Question

descriptive statistics
approximating the mean of a data set given a histogram
the moon illusion refers, roughly speaking, to the common experience that the moon appears larger (and closer to us) when it is near the horizon than when it
is at its zenith. to investigate, we have designed an experiment in which participants must adjust the diameter of a variable circular light placed at eye level to
match the diameter of a fixed circular light placed well above eye level. (see, for example, kaufman, l. & rock, i. (1989), \the moon illusion thirty years later,\
chapter 8 in m. hershenson (ed.) the moon illusion, hillsdale, nj: l. earlbaum.)
we have tested 26 participants and recorded for each participant her experimental \adjustable/ fixed ratio.\ our data are summarized in the following histogram
based on the histogram, estimate the mean adjustable/ fixed ratio for the participants. carry your intermediate computations to at least four decimal places,
and round your answer to at least two decimal places.

Explanation:

Step1: Find the mid - point of each class

Assume the class intervals:

• For the first bar (frequency = 2): Let the class interval be \(0 - 0.2\), mid - point \(x_1=\frac{0 + 0.2}{2}=0.1\)
• For the second bar (frequency = 3): Class interval \(0.2 - 0.4\), mid - point \(x_2=\frac{0.2+0.4}{2}=0.3\)
• For the third bar (frequency = 10): Class interval \(0.4 - 0.6\), mid - point \(x_3=\frac{0.4 + 0.6}{2}=0.5\)
• For the fourth bar (frequency = 9): Class interval \(0.6 - 0.8\), mid - point \(x_4=\frac{0.6+0.8}{2}=0.7\)
• For the fifth bar (frequency = 2): Class interval \(0.8 - 1.0\), mid - point \(x_5=\frac{0.8 + 1.0}{2}=0.9\)

Step2: Use the formula for the mean of a grouped data \(\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\)

Here, \(f_1 = 2\), \(f_2=3\), \(f_3 = 10\), \(f_4=9\), \(f_5 = 2\) and \(x_1 = 0.1\), \(x_2=0.3\), \(x_3=0.5\), \(x_4=0.7\), \(x_5=0.9\)

\(\sum_{i = 1}^{n}f_ix_i=f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5\)
\(=2\times0.1+3\times0.3 + 10\times0.5+9\times0.7+2\times0.9\)
\(=0.2+0.9+5+6.3+1.8\)
\(=14.2\)

\(\sum_{i=1}^{n}f_i=2 + 3+10+9+2=26\)

Step3: Calculate the mean

\(\bar{x}=\frac{14.2}{26}\approx0.55\)

Answer:

\(0.55\)