Question

Use SPSS, Excel or Minitab software to answer the following questions.

Generate normal random data with sample size, n = 40, mean = 20 and standard deviation = 11. Write the generated data in the following table.

Answer:

1		11		21		31
2		12		22		32
3		13		23		33
4		14		24		34
5		15		25		35
6		16		26		36
7		17		27		37
8		18		28		38
9		19		29		39
10		20		30		40

1. Conduct the descriptive statistics for center and dispersion. Interpret your results!

Answer:

2. Construct a 90% confidence interval for the average. Interpret your results!

Answer

3. Construct the histogram, boxplot, and stem and leaf graphical charts. Is the data approximately bell-shaped? Are there outliers? Explain!

Answer

Answer 1

We will use Minitab-software to genrate random sample and creat histogram , similar can be obtained from SPSS, Excel

Steps Generate normal random data with sample size, n = 40, mean = 20 and standard deviation = 11.

Go to Cacl Random data Normal

Now use will get a box of Normal distibution , selcect mean = 20 , standard deviation = 11 and number of rows to genrate n =40

So genrated data is as follow

1   13.9769
2   10.3455
3   37.4762
4   19.2873
5   8.1485
6   12.6236
7   29.0664
8   42.0034
9   17.4957
10   13.1276
11   38.2036
12   13.9106
13   7.8937
14   11.9925
15   7.1173
16   -0.6491
17   16.7038
18   24.6159
19   17.9619
20   3.2336
21   3.3125
22   36.2256
23   39.5352
24   20.0319
25   17.0542
26   33.7631
27   42.6232
28   40.3384
29   17.5506
30   21.7886
31   24.2300
32   30.0854
33   42.9405
34   25.5639
35   25.6088
36   20.5326
37   18.1205
38   27.1002
39   24.1429
40   20.4310

Q Conduct the descriptive statistics for center and dispersion. Interpret your results!

To calculated mean and standard deviation

Step - Go to Cacl Column Statistics ( then select mean , standard deviation as other required things )

Mean of Genrated Random Variable

Mean of Genrated Random Variable = 21.8878

Standard Deviation of Genrated Random Variable

Standard deviation of Genrated Random Variable = 11.8023

Median of Genrated Random Variable

Median of Genrated Random Variable = 20.2315

Q.Construct a 90% confidence interval for the average. Interpret your results!

90% confidence interval is given by

C.I = ( - * ,   + * )

Now = 21.8878   , s = 11.8023 , n=40

Now is t-distributed with n-1 = 39 degree of freedom , at = 0.10

which is calculated as follow

Steps - Calc Probability distribution t distribution

then selct degree of freedom = 39 ,, and for = 0.10 , critivcal value to be 1-0.10/2

Inverse Cumulative Distribution Function

Student’s t distribution with 39 DF

P( X ≤ x )        x
      0.95 1.68488

Thus 90% confidence interval for the average. is

C.I = ( - * ,   + * )

      = ( 21.8878 - 1.68488 * ,   21.8878 + 1.68488 * )

      = ( 21.8878 - 1.68488 * 1.866107 ,   21.8878 + 1.68488 * 1.866107 )

      = ( 18.74363 , 25.03197 )

90% confidence interval for the average. is ( 18.74363 , 25.03197 )

By 90% confidence we say that we are 90% sure that population average will lie within this interval ( 18.74363 , 25.03197 )

Construct the histogram, boxplot, and stem and leaf graphical charts. Is the data approximately bell-shaped? Are there outliers?

To plot histogram select - Graph Histograme

To genrate boxplot select - Graph Boxplot

To make stem and leaf graphical charts select - Graph Stem-and-Leaf

Stem-and-Leaf Display: Genrated Random Variable

Stem-and-leaf of Genrated Random Variable N = 40
Leaf Unit = 1.0

3   0 033
6   0 778
12 1 012333
19 1 6777789
(7) 2 0001444
14 2 5579
10 3 03
8   3 6789
4   4 0222
0   4

We can't sat that sample data is bell-shaped since in histograme we can see that there are some peak at right extreame points , althoug it have peak in center but just it have fewer observation in range 25-30 , it can be consider approximately bell-shaped .

Are there outliers? Explain!

Now summary of ablove random nubers is as follow

summary(x)
   Min.      1st Qu.     Median      Mean 3rd Qu. Max.
-0.6491   13.3234 20.2314   21.8879 29.8306   42.9405

IQR = Q3 - Q1 = 29.8306 - 13.3234 = 16.5072

Outliers are

any number greater than 1.5*IQR + Q3 = 1.5* 16.5072+ 29.8306 = 54.5914

any number less than Q1-1.5*IQR - Q1 = 13.3234 - 1.5* 16.5072 = -11.4374

Thus any observation greater than 54.5914 and any observation less than -11.4374 are outliers

Now from Minitab output we have

Minimum of Genrated Random Variable

Minimum of Genrated Random Variable = -0.649059

Maximum of Genrated Random Variable

Maximum of Genrated Random Variable = 42.9405

Now maximum Observation is 42.9405 < 54.5914

And minimum observation is -0.649059 > -11.4374

We can stay than there are there are no outliers in our samples as all are within IQR range .