The following 13 questions (Q1 to Q13) are based on the following example: Patients recovering from an appendix operation normally spend an average of 6.3 days in the hospital. The distribution of recovery times is normal with a σ = 2.2 days. The hospital is trying a new recovery program designed to lessen the time patients spend in the hospital. The first 25 appendix patients in this new program were released from the hospital in an average of 5.5 days. On the basis of these data, can the hospital conclude that the new program has a significant reduction of recovery time. Test at the .01 level of significance. Q1: The appropriate statistical procedure for this example would be a A.z-test B.t-test Q2: Is this a one-tailed or a two-tailed test? A.one-tailed B.two-tailed Q3: The most appropriate null hypothesis (in words) would be A.There is no statistical difference in the amount of time appendix patients spend in the hospital when comparing the new recovery program to the population of patients on the traditional recovery program. B.There is a statistical difference in the amount of time appendix patients spend in the hospital when comparing the new recovery program to the population of patients on the traditional recovery program. C.The new appendix recovery program does not significantly reduce the number of days spent in the hospital when compared to the population of patients on the traditional recovery program. D.The new appendix recovery program does significantly reduce the number of days spent in the hospital when compared to the population of patients on the traditional recovery program.Q4: The most appropriate null hypothesis (in symbols) would be A.μnew program = 6.3 B.μnew program = 5.5 C.μnew program  6.3 D.μnew program  6.3 Q5: Set up the criteria for making a decision. That is, find the critical value using an alpha = .01. (Make sure you are sign specific: + ; - ; or  ) (Use your tables) Summarize the data into the appropriate test statistic. Steps:Q6: What is the numeric value of your standard error? Q7: What is the z-value or t-value you obtained (your test statistic)? Q8: Based on your results (and comparing your Q7 and Q5 answers) would you A.reject the null hypothesis B.fail to reject the null hypothesis Q9: The best conclusion for this example would be A.There is no statistical difference in the amount of time appendix patients spend in the hospital when comparing the new recovery program to the population of patients on the traditional recovery program. B.There is a statistical difference in the amount of time appendix patients spend in the hospital when comparing the new recovery program to the population of patients on the traditional recovery program. C.The new appendix recovery program does not significantly reduce the number of days spent in the hospital when compared to the population of patients on the traditional recovery program. D.The new appendix recovery program does significantly reduce the number of days spent in the hospital when compared to the population of patients on the traditional recovery program. Q10: Based on your evaluation of the null in Q8 and your conclusion is Q9, as a researcher you would be more concerned with a A.Type I statistical error B.Type II statistical error Calculate the 99% confidence interval. Steps:Q11: The mean you will use for this calculation is A. 5.5 B. 6.3 Q12: What is the new critical value you will use for this calculation? Q13: As you know, two values will be required to complete the following equation: __________    __________ The following 4 questions (Q14 to Q17) are based on the following situation: If α = .10, and β = .30, complete the following questions by inserting the appropriate probability of each. Q14: The statistical decision is to reject the null, and H0 is really true (ie: a Type I error) Q15: The statistical decision is to fail to reject null, and H0 is really true (ie: a correct decision) Q16: The statistical decision is to reject the null, and H0 is really false (ie: Power) Q17: The statistical decision is to fail to reject the null, and H0 is really false (ie a Type II error) The following 14 questions (Q18 to Q31) are based on the following example: A researcher wants to determine whether high school students who attend an SAT preparation course score significantly different on the SAT than students who do not attend the preparation course. For those who do not attend the course, the population mean is 1050 (μ = 1050). The 16 students who attend the preparation course average 1150 on the SAT, with a sample standard deviation of 300. On the basis of these data, can the researcher conclude that the preparation course has a significant difference on SAT scores? Set alpha equal to .05. Q18:The appropriate statistical procedure for this example would be a A.z-test B.t-test Q19: Is this a one-tailed or a two-tailed test? A.one-tailed B.two-tailed Q20: The most appropriate null hypothesis (in words) would be A.There is no statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.B.There is a statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.C.The students who took the SAT prep course did not score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course. D.The students who took the SAT prep course did score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course. Q21: The most appropriate null hypothesis (in symbols) would be A. μSATprep = 1050 B.μSATprep = 1150 C.μSATprep  1050 D.μSATprep  1050 Q22: Set up the criteria for making a decision. That is, find the critical value using an alpha = .05. (Make sure you are sign specific: + ; - ; or  ) (Use your tables) Summarize the data into the appropriate test statistic. Steps:Q23: What is the numeric value of your standard error? Q24: What is the z-value or t-value you obtained (your test statistic)?Q25: Based on your results (and comparing your Q24 and Q22 answers) would you A.reject the null hypothesis B.fail to reject the null hypothesis Q26: The best conclusion for this example would be A.There is no statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.B.There is a statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.C.The students who took the SAT prep course did not score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course. D.The students who took the SAT prep course did score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course. Q27: Based on your evaluation of the null in Q25 and your conclusion is Q26, as a researcher you would be more concerned with a A.Type I statistical error B.Type II statistical error Calculate the 99% confidence interval. Steps:Q28: The mean you will use for this calculation is A. 1050 B. 1150 Q29: What is the new critical value you will use for this calculation? Q30: As you know, two values will be required to complete the following equation: __________    __________ Q31: Which of the following is a more accurate interpretation of the confidence interval you just computed? We are 99% confident that the scores fall in the interval _____ to _____. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval _____ to _____. We are 99% confident that the example above has correct values. We are 99% confident that the difference in SAT scores between the students who took the prep course and the students who did not falls in the interval _____ to _____. The following 2 questions (Q32 to Q33) are based on the following situation: The national average for the verbal section of the Graduate Record Exam (GRE) is 500 with a standard deviation of 100. A researcher uses a sampling distribution made up of samples of 100. Q32: According to the Central Limit Theorem, what is the mean of the sampling distribution of means? A.10 B.50 C.100 D.500 Q33:According to the Central Limit Theorem, what is the standard error of the mean? a. 10b.50 c.100 d.500 Q34:As you increase the number of subjects in your sample, the calculated value of a t-test will A.increase B.decrease C.remain the same Q35:As you decrease the true distance between the null and alternative hypotheses (μ1 – μ0), the likelihood of rejecting the null hypothesis A.increases B.decreases C.remains the same Q36:Keeping everything else the same, if you were to decrease your alpha level from .05 to .01, the likelihood of rejecting the null hypothesis A.increases B.decreases C.remains the same The following 4 questions (Q37 to Q40) are either “True” or “False”Q37: The single-most critical component of deciding whether you are to conduct a t-test versus a z-test for hypothesis testing is whether there is a ‘’.Q38: Predicting the characteristics of an entire group, after having measured a small group, is the major goal of inferential statistics. Q39: Degrees of freedom for a single sample z-test is/are ‘n-1’.Q40: Degrees of freedom for a single sample t-test is/are ‘n-1’.

read the attached article and write a four-paragraph (maximum) critique addressing the following questions:Do you believe the results? Why or why not?State the reported confidence interval(s) estimate in two ways: the upper and lower limitsthe sample proportion ± the error term(don’t forget to include the confidence level)State the meaning of the confidence interval(s).Can a reported mean of your article data be treated as a value from a population having a normal distribution? Why or why not?https://files.eric.ed.gov/fulltext/EJ1174957.pdf Link to articlehttps://eric.ed.gov/?q=Confidence+Interval&pr=on&f...