Question

7) Let's use the the data above for female Kestrel length. Here are the data (again), as well as the normal scores. Note that we added added one more female, and that the data are sorted): Length(cm): 29.1 25.1 25.4 25.5 25.6 25.8 26.6 -1.73 -1.15-0.81 -0.55 -0.32 -0.10 27.0 27.3 27.3 28.2 28.6 0.10 0.32 0.55 0.81 1.15 Z-scores: 1.73 (a) Construct a q- plot (normal probability plot) (b) Verify the first three normal scores (i.e., calculate the first three normal scores by hand (you want to verify the three smallest normal scores, given in bold above)).

Answer 1

Solution: 7a. Use MINITAB to plot the normal probability plot. MINITAB procedure: Step 1: Choose Graph> Probability Plot. Step 2: Choose Single, and then click OK. Step 3: In Graph variables, enter the column of Lengd (cm). Step 4: Click OK. MINIT AB output: Probability Plot of Length(cm) Narmal. 95% CI SAMSUN Percent 28 Length(cm) 30 32

Use MINITAB to obtain the value of mean andstandard deviation MINIT AB procedure: Step1: Choose Stat > Basic Statistics > Display Descriptive Statistics. Step 2: In Variables enter the column Length Step 3: Choose option statistics, andselect mean and standard deviation Step 4: Click OK MINIT AB output: Descriptive Statistics: Length(cm) Variable Length (cm) Mean 26.792 St Dev 1.350 From the MINITAB output, The value of meanis 26.792. The value of standard deviation is 1.350.

The nomal scores for the first three values are obtainedbelow: The normal score for x=25.1is obtained below , 25.1-26.792 1.350 -1.692 1.350 =-1.25 The nomal score for x= 25.4 is obtained below: 25.4-26.792 1.350 _-1.392 1.350 --1.03 The nomal score for x=25.5 is obtained below: , 25.5 - 26.792 1.350 -1.292 1.350 =-0.96 Thus, it is clear that the first three scores were not equal to the scores given in bold format.