MATH-533 Week 6 Course Project, Part B – Hypothesis Testing and Confidence Intervals – Download For An A+

Running Head: Course Project Part B MATH-533 Week 6 Course Project, Part B – Hypothesis Testing and Confidence Intervals – Download For An A+ Hypothesis Testing MATH – In the hypothesis testing, two type hypotheses are made, one is called Alternate hypothesis and second is null hypothesis. What would be researched, it is called Alter hypothesis and in against of alternate hypothesis called Null hypothesis. Ans. 1 The average (mean) sales per week exceed 41.5 per salesperson: In the given problem, it will test that the sales per week is more than 41.5 per sales person. It is important question which test will be applied. In this scenario, single tail test will be done because only one side of result will be checked. On the other hand, sample size is greater than 30. So Z score will be calculated to test the alternate hypothesis. Units are taken in $1000 (Anderson, Sweeney & Williams, 2011). Key Statics Sales Mean 42.34 0.41711 Standard Error 4 Median 42 Mode 44 4.17113 Standard Deviation Sample Variance 7 17.3983 8 Kurtosis -0.21708 0.07260 Skewness 2 Range 20 Minimum 32 Maximum 52 Sum 4234 Count 100 Step – 1- Hypotheses Ho: μ ≤ 41.5 Ha: μ > 41.5 Step -2 Level of Significance α = 0.05 Critical Value Z0.05 = 1.645 (taken from table) Step – 3 and 4- Calculation Z score Test – Single Tail (Result According to Minitab) One-Sample Z Check Snap Shoot of Minitab in Appendix I Test of μ = 41.5 vs > 41.5 The assumed standard deviation = 4.17 N Mean SE Mean 95% Lower Bound Z P 100 42.340 0.417 41.654 2.01 0.022 Step – 5- Decision Making Decision making rule according to p value is mentioned below as: p < α. Reject H0 and accept H1. or p > α. Fail to reject H0. In the given problem, the calculated p value is .022, so it is less than .05 (α). So reject the null hypothesis and accept alternate hypothesis. Calculated Z score value is 2.01 and value of α is 1.645. Z score value is more than critical value. So the null hypothesis is rejected. On the other hand, the lower bound value is 41.654 in comparison to average sales is more than 41.5 per sales per week. So it is evident that sales of every week are more than 41.5 is true. Ans. B :Testing to the true population proportion of salespeople that received online training is less than 55%. In this scenario, the Z score with single tail test will be used due the same reason as mentioned in first scenario. Step – 1- Hypotheses Ho: μ ≥ .55 Ha: μ < .55 Step -2 Level of Significance α = 0.05 Critical Value Z0.05 = 1.645 (taken from table) Critical Value: Step – 3 and 4 Calculations Z score Test – Single Tail (Result According to Minitab) One-Sample Z Check Snap Shoot of Minitab in Appendix II Online trained Sales Employess = 50 Total Sample =100 Portation of online trained Sales person from total sample = 50/100 = .5 Test and CI for One Proportion Test of p = 0.55 vs p < 0.55 Sample X N Sample p 95% Upper Bound Z-Value P-Value 1 50 100 0.500000 0.582243 -1.01 0.157 Using the normal approximation Step – 5 Decision Making Decision making rule according to p value is mentioned below as: p < α. Reject H0 and accept H1. or p > α. Fail to reject H0. In the given problem, the calculated p value is .157, so it is more than .05 (α). So fail to reject the null hypothesis. Calculated Z score value is -1.01 and value of α is 1.645. The Z-score value is less than critical value. So we are failing to reject the null hypothesis. On the other hand, the upper bound value is .58 in comparisons to sales who took online training is less than .55. So it can be said that sales persons who received online training portion is less than 55% is false (Anderson, Sweeney & Williams, 2011). Ans. C :To check the average (mean) number of calls made per week by salespeople that had no training is less than 145. In this scenario, the T score with single tail will be used because the sample size is less than 30. Key Values Calls made by Non traning person Mean 141.35 2.07018 Standard Error 3 Median 143 Mode 134 Standard Deviation 9.25814 85.7131 Sample Variance 6 Kurtosis -0.63781 Skewness -0.27811 Range 33 Minimum 124 Maximum 157 Sum 2827 Count 20 Step – 1- Hypotheses Ho: μ ≥ 145 Ha: μ < 145 Step -2 Level of Significance α = 0.05 Critical Value Z0.05 = 1.645 (taken from table) Critical Value: Step – 3 and 4 Calculations T test – Single Tail (Result According to Minitab) Check Snap Shoot of Minitab in Appendix III Sum of Calls made by Non Trained Sales Persons Sum of Calls made by Non Trained Sales Persons = 2827 Mean of Calls made by Non Trained Sales Persons Mean of Calls made by Non Trained Sales Persons = 141.35 Standard Deviation of Calls made by Non Trained Sales Persons Standard deviation of Calls made by Trained Sales Persons = 9.25814 One-Sample T Test of μ = 145 vs < 145 N Mean StDev SE Mean 95% Upper Bound T P 20 141.35 9.81 2.19 145.14 -1.66 0.056 Step – 5 Decision Making The p value of this problem is .056 and it is more than .05 (α). So there is enough evidence to fail the null hypothesis. The obtain value of Z-score is -1.66 and value of α is 1.645. Thus the Z-score value is less than, so o the null hypothesis is true.On the other hand, the upper bound value is of Z-score 145.14. It is more than average mean of 145. So it is false that sales persons who did not take any training made less than 145 calls per week (Anderson, Sweeney & Williams, 2011). Ans. D. To test the average (mean) time per call is greater than 15 minutes In this scenario, the Z score with single tail test will be used due the same reason as mentioned in first scenario. Step – 1- Hypotheses Ho: μ ≤ 145 Ha: μ > 145 Step -2 Level of Significance α = 0.05 Critical Value Z0.05 = 1.645 (taken from table) Critical Value: Step – 3 and 4 Calculations Z score Test – Single Tail (Result According to Minitab) One-Sample Z Check Snap Shoot of Minitab in Appendix IV One-Sample Z Test of μ = 15 vs > 15 The assumed standard deviation = 2.42 N Mean SE Mean 95% Lower Bound Z P 100 15.340 0.242 14.942 1.40 0.080 Step – 5 Decision Making The p value of this problem is .080 and it is more than .05 (α). So we are failed to reject the null hypothesis. The obtain value of Z-score is 1.40 and value of α is 1.645. Thus the Z-score value comes in critical value, so null hypothesis is true. It is also evident that the lower bound is 14.942. So it is not true that average call duration is more than 15 minutes . References Anderson, D., Sweeney,D. & Williams, T. (2011). Statistics for Business and Economics. Cengage Learning. Appendix I Calculation for Ans. A Appendix II Calculation Ans. B Appendix III Calculation for Answer C Appendix IV