can somone please check my work.

Question

The data from placebo 0mg is listed below.

57,27,32,31,34,38,71,33,34,53,36,42,26,52,36,55,36,42,36,54,34,29,33,33

 

 

What is the mean number of correct responses of the participants after taking the placebo (0 mg/kg)?

 

after adding all of the numbers i get 954 in which i then divide by the total of numbers which is 24 leaving me with

 

954/24 = 39.75 for the mean.

 

What are the standard deviation and the interquartile range of the d0 condition?

 

With this one its a little more complicated.

 

DATA   /  MEAN   /   ANSWER   /   SQUARED

 

57 - 39.75 = 17.25 = 297.5625

27 - 39.75 = -12.75 = -162.5625

32 - 39.75 = -7.75 = -60.0625

31 - 39.75 = -8.75 = -76.5625

34 - 39.75 = -5.75 = -33.0625

38 - 39.75 = -1.75 = -3.0625

71 - 39.75 = 31.25 = 976.5625

33 - 39.75 = -6.75 = -45.5625

34 - 39.75 = -5.75 = -33.0625

53 - 39.75 = 23.25 = 540.5625

36 - 39.75 = -3.75 = -14.0625

42 - 39.75 = 2.25 = 5.0625

26 - 39.75 = -13.75 = -189.0625

52 - 39.75 = 12.25 = 150.0625

36 - 39.75 = -3.75 = -14.0625

55 - 39.75 = 21.25 = 451.5625

36 - 39.75 = -3.75 = -14.0625

42 - 39.75 = 2.25 = 5.0625

36 - 39.75 = -3.75 = -14.0625

54 - 39.75 = 14.25 = 203.0625

34 - 39.75 = -5.75 = -33.0625

29 - 39.75 = -10.75 = -115.5625

33 - 39.75 = -6.75 = -45.5625

33 - 39.75 = -6.75 = -45.5625

 

squared total = 1730.5 (variance)

 

standard deviation

σ=√1730.5

=41.599279

=42

 

 

 

Answer 1

I didn't do the arithmetic but I don't think your sum of squares is correct (squares can't be negative) and for the variance don't forget to divide by n before taking the square root.

 

 

Answer 2

you added right but on your Data/Mean/ Answered / Squared table you left off the second number.

It should be 57,27,32  

not 57,32,31