This document provides study material on strategics for business. It includes topics like decision making, confidence intervals, sample size estimation, hypothesis testing, regression analysis, and more.
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Strategics for business Decision
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Table of Contents Week 7: Question 2..........................................................................................................................1 a. Provide 95% confidence interval for all patients....................................................................1 b. Sample size required to estimate proportion of all hospitals..................................................1 Week 8: Question 10........................................................................................................................2 Hypothesis...................................................................................................................................2 Test statistic formula...................................................................................................................2 Level of significance...................................................................................................................2 Decision rule...............................................................................................................................2 Calculation..................................................................................................................................2 Conclusion...................................................................................................................................3 Week 9: Question 3..........................................................................................................................3 a. Null and alternative hypothesis...............................................................................................3 b. Decision rule..........................................................................................................................3 c. Calculate the test statistic........................................................................................................3 d. Decision...................................................................................................................................5 Week 10: Question 2........................................................................................................................5 a. Estimated regression line and interpret the slope....................................................................5 b. Estimated total personal wealth when a person is 50 year old................................................5 c. What is value of coefficient of determination.........................................................................6 d. testing the value at 10% level of significance.........................................................................7 Week 11: Question 3........................................................................................................................8 a. Estimated regression equation.................................................................................................8 b. Compute the coefficient of determination...............................................................................9 c. Test to determine y is significantly related to independent variable.......................................9 d. Test whether x and y are significantly related......................................................................10
Week 7: Question 2 a. Provide 95% confidence interval for all patients The proportion is as mention below: Step 1: CI =p +- Z(a/2)√p (1 – p) / n in which P = sample proportion and n= sample size So, as per the question n= 400 and the level of significance is 0.05 which is fixed. Therefore, sample proportion = x / n So, sample proportion = 800 / 400 = 0.2 Step 2: In accordance with the table of standard normal distribution, there is a need to find the critical value and the level of significance is 0.05 in which z= 1.96 so, = CI =p +- Z(a/2)√p (1 – p) / n =0.2 +,- 1.96√0.2 (1 – 0.2) / 400 = 0.2 +- 1.96 (0.02) = 0.2 +- 0.0392 = (0.1608 and 0.2392) b. Sample size required to estimate proportion of all hospitals Confidence level95% Margin of error4% Population Proportion50% Sample size601 So, it is interpreted that when 0.04% of margin of error is given, then the sample size estimated for proportion of all hospitals referrals to the health care is 601. 1
Week 8: Question 10 Hypothesis H>0 Null Hypothesis: To determine there is no significant relationship between salaries and population H>1 Alternative hypothesis: To determine there is a significant relationship between salaries and population Test statistic formula Z score = x-μ/ Standard deviation Where x is raw score and is population mean. Level of significance The Z score value is 0.2 and on the other hand p value is 0.079 which is greater than 0.05 and when the value if higher than level of significance which means alternative hypothesis is accepted. So, it is reflected that there is a significantrelationship between salariesand population. Decision rule As per the decision rule, it is stated that if the test statistic follows a normal distribution then critical value from the standard normal distribution which is known as Z statistic. So the rule state that when the Z score value of greater than 0.05 then the alternative hypothesis is accepted and on the other other side when the value is less than 0.05, if means the null hypothesis is accepted while alternative hypothesis is rejected. Calculation Step 1: x-μ/ Standard deviation Step 2: 50000-48400 / 8000 Step 3: 0.2 So, p-value from the Z table shows that P (x<50000) = 0.57926 P(x>50000) = 1- P (x<50000) = 0.42075 P (48400 < x < 50000) = P (x< 50000) – 0.5 = 0.07926 2
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Conclusion So, from the above, it has been concluded that there is a relationship between the salaries and population because the value is greater than 0.05. So, it can be stated that there is an increase in the starting salaries when the employees are hire as compared to after some time. Week 9: Question 3 a. Null and alternative hypothesis H0: Null hypothesis: To analyze there is no relationship between average value of production of three lines and hourly quantity H1: Alternative hypothesis: To analyze there is a relationship betweenaverage value of production of three lines and hourly quantity b. Decision rule As per the decision rule, it is analyzed that there are two rules which proves the testing of an hypothesis such that, ï‚·When the value of p which is greater than 0.05, in that case null hypothesis is rejected while alternative hypothesis is accepted. (P > 0.05 = alternative hypothesis) ï‚·On the other side, the value of p is less than 0.05, so null hypothesis is accepted and on the other side, alternative hypothesis is rejected (P < 0.05= null hypothesis). c. Calculate the test statistic GroupCountSumAverageVariance Process 14120304.66 Process 241363411.33 Process 341283213.33 Anova Sourceof variation SSdfMSFP value Between Groups 32 (SST) 2161.630.24 3
Within Groups88 (SSE) 99.77 Total12011 Step 1: to determine the value of df df= k-1 where k = total number of group so, here total number of groups are 3 which process 1, process 2 and process 3 df= 3-1 df = 2 Step 2: to determine df value within a groups, so, df for second source = N-k here N is total number of process in anova table which is 12 So, df = 12 – 3 = 9 Therefore total value is 9+2 = 11 Step 3: To determine the value of MS, the below mention formula is used i.e. For first source, MS = SST / (k – 1) MS = 32 / (3-1) MS= 32 / 2 MST = 16 (1stsource) Next for second source = MS = SSE / (N – k) MS = 88 / (12 – 3) MS = 88 / 9 MSE = 9.77 Step 3: To determine the value of F, the formula is used which is as mention below: F = MST / MSE F = 16 / 9.77 F = 1.63 4
Step 4: to determine the value of p- significant value = the value if f- distribution of 1.63 is 0.24 d. Decision In accordance with the above table, it is interpreted that the value of significant factor is greater than 0.05 i.e. (P > 0.05) that means alternative hypothesis is accepted and on the other side,rejectingthenullhypothesis.So,thedecisionreflectedthatthereisasignificant relationship between the mean value of three lines of production process with the hourly quantity of production within a XYZ company. So when the working hours within a firm is increases it means that the production of all the three lines are also increases. Therefore, this shows a positive relationship between both the variables in the quoted organization. Week 10: Question 2 a. Estimated regression line and interpret the slope The regression line equation isÅ·=a+bx,here y is the dependent variable and x is an independent variable. Through the above graph, it is interpreted that there is a relationship between both variables I.e independent and dependent variables. So it is stated that with when the age of person is increases, there is an increase in chances of wealth as well. This reflect the positive interrelationship between the two variables. 5 30405060708090 0 100 200 300 400 500 Age Line Fit Plot Column B Column V Age Total wealth
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b. Estimated total personal wealth when a person is 50 year old X50 Regression equation5.3265*x+45.2159 266.325 Total personal wealth311.5409 c. What is value of coefficient of determination SUMMARY OUTPUT Regression Statistics Multiple R0.9547040663 R Square0.9114598542 Adjusted R Square0.8967031633 Standard Error28.9895420825 Observations8 ANOVA dfSSMSF Significance F Regression1 51907.6386 990981 51907.6386 990981 61.7658699 187 0.00022451 52 Residual6 5042.36130 09019 840.393550 1503 Total756950 6
Coefficients Standard Errort StatP-valueLower 95%Upper 95% Lower 95.0% Upper 95.0% Intercept 45.2159059 852 39.8049868 594 1.13593570 94 0.29931074 67 - 52.1833880 957 142.615200 0662 - 52.183388 0957 142.61 52000 662 Age 5.32659196 5 0.67775877 39 7.85912653 66 0.00022451 52 3.66817598 91 6.98500794 1 3.6681759 891 6.9850 07941 The coefficient of determination is the value of R square and the value is 0.911 which is approx 0.1. so it can be stated that there is a positive linear association and the dependent value is 10% fluctuate in positive or negative side of independent value. d. testing the value at 10% level of significance At 90% Confidence level Step 1 : Hypothesis H0: Null Hypothesis: To determine there is no relationship between wealth and age H1 : Alternative Hypothesis: To determine there is a relationship between wealth and age Step 2 : Standardized the test statistics SUMMARY OUTPUT Regression Statistics Multiple R0.9547040663 R Square0.9114598542 Adjusted R Square0.8967031633 Standard Error28.9895420825 Observations8 Step 3: Level of significance the level of significance is known as alpha which is clearly shows that probability of rejecting the null hypothesis or alternative hypothesis and in the case the level of confidence is 90% because 10% level o significance is chosen. So as per the decision rule, decision is taken in 7
order to determine there is actual difference or not. So, in accordance with the anova table it is interpreted that the value of p is 0.00 which is greater than 0.01 and that is why, alternative hypothesis is accepted. Step 4: Decision rule The hypothesis testing rule shows that when the value of level of significance is greater than 0.01 than alternative hypothesis is accepted and it is so because the level to check confidence is 90% and on the other side, when the value is less than 0.01 then null hypothesis is accepted while alternative hypothesis is rejected. Step 5: Calculation ANOVA dfSSMSF Significanc e F Regression1 51907.6 3869909 81 51907.638699 0981 61.76586 99187 0.00022451 52 Residual6 5042.36 1300901 9 840.39355015 03 Total756950 Coefficients Standard Errort StatP-valueLower 95%Upper 95% Lower 90.0% Upper 90.0% Intercept 45.2159059 852 39.8049868 594 1.13593570 94 0.29931074 67 - 52.1833880 957 142.615200 0662 - 32.132359 5461 122.56 41715 166 Age 5.32659196 5 0.67775877 39 7.85912653 66 0.00022451 52 3.66817598 91 6.98500794 1 4.0095844 807 6.6435 99449 3 8
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Step 6: Conclusion As per the above table, it is concluded that the level of significance or p value is greater than 0.01 i.e. 0.00 and that is why, alternative hypothesis is accepted and as a result, it is stated that there is a close relationship between the wealth and age of person. Like, when the age of a person increases, there is a definitely increase in the wealth. Week 11: Question 3 a. Estimated regression equation The estimated regression line is y= b0 + b1x1 + b2x2 + b3x3 In this, y is dependent variable while x1, x2 and x3 are independent variables So, with the help of data provided, the equation is b0 = 0.0136, b1 = 0.7992, b2 = 0.2280 and b3 = -0.5796 therefore, the regression equation is y = 0.0136 + 0.7992x1 + 0.2280x2 - 0.5796x3. b. Compute the coefficient of determination The coefficient of determination is also known as R square in which different rules are applicable. Also, as per the given situation, the coefficient of determination is Regression sum of squares is 45.936 while on the other side, error sum of squares is 2.6218 Therefore, the coefficient of determination is as mention below: R square = SSR / SSR +SSE = 45.9634 / 45.9634 + 2.6218 = 45.9634 / 48.5852 = 0.9460 So, the value is positive and high, so it is reflected that the relationship between the variables is strong. c. Test to determine y is significantly related to independent variable For testing variable x1 H0: Null hypothesis: There is no significant relationship between family spending and income 9
H1: Alternative hypothesis: There is a significant relationship between family spending and income. To determine the value of p the formula is used = Coefficient / standard error = 0.7992 / 0.074 = 10.8 The p value for the x1 is 1 which is greater than 0.05 and this shows that the alternative hypothesis is accepted and null hypothesis is rejected. For testing variable x2 H0: Null hypothesis: There is no significant relationship between family spending and family size H1: Alternative hypothesis: There is a significant relationship between family spending and family size. = Coefficient / standard error = 0.2280 / 0.190 = 1.2 The p value for this variable is 0.88 which is greater than 0.05 and this is clearly shows that the alternative hypothesis is accepted and that is why, it is stated that y independent variable i.e. family spending is directly related to family size. d. Test whether x and y are significantly related For testing variable x3 H0: Null hypothesis: There is no significant relationship between family spending and addition of savings H1: Alternative hypothesis: There is a significant relationship between family spending and addition of savings. To determine the p value, formula is used = Coefficient / standard error = -0.5796 / 0.920 = -0.63 10
Here, the p value for this is 0.2651 which clearly reflected that it is greater than 0.05 and that is why, it is stated that there is x3 and y are directly related or it has a significant relationship. 11