Question

Data on the level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) for each of 992 senior managers at CPA firms are saved in CORSUS.txt
a) Construct a histogram for the data and use it to evaluate the validity of the normality assumption.
b) Obtain the mean and standard deviation for the data and use these statistics to evaluate the validity of the normality assumption.
c) Obtain the interquartile rage for the data and use these statistics to evaluate the validity of the normality assumption

Answer 1

Answer:

A) attached below

B) mean value = 67.755,  std = 26.871

C) IQR( interquartile range ) = 37

Explanation:

A) Construct a histogram for the data and use it to evaluate the validity of normality assumption

Using Minitab to construct the Histogram from the shape of the Histogram we can see that the Normality assumption is valid because the shape is fairly symmetric

screenshot of Histogram is attached below

B) Obtain the mean and standard deviation for the data and use these statistics to evaluate the validity of the normality assumption.

still using Minitab to determine the std and mean values

mean value = 67.755,  std = 26.871

Next : find the percentage of the observation that lie within 1,2 and 3 std from the mean

For one(1) std from the mean the interval = ( 40.884, 94.626 )

percentage of observation = 665 / 992 = 67.04

For two(2) std from the mean; The interval = ( 14.013 , 121.497 )

percentage of observation = 946 / 992 = 95.36%

For three(3) std from the mean ; The interval = ( -12.858, 148.368 )

percentage of observations = 991 / 992 = 99.90%

The percentages from the above calculations indicates the validity of the normality assumption

C) Obtain the interquartile rage for the data and use these statistics to evaluate the validity of the normality assumption

using MINITAB

since the data are assumed Normal; Ratio = $\frac{IQR}{S} = 1.3$

std (s) = 26.871,   IQR( interquartile range ) = 37

Next check if     IQR / S will be = 1.3

= 37 / 26.871 = 1.377 ( This validates the normality assumption )