Question

An article in Human Factors (June 1989) presented data on visual accommodation (a function of eye movement) when recognizing a speckle pattern on a high-resolution CRT screen. The data are as follows: 36.45, 68.71, 37.43, 42.18, 26.72, 50.77, 39.30, and 49.71. Construct a normal probability plot. Does it seem reasonable to assume that visual accommodation is normally distributed?

Answer 1

Solution: Follow the below steps to derive z-score for each x-value: . Sort the data for x in ascending order Create a Position variable i which is the position of the observation in the dataset Calculate fias below: i 0.375 n 0.25 Calculate z-score for each f value using normal distribution table Hence, the table can be constructed as below: 26.72 36.45 37.43 39.30 42.18 49.71 50.77 68.71 0.076 0.197 0.318 0.439 0.561 0.682 0.803 0.924 Z-score 1.434 0.852 0.473 0.153 0.153 0.473 0.852 1.434 4 5 6 7 The normal probability plot is as below:

2.00 1.50 1.00 0.50 8 0.00 0.50 2 30 50 60 70 80 -2.00 Assessing the normal probability plot above, we can say thatthe sample data could have come from a population that is normally distributed since the relationship between the expected z-scores and the observed values is approximately linear