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SDA 101 Students  Chi-Square Test of Independence c
SPSS output for Regression  Tests of Significance   Normal distribution  
 SPSS Instructions  Exam questions More Questions

Question One

A SRS of a certain substance was obtained and each specimen in the sample was weighed. The weights (in mg) are given in the stem plot below

    Stem-and-Leaf Plot of weights

     Stem &  Leaf

             0     11123333

            0    678

             1     12

             1    799

             2     4

             2     56

             3

             3

             4

             4

             5      3

 Stem width:   100.00

 

 Test for Normality     Table A

WEIGHT

Kolmogorov-Smirnov 

Shapiro-Wilk

  

Statistic

df

Sig

Statistic

df

Sig

  

.193

20

.049

.803

20

.001

Note you will not use all of the words and each cell is considered as one dash in the paragraph
For example    is not   would fit in the space             

Cube

Cube root

df

higher

is not

is

left

larger

lower

negative

p-value

positive

random

right

rough

sample

size

simple

significance

smaller

space

square

square root

statistic

Statistics

would

would not

m

s

s

0.001

0.05

0.5

0.803

12.2

12.931

13.2

13.931

20

123

130.31

x

 

 

 

 

 

 

            The stem plot above shows that the data from the Simple Random Sample is skewed to the right. The mean would therefore be larger than the median. One possible transformation to eliminate the skewness is the log transformation, another possible transformation is cube root.

            The table A shows that the data is not normally distributed. This can be seen because the p-value is equal to 0.001 and this is less than the level of significance which is usually the value of 0.05.

The mean of the data set is 122 and the sample standard deviation is 129.31. The symbol for the sample mean is `x   and for the sample standard deviation is s

If a linear transformation was applied to the data it would not change the shape of the stem plot. If the transformation involved dividing each observation by 10 and then adding 1, the mean and standard deviation of the transformed data would be
13.2 and 12.931.  

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