Question

Question 24:

-create a frequency table

-Identify the IV & DV

-Graph it

A researcher conducts a study comparing two different treatments with a sample of n = 16 participants in each treatment. The study produced the following data: 

Treatment 1: 6 7 11 4 19 17 2 5 9 13 6 23 11 4 6 1

Treatment 2: 10 9 6 6 2 11 1 12 7 10 9 1 11 8 6 3 

a. Calculate the mean for each treatment. Based on the two means, which treatment produces the higher scores? 

b. Calculate the median for each treatment. Based on the two medians, which treatment produces the higher scores? 

c. Calculate the mode for each treatment. Based on the two modes, which treatment produces the higher scores?

Answer 1

Anso Formulae o Mean Exif Exit Efir Medians Average of th}, +th obs" Mode: The item are repeated maximum times, 1) Treatment & Table 1 value Tally Frequency (fi) ascending Corresponding ordes xi frequency (fi.) -5 - - = = - - - - - 23 EH-16. SHEN - 16 Exifi= 144 .: Mean for Treatment 1=X exifi Efi 144 16 . Median for Treatment a - My = Average of (4) the (4+1)th obsn = Average of (16) the ( 16 / + 1)th obs". - Average of C8th & gth) obs" Shot on Y95 - Average of ( 1413)

i Median M = 11 +13 IM, =12 . Mode - maximum frequency of correspounding values (xi). = maximum bequency is 3 then corresponding values is 6 [ Mode 1= 67 Treatment 28 values Tally values CE' ascending orders brequency xit To 2- 18 20 12 2 22 12 - A Total = 12 112 N-16. E= N=16. treatment 2 a. Mean for = 72 = Exiti N 1/2 X2 = 7 Median= M2 = Average of (4]th R (A + 12th obs? = Average of ( 16 th & ( 16 +15th obs? - Average or (8th & gth) obs". Average of ( 10 f11) Shot on Y95

Median= M2 = 10+11 21 = 10.5. .: Modez-maximum frequen [M2 = 10.5 dez-maximum frequency of correspounding class or values (xi) - maximum freg. is a 4 coresponding value is 6 Mode 2=6. Graph 1 for treatment 1:- Hishogram, yanin Histogram hovanbay Mode. + - - - - - . ou à 2 x-axis s. 6 Ag is in 17 ig 23 x-values treatment 2% Histogram Gerraph 28 for yaxis frequency Mode 2 + - - HI 3 6 10 11 12 1 2 x-axis 7 8 9 x-values

Conclusion i) Mean for treatment 1 = x=9 Mean for treatment 2= 72 = 7 . The treatment f is highly scores of treatment 2. j.e x17 32. 1 Median for treatment 1 = M, = 12 Median for treatment 2= Mac 10.5. ". The weatment I is higher scores of treatment 2. je. Mi > M2 iii) Mode for treatment 1= 6 . Mode for beatment 2 = 6 : Mode for both beatments are same. The both weatments are not produces higher score.