. Note that the histogram has most of its values concentrated on the left, with several much larger values on the right. Therefore, the data are skewed right.

An independent census agency polls a random sample of 30 households in a particular neighborhood to find how many people live in each household. Using Excel, calculate the mode(s) of the dataset provided below. Select the correct answer below: There are two modes. The modes are 2 and 4. There is one mode. The mode is 2. There is one mode. The mode is 4. There is no mode.

There are two modes. The modes are 2 and 4. The mean, median, and mode can be calculated quickly and easily in Excel using the built-in functions for these calculations. In C1 type "Mode", highlight cell range C2:C5, while highlighted type "=MODE.MULT(A2:A31)", and then press CTRL+SHIFT+ENTER. If there is one mode, each cell in cell range C2:C5 will display that mode's value. If there is more than one mode, each mode is displayed in one of the cells in the cell range C2:C5. If each cell in the cell range C2:C5 has a unique mode value, then there may be more modes and step 1 should be repeated with a taller cell range, for example C2:C10. If there are more cells in the range than there are modes but there is more than one mode, the remaining cells in the range will display #N/A to indicate there were no other modes found. Cell C2 should display 4 and cell C3 should display 2. Cells C4 and C5 both should display #N/A, indicating that there are two modes: 0 and 4.

The following data values represent the daily amount spent by a family each day during a 7 day summer vacation. $96,$125,$80,$110,$75,$100,$121 To determine the "spread" of the data, would you employ calculations for the sample standard deviation, or population standard deviation for this data set? Select the correct answer below: Use calculations for sample standard deviation Use calculations for population standard deviation

Which of the following frequency tables shows a skewed data set? Select all that apply. Select all that apply: Value 5,6,7,8,9,10,11,12,13,14 Frequency 2,5,3,15,11,15,13,5,3,4 Value 13,14,15,16,17,18,19,20 Frequency 2,5,1,13,23,26,15,12 Value 5,6,7,8,9,10,11,12,13,14,15 Frequency 1,1,9,20,24,20,6,11,5,2,1 Value 0,1,2,3,4,5,6,7 Frequency 4,12,23,28,17,7,6,3

Value 13,14,15,16,17,18,19,20 Frequency 2,5,1,13,23,26,15,12 Value 0,1,2,3,4,5,6,7 Frequency 4,12,23,28,17,7,6,3 Remember that data are left skewed if there is a main concentration of large values with several much smaller values. Similarly, right skewed data have a main concentration of small values with several much larger values. We can see that the following is left skewed because of the concentration of large values with many smaller values: Value 13,14,15,16,17,18,19,20 Frequency 2,5,1,13,23,26,15,12 And the following is right skewed because of its concentration of small values with many larger values: Value 0,1,2,3,4,5,6,7 Frequency 4,12,23,28,17,7,6,3 The other frequency tables are more balanced and symmetrical.

A researcher with an organization dedicated to studying trends in crime and the causes underlying crime has been focusing on the effects of unemployment. Data on the unemployment rates for only 10 states are provided below. Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Wes is the owner of a real estate agency and is analyzing the time the agents take to sell houses. He reviews each house sold by his agency to determine the number of days each house was on the market before it was sold. The data for a random sample of 20 houses sold in the last year are provided below. Use Excel and the QUARTILE.INC function to construct a box and whisker plot for the dataset. What is the value of the third quartile?

165 1. Open the dataset in Excel. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Steps (2) through (6) have you construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1, copy the header for the dataset from A1. 5. In cell C2, write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 65, Q1 is 111.75, the median is 140, Q3 is 165, and the maximum is 199. Continue on to construct the box plot. Steps (7) through (10) have you construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9, write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Steps (11) through (15) have you construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then the Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment, and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and then click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot, and click Format Selection on the toolbar above. In the Format Error Bars window, click the Vertical Error Bars tab, click the Plus button under Direction, and click the Cap button under End Style. Under Error Amount, click Custom, and then press Specify Value. In the resulting pop-up, click on the Positive Error Value field and select the data in cell C12. 15. Create error bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount, click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color. Looking at the box and whisker plot, the third quartile, where the top whisker intersects with the top side of the box, is 165.

A large company has two major departments, Development and Marketing. 100 employees are randomly selected from each department, and the age of each employee, in years, is recorded in the accompanying samples. Both departments have an employee who is 22 years old. Based on the z-scores you calculated above, would it be more likely for the Development or Marketing department to have a 22 year old employee? Select the correct answer below: A 22 year old employee would more likely be found in the Development department, because the absolute value of the z-score for Development is less than for Marketing. A 22 year old employee would more likely be found in the Development department, because the absolute value of the z-score for Development is greater than for Marketing. A 22 year old employee would more likely be found in the Marketing department, because the absolute value of the z-score for Development is less than for Marketing. A 22 year old employee would more likely be found in the Marketing department, because the absolute value of the z-score for Development is greater than for Marketing.

A 22 year old employee would more likely be found in the Marketing department, because the absolute value of the z-score for Development is greater than for Marketing. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for the Development department, 1.96, is more than the absolute value of the z-score for the Marketing department, 1.70, a 22 year old employee would more likely be found in the Marketing department.

Natasha records the price for a gallon of home-heating oil from 10 randomly selected providers in her region on May 15. She does it again with another 10 randomly selected providers on November 15. The results (in dollars) are shown in the accompanying samples. Based on the z-scores you calculated above, would a price of $2.899 be more likely on May 15 or on November 15? Select the correct answer below: A price of $2.899 would be more likely on May 15, because the absolute value of the z-score for this price on May 15 is less than on November 15. A price of $2.899 would be more likely on May 15, because the absolute value of the z-score for this price on May 15 is greater than on November 15. A price of $2.899 would be More likely on November 15, because the absolute value of the z-score for this price on May 15 is less than on November 15. A price of $2.899 would be more likely on November 15, because the absolute value of the z-score for this price on May 15 is greater than on November 15.

A price of $2.899 would be more likely on November 15, because the absolute value of the z-score for this price on May 15 is greater than on November 15. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for November 15, 1.74, is less than the absolute value of the z-score for May 15, 3.27, a price of $2.899 would be more likely on November 15.

Austin is looking to rent a two-bedroom apartment in one of two towns, Gardiner or Augusta. He randomly selects 12 two-bedroom apartments from both towns and records the area of each apartment. The area of each apartment, in square feet, is provided in the samples shown below. Both towns have a two-bedroom apartment that is 635 square feet. Based on the z-scores you calculated above, for which town would it be more likely to find a two-bedroom apartment that is 635 square feet. Select the correct answer below: A two-bedroom apartment in Gardiner would be more likely have an area of 635 square feet, because the absolute value of the z-score for Gardiner is greater than for Augusta. A two-bedroom apartment in Augusta would be more likely have an area of 635 square feet, because the absolute value of the z-score for Gardiner is greater than for Augusta. A two-bedroom apartment in Gardiner would be more likely have an area of 635 square feet, because the absolute value of the z-score for Gardiner is less than for Augusta. A two-bedroom apartment in Augusta would be more likely have an area of 635 square feet, because the absolute value of the z-score for Gardiner is less than for Augusta.

A two-bedroom apartment in Augusta would be more likely have an area of 635 square feet, because the absolute value of the z-score for Gardiner is greater than for Augusta. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score compared to Augusta apartments, 1.81, is less than the absolute value of the z-score compared to Gardiner apartments, 1.86, a two-bedroom apartment in Augusta would be slightly more likely have an area of 635 square feet.

Isabel is looking at the prices for round-trip airfare from Setauket to Orchard Park where both flights occur on Wednesday or both flights occur on Sunday. She randomly selected 12 round-trips where both flights occur on Wednesday and 12 round-trips where both flights occur on Sunday. Isabel records the prices for each round-trip airfare in dollars as shown in the samples provided. One of the round-trips on Wednesday and one of the round-trips on Sunday both cost $205. Based on your calculations of the z-scores, would airfare be more likely to costs $205 on Wednesday or on Sunday? Select the correct answer below: Airfare would be more likely to cost $205 on Sunday, because the absolute value of the z-score for Wednesday is greater than for Sunday. Airfare would be more likely to cost $205 on Sunday, because the absolute value of the z-score for Wednesday is less than for Sunday. Airfare would be more likely to cost $205 on Wednesday, because the absolute value of the z-score for Wednesday is greater than for Sunday. Airfare would be more likely to cost $205 on Wednesday, because the absolute value of the z-score for Wednesday is less than for Sunday.

Airfare would be more likely to cost $205 on Sunday, because the absolute value of the z-score for Wednesday is greater than for Sunday. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for Wednesday, 1.77, is greater than the absolute value of the z-score for Sunday, 1.38, airfare would be more likely to cost $205 on Sunday.

A manufacturing plant produces custom hardware for specific applications in construction. One particular kind of bolt that is intended to have a length of 84mm is produced by two different machines, A and B. Tristan is an employee in the department that produces this bolt. He randomly selects samples of 100 bolts from each machine and measures the length of each one. The lengths of the bolts, in millimeters, are shown in the table below. Each machine produced a bolt that has a length of 84.05mm. Based on the z-scores you calculated above, would an 84.05mm bolt more likely be produced by Machine A or Machine B? Select the correct answer below: An 84.05mm bolt would more likely be produced by Machine A, because the absolute value of the z-score for Machine A is greater than for Machine B. An 84.05mm bolt would more likely be produced by Machine A, because the absolute value of the z-score for Machine A is less than for Machine B. An 84.05mm bolt would more likely be produced by Machine B, because the absolute value of the z-score for Machine A is greater than for Machine B. An 84.05mm bolt would more likely be produced by Machine B, because the absolute value of the z-score for Machine A is less than for Machine B.

An 84.05mm bolt would more likely be produced by Machine A, because the absolute value of the z-score for Machine A is less than for Machine B. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for Machine A, 1.74 is less than the absolute value of the z-score for Machine B, 3.57, an 84.05mm bolt would more likely be produced by Machine A.

Coastal z -score: 1.80 Inland z -score: -1.57 Step 1. Press STAT and then EDIT. Enter the Coastal data into list L1 and the Inland data into list L2. Step 2. Press STAT, then CALC, and then 1-VAR STATS. Step 3. Enter 1-VAR STATS, 2ND, and then the 1 key for the list L1. Enter 1-VAR STATS, 2ND, and then the 2 key for the list L2. Step 4. Read the sample mean, x¯¯¯, and the sample standard deviation, sx, for each list from the output. The sample mean for the Coastal data is 37.51 with a sample standard deviation of 6.157, rounded to three decimal places. The sample mean for the Inland data is 57.16 with a sample standard deviation of 5.436, rounded to three decimal places. Step 5. Compute the z-score for the coastal location having 48.6 inches of total seasonal snowfall, rounding to two decimal places. z≈48.6−37.51/6.157≈1.80 Compute the z-score for the inland location having 48.6 inches of total seasonal snowfall, rounding to two decimal places. z≈48.6−57.16/5.436≈−1.57

Casey is looking to rent a two-bedroom apartment in one of two towns, Gardiner or Augusta. He randomly selects 100 two-bedroom apartments from both towns and records the area of each apartment. The area of each apartment, in square feet, is provided in the samples shown below. Both towns have a two-bedroom apartment that has an area of 645 square feet. Which of the two towns is more likely to have a two-bedroom apartment with 645 square feet? Use Excel to calculate the z-scores corresponding to a 645 square foot apartment in each city. Round each z-score to two decimal places. Provide your answer below: Gardiner z -score: Augusta z -score:

Gardiner z -score: -1.46 Augusta z -score: -1.38 1. Enter the Gardiner data into column A and the Augusta data into column B in Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, make sure Columns is selected under Group By, tick the Summary statistics check box, and press OK. 4. Read the sample mean and sample standard deviation of each group from the output. The sample mean for the Gardiner data is 701.8 with a sample standard deviation of 39.013, rounded to three decimal places. The sample mean for the Augusta data is 680.4 with a sample standard deviation of 25.713, rounded to three decimal places. 5. Use the sample mean and sample standard deviation of each data set to compute the z-score for the given values. Compute the z-score for the apartment in Gardiner having an area of 645 square feet, rounding to two decimal places. z≈645−701.8/39.013≈−1.46 Compute the z-score for the apartment in Augusta having an area of 645 square feet, rounding to two decimal places. z≈645−680.4/25.713≈−1.38

When Stephan moved from Illinois to Florida, his average monthly electric bill increased from $83 to $102. He is curious to know whether his IL or FL electric bill is relatively more or less expensive, when compared to the distribution of electric bills for each state. In Illinois, the mean monthly electric bill is $85, with a standard deviation of $3.20. In Florida, the mean monthly electric bill is $105, with a standard deviation of $4.00. Compute the z-scores for Stephan's IL and FL electric bills. Round to three decimal places if necessary. Provide your answer below: Illinois: z= Florida: z=

Illinois: z= -0.625 Florida: z= -0.75 The mean monthly electric bill in Illinois is $85, with a standard deviation of $3.20. The z-score corresponding to Stephan's IL electric bill of $83 is z=x−μ/σ=83−85/3.2=−0.625 The mean monthly electric bill in Florida is $105, with a standard deviation of $4.00. The z-score corresponding to Stephan's FL electric bill of $102 is z=x−μ/σ=102−105/4.00=−0.75

Based on the z scores found above, is Kathy or Linda's starting salary higher, when compared to the salary distributions of each company? Select the correct answer below: Kathy Linda Kathy and Linda's starting salaries are equivalent Not enough information

Kathy Kathy's starting salary of $31,500 has a z-score of z = −1.5, which means her salary 1.5 standard deviations below her company's mean salary. Linda's starting salary of $33,000 has a z-score of z = −2, which means her salary is 2 standard deviations below her company's mean salary. Kathy's salary corresponds to a greater z − -score, when compared to their respective company's salary distributions, Kathy's has the better starting salary.

Kathy and Linda both accepted new jobs at different companies. Kathy's starting salary is $31,500 and Linda's starting salary is $33,000. They are curious to know who has the better starting salary, when compared to the salary distributions of their new employers. A website that collects salary information from a sample of employees for a number of major employers reports that Kathy's company offers a mean salary of $42,000 with a standard deviation of $7,000. Linda's company offers a mean salary of $45,000 with a standard deviation of $6,000. Find the z-scores corresponding to each woman's starting salary. Provide your answer below: Kathy's z-score = Linda's z-score =

Kathy's z-score = -1.5 Linda's z-score = -2 Whether the data is from a sample or population, the formulas for the z-score remains the same: z = data value−mean/standard deviation Kathy's starting salary is $31,500, for a company with a mean salary of $42,000 and standard deviation $7,000. The corresponding z-score is z = x−μ/σ = 31,500−42,000/7,000 = −1.5 So Kathy's starting salary is 1.5 standard deviations below her company's mean salary. Linda's starting salary is $33,000, for a company with a mean salary of $45,000 and standard deviation $6,000. The corresponding z-score is z = x−μ/σ = 33,000−45,000/6,000 = −2.0 So Linda's starting salary is 2 standard deviations below her company's mean salary.

Machine A z -score: -1.12 Machine B z -score: -1.33 Step 1. Press STAT and then EDIT. Enter the Machine A data into list L1 and the Machine B data into list L2. Step 2. Press STAT, then CALC, and then 1-VAR STATS. Step 3. Enter 1-VAR STATS, 2ND, and then the 1 key for the list L1. Enter 1-VAR STATS, 2ND, and then the 2 key for the list L2. Step 4. Read the sample mean, x¯¯¯, and the sample standard deviation, sx, for each list from the output. The sample mean for the Machine A data is 44.003 with a sample standard deviation of 0.030, rounded to three decimal places. The sample mean for the Machine B data is 44.001 with a sample standard deviation of 0.023, rounded to three decimal places. Step 5. Compute the z-score for a bolt produced by Machine A that is 43.97mm long, rounding to two decimal places. z≈43.97−44.003/0.030=−1.10 Compute the z-score for a bolt produced by Machine B that is 43.97mm long, rounding to two decimal places. z≈43.97−44.001/0.023≈−1.35

The midterm and final exam grades for a statistics course are provided in the data set below. Jaymes, a student in the class, scored 86 on both exams. Treat the given data sets as samples. Jaymes's wants to know which grade is more unusual, the midterm grade or the final exam grade. Use Use a TI-83, TI-83 Plus, or TI-84 calculator to calculate the z-scores corresponding to each grade. Round your answer to three decimal places. Provide your answer below: Midterm z-score: Final exam z-score:

Midterm z-score: 1.655 Final exam z-score: 1.188 Step 1. Press STAT and then EDIT. Enter the Midterm data into list L1 and the Final data into list L2. Step 2. Press STAT, then CALC, and then 1-VAR STATS. Step 3. Enter 1-VAR STATS, 2ND, and then the 1 key for the list L1. Enter 1-VAR STATS, 2ND, and then the 2 key for the list L2. Step 4. Read the sample mean, x¯¯¯, and the sample standard deviation, sx, for each list from the output. The sample mean for the Midterm data is 81.1 with a sample standard deviation of 2.961, rounded to three decimal places. The sample mean for the Final data is 77.1 with a sample standard deviation of 7.490, rounded to three decimal places. Step 5. Compute the z-score for Jaymes scoring 86 points on the midterm, rounding to three decimal places. z≈86−81.1/2.961≈1.655 Compute the z-score for Jaymes scoring 86 points on the final, rounding to three decimal places. z≈86−77.1/7.490≈1.188

A statistics professor gives a survey to each of the 100 students in an introductory statistics lecture. The survey asks the students how many text messages they think they sent yesterday. Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Population standard deviation The professor surveys all 100 students in the class - that is, the entire population of the class. So a population standard deviation would be more appropriate.

An education reform lobby is compiling data on the state of education in the United States. In their research they looked at the percent of people who graduate high school in 20 different states. Would it be more appropriate to find a sample standard deviation or population standard deviation in this situation? Select the correct answer below: Sample standard deviation Population standard deviation

Sample standard deviation Since data is available for only 20 of the 50 states, a sample standard deviation would be more appropriate.

A researcher for an organization that collects and reports on crime has data on the murder rates for 10 states from a certain year. The murder rate is the number of murders per 100,000 inhabitants. The murder rate sample data are reproduced below. Use a TI-83, TI-83 Plus, or TI-84 to calculate the sample standard deviation and the sample variance of the murder rates. Round your answers to one decimal place. Standard Deviation = Variance =

Standard Deviation = 2.2 Variance = 4.7 To determine the sample standard deviation and sample variance for a data set {x1,x2,...,xn} using a graphing calculator (such as a TI-83, TI-83 Plus, or TI-84), follow these steps: Enter data into List L1. Press STAT and then ENTER to get to the list edit screen. Run 1-Var Stats. Press STAT, then press RIGHT, and then press ENTER to paste 1-Var Stats to the home screen. Press ENTER again. This shows the sample standard deviation Sx. Access and square Sx from the statistics variable screen to get the sample variance. Press VARS, then 5, then 3, then x2, and then ENTER. The sample standard deviation is s≈2.2 and the sample variance is s2≈4.7, rounding each to one decimal place.

A statistics professor gives a survey to each of the 100 students in an introductory statistics lecture. The survey asks the students how many text messages they think they sent yesterday. The data are included below. Use Excel to calculate the population standard deviation and the population variance. Round your answers to one decimal place. Do not round until you've calculated your final answer Provide your answer below: Standard Deviation = Variance =

Standard Deviation = 53.1 Variance = 2819.1 To determine the population standard deviation and population variance for a data set {x1,x2,...,xN} using Excel, follow these steps: Open the included data with Excel. The data should occupy cell range A2:A21. Select cell B3 and type "=STDEV.P(", select the range A2:A101, and then hit ENTER. This gives the population standard deviation. Select cell B4 and type "=VAR.P(", select the range A2:A101, and then hit ENTER. This gives the population variance. The population standard deviation is σ≈53.1 and the population variance is σ2≈2819.7, rounding each to one decimal place.

An education reform lobby is compiling data on the state of education in the United States. In their research they looked at the percent of people who graduate high school in 20 different states. The data are provided below. Use Excel to calculate the sample standard deviation and the sample variance. Round your answers to one decimal place. Provide your answer below: Standard Deviation = Variance =

Standard Deviation = 6.5 Variance = 42.4 To determine the sample standard deviation and sample variance for a data set {x1,x2,...,xn} using Excel, follow these steps: Open the included data with Excel. The data should occupy cell range A2:A21. Select cell B3 and type "=STDEV.S(", select the range A2:A21, and then hit ENTER. This gives the sample standard deviation. Select cell B4 and type "=VAR.S(", select the range A2:A21, and then hit ENTER. This gives the sample variance. The sample standard deviation is s≈6.5 and the sample variance is s2≈42.4, rounding each to one decimal place.

A budding collector is looking over her growing coin collection, recording various details about each coin in a notebook. The collector has 17 rare coins in her collection. The data for the age of each coin in years are included below. Use a TI-83, TI-83 Plus, or TI-84 to calculate the population standard deviation and the population variance. Round your answers to one decimal place. Standard Deviation = Variance =

Standard Deviation = 9.7 Variance = 88.6 To determine the population standard deviation and population variance for a data set {x1,x2,...,xn} using a graphing calculator (such as a TI-83, TI-83 Plus, or TI-84), follow these steps: Enter data into List L1. Press STAT and then ENTER to get to the list edit screen. Run 1-Var Stats. Press STAT, then press RIGHT, and then press ENTER to paste 1-Var Stats to the home screen. Press ENTER again. This shows the population standard deviation σx. Access and square σx from the statistics variable screen to get the population variance. Press VARS, then 5, then 4, then x^2, and then ENTER. The population standard deviation is σ≈9.4 and the population variance is σ^2≈88.6, rounding each to one decimal place.

Tyler is a student in an environmental science class looking into the seasonal snowfall totals for two nearby locations. One location is on the coastline, while a second location is 30 miles inland. He randomly selects 10 seasons for the coastal location and another 10 seasons for the inland location, recording the annual snowfall total in inches for each season. The results of Tyler's survey are shown in the samples provided. Both locations have a season that had 48.6in. of total snowfall. Based on the z scores calculated above, is the 48.6in. snowfall amount more unusual for the coastal location or the inland location? Select the correct answer below: The 48.6 inches of snowfall at the coastal location is more unusual because the absolute value of the z-score for the coastal location is less than the absolute value of the z-score for the inland location. The 48.6 inches of snowfall at the coastal location is more unusual because the absolute value of the z-score for the coastal location is greater than the absolute value of the z-score for the inland location. The 48.6 inches of snowfall at the inland location is more unusual because the absolute value of the z-score for the inland location is greater than the absolute value of the z-score for the coastal location. The 48.6 inches of snowfall at the inland location is more unusual because the absolute value of the z-score for the inland location is less than the absolute value of the z-score for the coastal location.

The 48.6 inches of snowfall at the coastal location is more unusual because the absolute value of the z-score for the coastal location is greater than the absolute value of the z-score for the inland location. A z-score with a greater absolute value means that the data value is more unusual. Since the absolute value of the z-score for the coastal location, 1.80, is greater than the absolute value of the z-score for the inland location, 1.57, the 48.6 inches of snowfall is more unusual at the coastal location.

Casey is looking to rent a two-bedroom apartment in one of two towns, Gardiner or Augusta. He randomly selects 100 two-bedroom apartments from both towns and records the area of each apartment. Both towns have a two-bedroom apartment that has an area of 645 square feet. Based on the z-scores you calculated above, which of the two towns is more likely to have a two-bedroom apartment with 645 square feet? Select the correct answer below: The absolute value of the z-score for the apartment in Gardiner is greater than for the apartment in Augusta, so the apartment in Gardiner having an area of 645 square feet is more likely. The absolute value of the z-score for the apartment in Gardiner is less than for the apartment in Augusta, so the apartment in Gardiner having an area of 645 square feet is more likely. The absolute value of the z-score for the apartment in Augusta is greater than for the apartment in Gardiner, so the apartment in Augusta having an area of 645 square feet is more likely. The absolute value of the z-score for the apartment in Augusta is less than for the apartment in Gardiner, so the apartment in Augusta having an area of 645 square feet is more likely.

The absolute value of the z-score for the apartment in Augusta is less than for the apartment in Gardiner, so the apartment in Augusta having an area of 645 square feet is more likely. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for Augusta, 1.38, is less than the absolute value of the z-score for Gardiner, 1.46, the apartment that has an area of 645 square feet is more likely to occur in Augusta than in Gardiner

The midterm and final exam grades for a statistics course are provided in the data set below. Jaymes, a student in the class, scored 86 on both exams. Treat the given data sets as samples. Based on the z-scores calculated above, which of Jaymes's grades is more unusual, the midterm grade or the final exam grade? Select the correct answer below: The absolute value of the z-score for the final exam grade is greater than for the midterm grade, so the final exam grade is more unusual. The absolute value of the z-score for the midterm exam grade is greater than for the final grade, so the midterm grade is more unusual. The absolute value of the z-score for the midterm exam grade is less than for the final grade, so the midterm grade is more unusual. The absolute value of the z-score for the final exam grade is less than for the midterm grade, so the final exam grade is more unusual.

The absolute value of the z-score for the midterm exam grade is greater than for the final grade, so the midterm grade is more unusual. A z-score with a greater absolute value means that the data value is more unusual. Since the absolute value of the z-score for the midterm grade, 1.655, is greater than the absolute value of the z-score for the final grade, 1.188, Jaymes' midterm grade is more unusual.

The z -score for May 15 is 3.25 and the z -score for November 15 is -1.74. Step 1. Press STAT and then EDIT. Enter the data for May 15 into list L1 and the data for November 15 into list L2. Step 2. Press STAT, then CALC, and then 1-VAR STATS. Step 3. Enter 1-VAR STATS, 2ND, and then the 1 key for the list L1. Enter 1-VAR STATS, 2ND, and then the 2 key for the list L2. Step 4. Read the sample mean, x¯¯¯, and the sample standard deviation, sx, for each list from the output. The sample mean for May 15 is 2.755 with a sample standard deviation of 0.044, rounded to three decimal places. The sample mean for November 15 is 3.017 with a sample standard deviation of 0.068, rounded to three decimal places. Step 5. Compute the z-score for $2.899 on May 15, rounding to two decimal places. z≈2.899−2.755/0.044≈3.27 Compute the z-score for $2.899 on November 15, rounding to two decimal places. z≈2.899−3.017/0.068≈−1.74

The z -score for machine A is 1.74 and the z -score for machine B is 3.57. 1. Enter the data for Machine A into column A and the data for Machine B into column B in Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, make sure Columns is selected under Group By, tick the Summary statistics check box, and press OK. 4. Read the sample mean and sample standard deviation of each group from the output. The sample mean for machine A is 84.02 with a sample standard deviation of 0.0172, rounded to four significant figures. The sample mean for machine B is 83.98 with a sample standard deviation of 0.0196, rounded to four significant figures. 5. Use the sample mean and sample standard deviation of each data set to compute the z-score for the given values. Compute the z-score for machine A producing the bolt that has a length of 84.05mm, rounding to two decimal places. z≈84.05−84.02/0.0172≈1.74 Compute the z-score for machine B producing the bolt that has a length of 84.05mm, rounding to two decimal places. z≈84.05−83.98/0.0196≈3.57

The z -score for the Development department is -1.96 and the z -score for the Marketing department is -1.70. 1. Enter the data for the Development department into column A and the data for the Marketing department into column B in Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, make sure Columns is selected under Group By, tick the Summary statistics check box, and press OK. 4. Read the sample mean and sample standard deviation of each group from the output. The sample mean for the Development department is 40.6 with a sample standard deviation of 9.494, rounded to three decimal places. The sample mean for the Marketing department is 34.4 with a sample standard deviation of 7.291, rounded to three decimal places. 5. Use the sample mean and sample standard deviation of each data set to compute the z-score for the given values. Compute the z-score for the 22-year-old employee in the Development department, rounding to two decimal places. z≈22−40.6/9.494≈−1.96 Compute the z-score for the 22-year-old employee in the Marketing department, rounding to two decimal places. z≈22−34.4/7.291≈−1.70

The z -score for the apartment in Gardiner is -1.86 and the z -score for the apartment in Augusta is -1.81 Step 1. Press STAT and then EDIT. Enter the data for Gardiner into list L1 and the data for Augusta into list L2. Step 2. Press STAT, then CALC, and then 1-VAR STATS. Step 3. Enter 1-VAR STATS, 2ND, and then the 1 key for the list L1. Enter 1-VAR STATS, 2ND, and then the 2 key for the list L2. Step 4. Read the sample mean, x¯¯¯, and the sample standard deviation, sx, for each list from the output. The sample mean for Gardiner is 700 with a sample standard deviation of 35.014, rounded to three decimal places. The sample mean for Augusta is 675.917 with a sample standard deviation of 22.573, rounded to three decimal places. Step 5. Compute the z-score for the apartment in Gardiner that has 635 square feet, rounding to two decimal places. z≈635−700/35.014≈−1.86 Compute the z-score for the apartment in Augusta that has 635 square feet, rounding to two decimal places. z≈635−675.917/22.573≈−1.81

The z -score for the round-trip airfare on Wednesday is 1.77 and the z -score for the round-trip airfare on Sunday is -1.38. Step 1. Press STAT and then EDIT. Enter the data for Wednesday into list L1 and the data for Sunday into list L2. Step 2. Press STAT, then CALC, and then 1-VAR STATS. Step 3. Enter 1-VAR STATS, 2ND, and then the 1 key for the list L1. Enter 1-VAR STATS, 2ND, and then the 2 key for the list L2. Step 4. Read the sample mean, x¯¯¯, and the sample standard deviation, sx, for each list from the output. The sample mean for Wednesday is 170.083 with a sample standard deviation of 19.783, rounded to three decimal places. The sample mean for Sunday is 229.5 with a sample standard deviation of 17.692, rounded to three decimal places. Step 5. Compute the z-score for the airfare that occurs on Wednesday and costs $205, rounding to two decimal places. z≈205−170.083/19.783≈1.77 Compute the z-score for the airfare that occurs on Sunday and costs $205, rounding to two decimal places. z≈205−229.5/17.692≈−1.38

An event coordinator for a particular marathon held yearly is reviewing the data from the top 30 race finish times from the last race. Using Excel, calculate the mode(s) of the dataset provided below. Select the correct answer below: There are two modes. The modes are 2.47 and 4.14. There is one mode. The mode is 2.47. There is one mode. The mode is 4.14. There is no mode.

There are two modes. The modes are 2.47 and 4.14. The mean, median, and mode can be calculated quickly and easily in Excel using the built-in functions for these calculations. In C1 type "Mode", highlight cell range C2:C5, while highlighted type "=MODE.MULT(A2:A31)", and then press CTRL+SHIFT+ENTER. If there is one mode, each cell in cell range C2:C5 will display that mode's value. If there is more than one mode, each mode is displayed in one of the cells in the cell range C2:C5. If each cell in the cell range C2:C5 has a unique mode value, then there may be more modes and step 1 should be repeated with a taller cell range, for example C2:C10. If there are more cells in the range than there are modes but there is more than one mode, the remaining cells in the range will display #N/A to indicate there were no other modes found. Cell C2 should display 2.47, and cell C3 should display 4.14. Cells C4 and C5 should both display #N/A, indicating that there are two modes: 2.47 and 4.14.

The following data set provides information about the City of Somerville Assessors Valuation for the fiscal year 2016. Is there a direct correlation between the commercial living area's standard deviation and number of offices in a building? Select the correct answer below: There is a direct correlation, and it leads to more offices. There is a direct correlation, and it leads to fewer offices. There is not enough information to determine direct correlation.

There is not enough information to determine direct correlation. There is not enough information to determine direct correlation between the commercial living area's standard deviation and number of offices in a building.

4. Calculate the quartile differences for each dataset: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. Set up this table similarly to the quartile table, title it "Differences," and have the row labels match the image below. The first row of data below the headers should be the contents of the corresponding row of the quartile table.

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Example Question: A doctor decides to look at a sample of temperature data gathered during routine checkups of her patients. She wants to examine the spread of the data. In addition she wants to know what temperatures should be considered unusual, given the natural variation she sees in her patients' temperatures. The temperatures of 20 patients, recorded in degrees Fahrenheit, are given in the table below. Construct the box and whisker plot for the dataset using Excel. Find the five-number summary and use it to construct the box and whisker plot.

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How To Construct a box and whisker plot using Excel. To construct a box and whisker plot using Excel, use the following steps. 1. Arrange data in columns, one column for each dataset.

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The following set of data represents stock prices of a pharmaceutical company, find the sample variance of the: 14, 7, 10, 9. Round your answer to ONE decimal place. Provide your answer below: sample variance = price^2

sample variance = 8.7 price^2 First, we find that the mean is 14+7+10+9/4=40/4=10 Now, we need to take the deviations from the mean and square them: Value 14,7,10,9 Deviation 4.0,-3.0,0.0,-1.0 Deviation 16.0,9.0,0.0,1.0 Finally, we add up the squared deviations and divide by the number of data values minus one (4−1=3). 16.0+9.0+0.0+1.0/3=8.7 The sample variance shows how the stock values differ from the stock's mean. The larger the difference, the riskier the stock. This would say that the stock varies from 10(mean), by 8.7(units^2).

11. Create errors bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount click Percentage: and set its value to 100.0% without typing the % symbol.

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2. Calculate the quartile values for each dataset: Minimum, Q1, Median, Q3, Maximum. Set up a new table with two more columns than there are datasets. The first column will have a title for the table as in the example below. In the five cells below this, use fill series or manually enter 0, 1, 2, 3, 4. Row labels identifying the values can be placed in the last column so that 0 is in the same row as Minimum and going down the labels read: Minimum, Q1, Median, Q3, Maximum. Then copy the headers to the first row of the table between the two new columns. This should create a setup like the blue table in the image below in cell range D1:H6.

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3. Under the cell below the first header in the new table write "=Quartile.Inc(A$2:A$12,$D2)", then copy this cell to the remaining four cells underneath, and then copy the column of five cells to every column corresponding to a dataset. The first argument should be the cell range of the first dataset, excluding the header. The second argument is the cell containing 0 below the table title. The placement of $ signs is important: Without them the formula cannot be correctly copied to adjacent cells in the quartile table.

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Solution: 1. Arrange data in a column. The dataset should occupy A1:A21, where A1 is the header and A2:A21 contains the data. Use steps (2) through (6) to construct a table for the quartile values. 2. Use fill series or manually enter the numbers 0, 1, 2, 3, and 4 into cell range B2:B6 from top to bottom, one number per cell. 3. Write the following entries into cell range D2:D6 from top to bottom, one entry per cell: Minimum, Q1, Median, Q3, Maximum. 4. In cell C1 copy the header for the dataset from A1. 5. In cell C2 write "=QUARTILE.INC(A$2:A$21,$B2)." 6. Then copy and paste this cell to C3:C6. Cells C2:C6 contain the five-number summary. The minimum is 97.7, Q1 is 98.3, the median is 98.5, Q3 is 99.15, and the maximum is 99.9. Continue on to construct the box plot. Use steps (7) through (10) to construct a table for the quartile differences. 7. Copy and paste cell range B1:C6 into cell range B7:C12. 8. Write the following entries into cell range D8:D12 from top to bottom, one entry per cell: Minimum, Q1−Minimum, Median−Q1, Q3−Median, Maximum−Q3. 9. In cell C9 write "=C3-C2" where C3 is the corresponding cell in the quartiles table. 10. Copy and paste this new cell to the remaining three cells underneath. Use steps (11) through (15) to construct the box plot from the quartile differences. 11. Create a stacked column chart type from the quartile differences. Select cells C7:C12. Then at the top of the Excel window, click the Insert tab, then the Charts button labeled Column, and then Stacked Column chart type. The Stacked Column chart type is the top row and second column of the Columns chart types. 12. Switch the row and column data on the resulting chart. Click the Design tab at the top of the Excel Window, and then press the button labeled Switch Row/Column. 13. Set the Fill to blank for the bottom two sections of the bar and also the top section. Click the bottom segment of one of the rectangles in the plot, click the Format tab, and below that click the Format Selection button. In the Format Data Series window that pops up, click the Fill tab, and then click the No Fill radio button. 14. Select the top visible segment and give it error bars that stretch from the top of the visible segment to the maximum of each dataset. Select the top visible segment. Then click on the Layout tab at the top of the Excel window, click on the Error Bars button, and click on the second option from the top labeled Error Bars With Standard Error. Then select the error bars that appear on the plot and click Format Selection on the tool bar above. In the Format Error Bars window, click the Vertical Error Bars tab, click Plus button under Direction, and click the Cap button under End Style. Under Error Amount click Custom, and then press Specify Value. In the resulting pop up click on the Positive Error Value field and select the data in cell C12. 15. Create errors bars for the bottom visible segment that extend to the minimum of each dataset. Click the segment below the bottom visible segment, being careful not to select the very bottom segment by accident. Select or click Format Selection again. In the Format Error Bars window, click the Vertical Error Bars tab, click the Minus button under Direction, and click the Cap button under End Style. Under Error Amount click Percentage:, and then set its value to 100.0% without typing the % symbol. Steps (16) through (20) are optional and deal with beautifying the box and whisker plot. 16. Delete the legend. 17. Modify the vertical axis to more tightly contain the data. For this data set the vertical axis was changed to run from 97.5 to 100.0 with major ticks in increments of 0.5 and minor ticks marks in increments of 0.1. 18. Set the fill of both visible box segments to solid color. 19. Set the border color of the visible box segments to solid color. 20. Set the line color of the error bars to solid color.

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The following data set provides information about the City of Somerville Assessors Valuation for the fiscal year 2016. For residential buildings, what is the sample standard deviation of the land area? Round your answer to ONE decimal places. 2296,1620,3343,4032,3438

standard deviation = 969.5 The mean is 2,945.8. The sample variance is the sum of the squares of the difference of the specific data value from the mean divided by 4, which is n−1. Then, the sample standard deviation is the square root of the sample variance. For this sample, the sample standard deviation is 969.5.

The following data set represents the ages of all 6 of Nancy's grandchildren. 11,8,5,6,3,9 Given that the variance of this data set is 7, what is the standard deviation of their ages? Round the final answer to one decimal place. Provide your answer below: std =

std = 2.6 Remember that the standard deviation is the square root of the variance. Since we just found that the variance is 7, we find that the standard deviation is 7-√≈2.6457..., or 2.6 to the nearest tenth.

The following data values represent the daily amount spent by a family each day during a 7 day summer vacation. Find the variance of this dataset: $96,$125,$80,$110,$75,$100,$121 Round the final answer to one decimal place. Provide your answer below: variance =

variance = 314.3 Correct answers: 314.3 In this case, the population standard deviation should be used because the data set represents the amount spent each day of the 7 day vacation. First, we find that the mean is 96+125+80+110+75+100+121/7= 707/7=101 Now, we need to take the deviations from the mean and square them: Value 96,125,80,110,75,100,121 Deviation −5,24,−21,9,−26,−1,20 Deviation 25,576,441,81,676,1,400 The amounts spentars listed for every day of the vacation. Since we have data for the total population of vacation days, we will find the population variance and standard deviation. The population variance is the sum of the squared deviations, divided by the number of data values. 25+576+441+81+676+1+400/7= 2200/7≈314.28 So to one decimal place, the variance is 314.3

The following data set represents the ages of all 6 of Nancy's grandchildren. 11,8,5,6,3,9 Find the variance of their ages. If necessary, round the final answer to one decimal place. Provide your answer below: variance=

variance= 7.0 First, we find that the mean is 11+8+5+6+3+9 = 6 42 = 7 6 Now, we need to take the deviations from the mean and square them: Value 11,8,5,6,3,9 Deviation 4.0,1.0,−2.0,−1.0,−4.0,2.0 Deviation2 16.0,1.0,4.0,1.0,16.0,4.0 Since the total population of grandchildren is included, we find the population variance. Add up the squared deviations and divide by the number of data values: 16+1+4+1+16+4 = 7 6 So the variance is 7.

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