Coffee Shop Sales Analysis: Statistical Analysis
A coffee shop records the number of cups of coffee sold per day over a 20-day period: 120, 135, 110, 125, 140, 130, 115, 120, 145, 150, 130, 125, 135, 140, 120, 110, 130, 125, 140, 135. The owner wants to analyze the data to make staffing decisions. They calculate the mean, median, and standard deviation. They also want to know if there is an outlier using the 1.5×IQR rule. The owner plans to give a bonus to the employee who sold the most cups in a day, but only if that day's sales are more than one standard deviation above the mean. Determine the mean, median, standard deviation (population), and identify any outliers. Also, decide if a bonus should be awarded based on the highest sales day.
1. Calculate the mean, median, and population standard deviation of the daily coffee sales. Show your working.
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Data sorted: 110, 110, 115, 120, 120, 120, 125, 125, 125, 130, 130, 130, 135, 135, 135, 140, 140, 140, 145, 150. Sum = 110+110=220; +115=335; +120=455; +120=575; +120=695; +125=820; +125=945; +125=1070; +130=1200; +130=1330; +130=1460; +135=1595; +135=1730; +135=1865; +140=2005; +140=2145; +140=2285; +145=2430; +150=2580. Sum = 2580. Mean = 2580/20 = 129. Median: average of 10th and 11th values: 10th=130, 11th=130, so median = 130. Population standard deviation: compute variance = Σ(x - μ)² / N. μ=129. Deviations squared: (110-129)²=361, (110-129)²=361, (115-129)²=196, (120-129)²=81, (120-129)²=81, (120-129)²=81, (125-129)²=16, (125-129)²=16, (125-129)²=16, (130-129)²=1, (130-129)²=1, (130-129)²=1, (135-129)²=36, (135-129)²=36, (135-129)²=36, (140-129)²=121, (140-129)²=121, (140-129)²=121, (145-129)²=256, (150-129)²=441. Sum of squares = 361+361=722; +196=918; +81=999; +81=1080; +81=1161; +16=1177; +16=1193; +16=1209; +1=1210; +1=1211; +1=1212; +36=1248; +36=1284; +36=1320; +121=1441; +121=1562; +121=1683; +256=1939; +441=2380. Variance = 2380/20 = 119. Standard deviation = √119 ≈ 10.91.
2. Determine if there are any outliers using the 1.5×IQR rule. Also, decide if the bonus should be awarded based on the highest sales day being more than one standard deviation above the mean.
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First, find quartiles. Data sorted (20 values). Q1 is median of first 10 values: first 10: 110,110,115,120,120,120,125,125,125,130. Median of these = average of 5th and 6th = (120+120)/2=120. So Q1=120. Q3 is median of last 10: 130,130,135,135,135,140,140,140,145,150. Median = (135+140)/2=137.5. IQR = 137.5 - 120 = 17.5. Lower fence = Q1 - 1.5×IQR = 120 - 26.25 = 93.75. Upper fence = Q3 + 1.5×IQR = 137.5 + 26.25 = 163.75. Any data below 93.75 or above 163.75 are outliers. All data are between 110 and 150, so no outliers. Highest sales day = 150. Mean = 129, standard deviation ≈ 10.91. One standard deviation above mean = 129 + 10.91 = 139.91. Since 150 > 139.91, the highest sales day is more than one standard deviation above the mean. Therefore, the bonus should be awarded.