B. Ha: the mean daily order is different for each type of Pizza.D. Ha: the mean daily order is different for type 1 and type 3 Pizzas.E. None aboveB. 0.05 < p < 0.10C. 0.025 < p < 0.05D. 0.01 <p < 0.025E. p < 0.01*COURSE: DSCI 3710 Print Name: Exam 1 Version A Signature: Spring 2016 Student ID#: INSTRUCTIONS:- Please print your name and student ID number on this exam. Also, put your signature on this exam.- On your scantron PRINT your name and exam version. - This exam has 25 questions. You have 75 minutes to complete this exam. The exam is open book and open notes. You may use a laptop computer or any type ofhand calculator but please show all your work on the exam and mark all answers on the scantron. Usage of cell phones, digital cameras and other communication devices is prohibited.- Please DO NOT pull this exam apart. When you have completed the exam, pleaseturn your scantron and exam booklet into your instructor, at the front desk.- Good luck and we wish you well on the exam.Note: Whenever question(s) are connected you may be asked to assume a result (givena value) as an answer for the previous question but this result (value) may or may not be correct. The procedure is set in place to prevent you from losing points on a subsequent question because you made a mistake on some previous question/s.Use the information given in the following paragraph to answer the first four questions.225 people are randomly chosen at a shopping mall to taste-test a new brand of fruit drink. They are asked to rate the drink on a scale from 1 to 7, with 1 being very bad and 7 being very good. The results of the survey reveal that the average rating is 5.60 with a standard deviation of 1.35. The marketing division of the fruit drink distributor is only interested in selling this drink if the true mean rating is more than 5.15. 1. What is the alternative hypothesis for testing whether the fruit drink distributor should sell this drink?A. Ha: µ ≠ 5.15B. Ha: µ = 5.60C. Ha: µ > 5.60D. Ha: µ < 5.15E. Ha: µ > 5.15*2. What is the calculated value of the test statistic?A. 1.28B. 1.96C. 5.00*D. 2.86E. 1.653. What is the rejection region for testing at the 0.025 level of significance whether the fruit drink distributor should sell this drink?A. Z > 1.645B. Z > 1.96*C. Z < -1.28D. Z < -1.96E. Z > 1.284. Assuming the calculated value of the test statistic is 3.25, what is the conclusion of testing at the .10 level of significance whether the fruit drink distributor should sell this drink?A. Based on the sample data, there isn't sufficient evidence to conclude that the average rating is more than 5.15.B. Based on the sample data, there is sufficient evidence to conclude that the average rating is more than 5.15.*C. Based on the sample data, there isn't sufficient evidence to conclude that the average rating is no more than 5.15.D. Based on the sample data, there is sufficient evidence to conclude that the average rating is no more than 5.15.E. None of the above.Use the information given in the next paragraph to answer the next four questions.The table below shows the productivity index scores of 10 employees who participated in a productivity improvement program during one month period. The productivity index scores shown are those before and after completing the program, for each employee. Test whether there is significant improvement in the productivity after the training (an increase in the productivity index score). Excel analysis at the 1% significance level is shown.EmployeeBefore(1)After(2) t-Test: Paired Two Sample for Means196 94277 83 Before After364 75Mean 71.3 76.2486 89Variance 185.5667 157.95565xx yyObservations 10 10652 51Pearson Correlation 0.952981761 68Hypothesized Mean Difference 08xx yydf 9963 68t Stat -3.735011060 70P(T<=t) one-tail 0.002331t Critical one-tail 2.821438P(T<=t) two-tail 0.004661t Critical two-tail 3.249836 5. What is the alternative hypothesis for testing the belief that there was an increase in the mean of the productivity index scores after the productivity improvement program? A. Ha: µ2 > 0B. Ha: µ1 > µ2C. Ha: µ1 ≠ µ2D. Ha: µ1 < µ2 *E. Ha: µ1 = µ26. What is the table value (at the 1% level) of the appropriate critical test statistic to test the belief that there is an increase in the mean productivity index scores of employees who go through the productivity improvement program?A. 3.2498 B. 0.0023 C. -3.7350 D. 2.8214* E. 0.00477. What is the p-value for testing the hypothesis that there is an increase in the mean of the productivity index scores of employees who participated in the productivity improvement program?A. -3.7350 B. 0.0047 C. 0.00233* D. 2.8214 E. 0.01228. Based on the excel output and the significance level of 1%. What would be your conclusion about whether the training program increased the mean productivity index scores of employees who participated in the productivity improvement program?A. Conclude that there is insufficient evidence that the productivity improvement program increased the mean productivity index scores since p-value is less than α.B. Conclude that there is insufficient evidence that the productivity improvement program increased the mean productivity index scores since p is more than α.C. Conclude that there is sufficient evidence that the productivity improvement program increased the mean test productivity index since p-value is more than α.D. Conclude that there is sufficient evidence that the productivity improvement program increased the mean productivity index scores since p-value is less than α.*E. none of the above Please use the information given in the paragraph below to answer the four questions that follow. 1Starbucks wants to investigate the proportion of customers in Dallas area who would like Evenings program which features a thoughtful selection of wine and craft beer alongside the signature coffee and tea beverages. If over half of the customers favor it, Starbucks will approve the Evenings program. A random sample of 625 students was obtained and 350 students indicated they would like the new services. A hypothesis test was conducted using the KPK macro-based MS Excel output provided below to determine if a majority of students favor the Evenings program. A significance level of 5% was used.Z Test for One Proportion Sample Proportion 0.560000 Number of Observations 625 Ho:XXX Ha:XXX Z* 3.000000 P[Z  Z*] XXX Z