Please show all work and formulas in excel. Health Administration wants that the

Question

Please show all work and formulas in excel.  

Health Administration wants that the Chair of a Master's program in determine if there is a difference in GPAs between Male students and Female students who have completed all of the required core courses in the program Suppose, further, that the Chair has obtained this data from the Registrar and has promised to keep the information confidential. You have been asked to analye the data using your Excel skills. Assume that you have been working on thi analysis, and you have determined that the 47 Males in the program have average GPA of 3.15 with a standard deviation of 0.42, while the 56 Females in the program have an average GPA of 3.45 with a standard deviation of 0.37 You now want to analyze the these data. (a) State the hypothesis and the research hypothesis on an Excel (b) Find the standard error of the difference between the means using ExceL (c) Find the critical t value using Appendix E, and enter it on your spreadsheet. (d) Perform a t-test using Excel. What is the value of t that you obtain? (e) Use 3 decimals for all figures that require a formula () State your result on your spreadsheet. (g)State your conclusion in plain English on your spreadsheet (h) Save the file as: HSM3

Explanation / Answer

here we use t-test with

(a) null hypothesis H0:mean1=mean2 and

alternate hypothesis H1:mean1?mean2

(b)(sp*(1/n1 +1/n2)1/2)=0.08

Difference between mean=0.03

(c) critical t(0.05,101)=1.9837

(d)statistic t=difference of mean/SE(difference of mean)=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2)=0.3/0.08=3.85

and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2

(f) |t|>t-critical

(g) we fail to accept null hypothesis and conclude that there is differenem between male and female GPA

following informaiton has been generated.........

t-test sample mean s s2 n (n-1)s2 male 3.15 0.42 0.1764 47 8.1144 female 3.45 0.37 0.1369 56 7.5295 difference= 0.3 0.3133 103 15.6439 sp2= 0.15 sp= 0.39 SE= 0.08 t= 3.85 one tailed p-value= 0.0001 two tailed p-value= 0.0002 critical t(0.05, 101) 1.9837

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