This analysis explores the correlation between the cost of flowers and the number of flowers in a shop. It includes calculations for mean, median, mode, standard deviation, and quartiles. The results show that there is no significant correlation between the cost of flowers and the number of flowers in the shop.
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Amount spent Lower B Mid- point xfCfx.fμx-μ(x-μ)^2f. ((x-μ)^2) 0-10 -0.5 5552550.3 - 45.3 2052.0 910260.45 10-20 9.5 1571210550.3 - 35.3 1246.0 98722.63 20-30 19.5 25102225050.3 - 25.3640.096400.9 30-40 29.5 35123442050.3 - 15.3234.092809.08 40-5039.545144863050.3-5.328.09393.26 50-6049.555156382550.34.722.09331.35 60-7059.565147791050.314.7216.093025.26 70-8069.575139097550.324.7610.097931.17 80-90 79.5 8569651050.334.7 1204.0 97224.54 90-100 89.5 95410038050.344.7 1998.0 97992.36 ∑f=100 ∑x.f=50 30 ∑x.f((x- μ)^2)=55091 Question 1 Part a. Mean, μ=∑x.f/∑f μ=5030 100=50.30 Median, is the number in the middle Therefore, medium 0.5(100) =50
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Bur 50 falls within the 50-60 class and the value is 63. Mode, Occurs where the frequency is highest. And from our table, we can see that the highest frequency is 15. Now the modal class is 50-60 In conclusion, the average cost spent on the flowers is $ 50.30, median number of flower ranges is 63 and the modal money spent is between $50 to $60. Fromthedataabove,Elena’ssalesandexpenditurearenormally distributed hence she experiences no lost at the moment. Part b Standard deviation s=√¿ s=√(55091 99) s=23.5897 Part c We will use the formula, Qk=LB−¿]
Where Qkis the quartile you are computing Cfk=the cumulative frequency before the kthnumber. fQk=frequency where the Qk falls. 1stQuartile. Q1=25/100*100=25 Given N=100, LB=29.5, Cfb=22, fQ1=12, i=10 Q1=29.5+[(25-35)/12]*10=29.5+2.5=32.0 Therefore, the Q1= 32.0 3rdQuartile. Q3=75/100*100=75 Given N=100, LB=59.5, Cfb=63 , fQ3=14, i=10 Q3=59.5+ [(75-63)/14)*10=59.5+8.57=68.07 Therefore, Q3=68.07 The lower quartile shows the least number of customers Elena can serve in a day while theupper quartile showsthe maximum number of customers that Elena serves in a busy day. Question 2 Loan SourceLss than $ 20000$20000-$60000 More than $60,000Total Building Society406545150
Banks2633564 Other158326 Total8110653240 We will start by defining our Null and Alternative hypothesis Ho: For any loan issued to a client, the size of the loan given is not related to the source of the loan. Ha: For any loan issued to a client, the size of the loan given and the loan source are related. We will use theα=0.05which is the 5% percent confidence level. Nowcalculatingthedegreesoffreedomusingtheaboveprovided table. Df=(row-1)(column-1) Df=(3-1)(3-1)=(2)(2)=4 Stating our decision rule; We check from the chi-square table using the values df=4, a=0.05. If the computed X^2 is greater than 9.4877, then we will reject the Ho unless otherwise. Let us now calculate the test statistics x2=Σ(fo−fe)2 fe
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Where x^2= is the chi-square fo=observable values fe=expected values fe=fcfr n To calculate these values effectively; we will generate another table for expected values. Expected<$20000$20000-$60000>$60000 Building Source50.6366.2533.13 Banks21.6028.2714.13 Other8.7811.485.74 x2=(40−50.63)2 50.63+(26−21.60)2 21.60+(15−8.78)2 8.78+(65−66.25)2 66.25+(33−28.27)2 28.27+(8−11.48)2 11.48+(45−33.13)2 33.13 x2=2.2318+0.8963+4.4064+0.02358+0.7914+1.0549+4.2528+5.8993+1.3079 x2=19.5565 Results Because our chi-square is 19.5565 which is greater than, 9.4877 we will REJECT the NULL hypothesis. Conclusion
For any loan issued to a client, the size of the loan given and the loan source are related. Problem 3 DateDJIA xS&P yx.yx^2y^2 11-Feb- 097939.53833.746619503.7463036136.62695122.3876 18-Feb- 097555.63788.425957009.857087544.7621606.0964 25-Feb- 097270.89764.95561503.7652865841.39585072.01 3-Mar- 096726.02696.334683529.5145239345.04484875.4689 10-Mar- 096926.49719.64984302.247976263.72517824.16 17-Mar- 097395.7778.125754742.0854696378.49605470.7344 24-Mar- 097660.21806.126175048.4958678817.24649829.4544 31-Mar- 097608.92797.87607092957895663.57636596.5369 7-Mar- 097789.56815.556352775.6660677244.99665121.8025 14-Apr- 097920.18841.56664831.4762729251.23708122.25 ∑x=74793. 13 ∑y=7842. 15 ∑xy=5882417 5.7 ∑x^2=560882 487 ∑y^2=6169640. 901 Part a Let the regression line be y=a +bx Where b= slope of the regression line A=y-intercept Y=the predicted y value for any given x value And
b=nΣxy−ΣxΣy n(Σx2)−(Σx)2 n=10 Now inserting these values in the b formula above b=10(58824175.7)−(74793.13)(7842.15) 10(560882487)−(74793.13)2 b=1702812.73 14812574.77 b=0.114957241 Now computing for the value of a, we use the formula b=y+bx a=(Σy n)−b(Σx n) a=(7842.15 10)−b(74793.13 10) a=(784.215)−b(7479.313) a=(784.215)−0.114957241(7479.313) a=−75.58619 y=(−75.58619)+(0.114957241)x Part b Given the values ofx=8000b=0.114957241∧a=−75.58619
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Uisngtheformulay=a+bxto estimate the values of y y=(−75.58619)+(0.114957241)8000 y=844.07 Therefore, the closing price for S&P500 is 844.07 Part c Using the correlation formula below, we would plug in the values of the given variables to compute r r=n∑xy−∑x∑y √[n(∑x2)−(∑x)2][n(∑y2)−(∑y)2] r=10(58824175.7)−(74793.13)(7842.15) √[10(560882487)−(74793.13)2][10(6169640.901)−(7842.15)2] r=1,702,812.571 √[5,608,824,870−5,594,012,295][616964.0901−61,499,316.62] r=1,702,812.571 √−3.4513E15 Introducing the complex identity i=√−1 r=1,702,812.571 √3.4513E15∗√−i r=0.028985∗−i r=−0.028985i Thevalueofrcanonlybecomputed∈acomplexformsincethedenominatorisanegativegivesanegativeva
This means that there isinsignificantor no correlation between the cost of flowers and the number of flowers present in the shop.