business stats

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Chandigarh University *

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101

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Statistics

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Nov 24, 2024

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xlsx

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Question 1 1. mean 10, 12, 16, 14, 10, 12, 16, 10, 19, 10, 11, 18, 28, 110, 27, 16, 144 Items(X) 10 10 12 16 14 10 12 16 10 11 18 28 110 27 16 144 19 483 mean 483/17= 28.41 c. Mode b. median Arranging the data in ascending order: 10, 10, 10, 10, 11, 12, 12, 14, 16, 16, 16, 18, 19, 27, 28, 1 The middle value is 16, so the median is 16. Mode = 10 10, 12, 16, 14, 10, 12, 16, 10, 19, 10, 11, 18, 28, 110, 27, 16, 144 Frequency of each number 10: 4 11: 1 12: 2 14: 1 16: 3 18: 1 19: 1 27: 1 28: 1 110: 1 144: 1 Mode = 10 (most frequently occurring number)

d. Range E. Variance F. Standard Deviation .d. Range = 134 To calculate the range, we subtract the smallest value from the greatest value. The greatest value is 144 and the smallest value is 10. Therefore, 144 – 10 = 134. Hence, the range of the given data is 134 .Step 1: Find the mean (average) 10 + 12 + 16 + 14 + 10 + 12 + 16 + 10 + 19 + 10 + 11 + 18 + 28 + 110 + 27 + 16 + 144 = 360 Mean = 360/16 = 22.5 Step 2: Subtract the mean from each value 10-22.5 = -12.5 12-22.5 = -10.5 16-22.5 = -6.5 14-22.5 = -8.5 10-22.5 = -12.5 12-22.5 = -10.5 16-22.5 = -6.5 10-22.5 = -12.5 19- 22.5 = -3.5 10-22.5 = -12.5 11-22.5 = -11.5 18-22.5 = -4.5 28-22.5 = 5.5 110-22.5 = 87.5 27-22.5 = 4.5 16-22.5 = -6.5 144-22.5 = 121.5 Step 3: Square the differences (-12.5)2 = 156.25 (-10.5)2 = 110.25 (-6.5)2 = 42.25 (- 8.5)2 = 72.25 (-12.5)2 = 156.25 (-10.5)2 = 110.25 (-6.5)2 = 42.25 (-12.5)2 = 156.25 (- 3.5)2 = 12.25 (-12.5)2 = 156.25 (-11.5)2 = 132.25 (-4.5)2 = 20.25 (5.5)2 = 30.25 (87.5)2 = 7612.25 (4.5)2 = 20.25 (-6.5)2 = 42.25 (121.5)2 = 14742.25 Step 4: Add all the squared differences 156.25 + 110.25 + 42.25 + 72.25 + 156.25 + 110.25 + 42.25 + 156.25 + 12.25 + 156.25 + 132.25 + 20.25 + 30.25 + 7612.25 + 20.25 + 42.25 + 14742.25 = 25962.25 Step 5: Divide the sum of the squared differences by the number of values 25962.25/16 = 1610.140625 Step 6: Take the square root of the answer √1610.140625 = 40.0145 HStep 1: Calculate the mean Total numbers = 16 Sum of numbers = 336 M 336/16 Mean = 21 Step 2: Calculate the difference between each number and the mean 10 - 2 - 21 = -9 16 - 21 = -5 14 - 21 = -7 10 - 21 = -11 12 - 21 = -9 16 - 21 = -5 10 - 21 - 21 = -2 10 - 21 = -11 11 - 21 = -10 18 - 21 = -3 28 - 21 = 7 110 - 21 = 89 27 - - 21 = -5 144 - 21 = 123 Step 3: Square the differences (-11)² = 121 (-9)² = 81 (-5)² = 25 (-7)² = 49 (-11 9)² = 81 (-5)² = 25 (-11)² = 121 (-2)² = 4 (-11)² = 121 (-10)² = 100 (-3)² = 9 7² = 7921 6² = 36 (-5)² = 25 123² = 15129 Step 4: Add all of the squared differences 121 + 81 + 25 + 49 + 121 + 81 + 25 + 121 + 100 + 9 + 49 + 7921 + 36 + 25 + 15129 = 22911 Step 5: Divide the sum of the squared differences by the number of value

g. 70th Percentile H. The IQR (interquartile range) i. Discuss whether or not an outlier exists in the data. Support your answer with mathe mparing it to the value of 110. Th Step 5: Divide the sum of the squared differences by the number of value 22911 ÷ 15 = 1527.4 Step 6: Take the square root of the answer √1527.4 = 39.14 Step 1: Arrange the data in order from least to greatest: 10, 10, 10, 10, 11, 12, 12, 14, 16, 16, 16, 18, 19, 27, 28, 110, 144 Step 2: Count the number of data points N = 17 Step 3: Find the 70th percentile 70th Percentile = (N + 1) × p = (17 + 1) × 0.7 = 18.9 Step 4: Find the value at the 18.9th rank 18.9th Rank = 19 Therefore, the 70th percentile of the given set of data is 19. .The IQR is the range of values between the first quartile (Q1) and the third quartile calculate the IQR, we first need to find the quartiles. Step 1: Order the data from lowest to highest: 10, 10, 10, 10, 11, 12, 12, 14, 16, 16, 16 27, 28, 110, 144 Step 2: Find the median (the middle value): 16 Step 3: Find the median of the lower half of the data (Q1): 10, 10, 10, 10, 11, 12, 12, 1 median of the lower half is 11. Step 4: Find the median of the upper half of the data (Q3): 16, 16, 18, 19, 27, 28, 110, median of the upper half is 19. Step 5: Calculate the interquartile range (IQR): IQR = Q3 - Q1 IQR = 19 - 11 IQR =

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j. The probability of drawing a number higher than 12 if one number was drawn at ra The probability of drawing a number higher than 12 if one number was drawn at random from th k. The probability of drawing a number higher than 12, not putting it back, and then dr a second number higher than 12 from the list is 0.36 (3 out of 8 numbers are higher than 12). l. Th QUESTION 2 l. Book A 197 $7.00 B 344 $9.25 C 250 $8.25 D 282 $8.75 E 184 $7.25 F 201 $7.50 G 144 $6.00 H 377 $9.99 A. Show the equation of the trendline on the s y = 0.04x + 5.06 B. Using Pearson’s Product Moment Correlatio The Pearson’s Product Moment Correlation C. A D. Acco Pages (x variable) Price (y variable) 1 2 3 0 50 100 150 200 250 300 350 400

According t E. nd positive, indicating t Question 3 gH0: μ = $2,375 Ha: μ ≠ $2,375 Test Statistic: t = ( - μ)/(s/√n) x̄ Calculations: = $2,222 x̄ μ = $2,375 s = $64 n = 20 t = (2,222 - 2,375)/(64/√20) t = -1.04 Degrees of Freedom: df = n - 1 df = 20 - 1

Question 4 A. P(650 boxes or less) = P(X ≤ 650) = P(z ≤ (650-725)/61) = P(z ≤ -2.21) = 0.0128 A. P(650 boxes or less) = P(X ≤ 650) = P(z ≤ (650-725)/61) = P(z ≤ -2.21) = 0.0129 B. C. df = 20 - 1 df = 19 Significance Level: α = 0.05 Critical Value: t-critical (α, df) = -1.729 P-value: p = 0.13 Conclusion: Since the p-value (0.13) is greater than the significance level (0.05), we fail to reject t This means that there is not enough evidence to suggest that the mean of all account from $2,375. = P[(x -) /View image!(760 - 774) / 51] = P(zView image!-0.27) Using z table, = 0.3936 P(700 ≤ X ≤ 800) = P(-0.98 ≤ z ≤ 1.31) = P(z ≤ 1.31) - P(z ≤ -0.98) = 0.0908 - 0.1144 = -0.0236 = 0.7788

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Question 5 Question 6 b90% Confidence Interval = (μ ± 1.645σ/√n) μ = 23.1 σ = 6.21 n = 89 90% Confidence Interval = (23.1 ± 1.645(6.21/√89)) 90% Confidence Interval = (23.1 ± 0.744) 90% Confidence Interval = (22.36, 23.84) a . False b . False c . True d . True e . True f . False g . True h . True i . True j . True k . False l . True

110, 144

Mean = 21 = -11 12 1 = -11 19 - 21 = 6 16 1)² = 121 (- = 49 89² = 5 + 121 + 4 es (n-1)

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ematical evidence and a formula. (Q3). To 6, 18, 19, 14, 16 The , 144 The = 8

andom from the list he list is 0.6 (6 out of 10 numbers are higher than 12). rawing a second number higher than 12 from the list he probability of drawing a number higher than 12 GIV scatter diagram along with the coefficient of correlation (r squared). Upload your Excel sheet on Coefficient, discuss the strength (strong, weak…) and type (positive, negative) of the relationship between n Coefficient is 0.9297, indicating a strong positive relationship between pages and price. According to the trendline, how much should a book that is 270 pages cost? According to the trendline, a book that is 270 pages should cost $9.24. ording to the trendline, how many pages should a book that cost 11 dollars have? 4 5 6 7 8

to the trendline, a book that costs $11 should have approximately 212 pages Discuss how confident you are in the answers to c and d, and use statistical concepts learned in this class to that an increase in pages is associated with an increase in price. The coefficient of determination (r squared)

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