Question

The dowj.txt file attached to this assignment contains adjusted daily data on the Dow Jones utilities...

The dowj.txt file attached to this assignment contains adjusted daily data on the Dow Jones utilities index from Aug 28 to Dec 18 1972. You should use SPSS to analyze the data.

  1. Use SPSS to Compute the Mean, median, Q1 (or 25th percentile), Q3 (or 75th percentile), Standard deviation. Copy your table over to your word document.
  2. Plot the histogram and Boxplot. Discuss if the distribution of the adjusted data is symmetric or skewed.
  3. Create the normal probability plot (Normality Plot from SPSS). Can we assume that the data are normally distributed?
  4. Use the normal approximation (Table or Excel method) defined by the sample average and standard deviation computed above and compute the probability that the adjusted index is larger than 0.75
  5. Compute the first and third quartiles using the normal approximation and compare them with the sample first and third quartiles (25th & 75th percentile) computed by SPSS. Write the answers in the Word document. You should also include the graphs with your answers. Just copy and paste the graphs in your word document.
dowj
0.25
0.26
-0.13
-0.19
-0.09
0.38
-0.1
0.1
0.41
0.45
0.29
0
0.06
0.23
0.48
-0.23
-0.25
-0.42
0.06
-0.15
-0.36
-0.67
0
-0.16
-0.51
-0.25
-0.1
-0.32
-0.29
0.13
0.1
0.28
0.32
-0.06
-0.58
-0.22
-0.48
-0.45
-1.21
-0.29
-0.41
-0.8
-0.54
-0.04
-0.44
-1.19
-0.44
-0.77
-0.51
0.09
0.42
-0.38
-0.48
-0.83
-0.16
-0.42
-0.41
-0.3
-1.53
-0.32
-0.03
0.77
0.35
0.16
-0.16
-0.09
0.06
0
0.22
-0.35
0.51
-0.06
-0.13
0.19
0.58
0.09
0.77
0 0
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Answer #1

1. Use SPSS to Compute the Mean, median, Q1 (or 25th percentile), Q3 (or 75th percentile), Standard deviation. Copy your table over to your word document.

Ans:

Statistics
dowj
N Valid 77
Missing 0
Mean -.1336
Median -.1300
Std. Deviation .42695
Percentiles 25 (Q1) -.4100
50 (median) -.1300
75 (Q3) .1450

2. Plot the histogram and Boxplot. Discuss if the distribution of the adjusted data is symmetric or skewed.

Ans:

20- Mean 0.13 Std. Dev. 0.427 N 77 10 -2.00 1.50 1.00 -0.50 0.00 0.50 1.00 dowj1.00 0.50 0.00 -0.50 1.00 1.50 59 -2.00 dowj

From the histogram and box-plot, we can conclude that the given data is slightly negative skew.

3. Create the normal probability plot (Normality Plot from SPSS). Can we assume that the data are normally distributed?

Ans:

Normal P-P Plot of dowj 1.0 0.8 E 0.6 O 0.4 0.2 0.0 0.6 0.8 1.C 0.0 0.2 0.4 Observed Cum Prob

Comment: From the above P-P plot, we can see that most of the values fall on or near to the straight line. Hence, we can assume that the data are normally distributed.

4. Use the normal approximation (Table or Excel method) defined by the sample average and standard deviation computed above and compute the probability that the adjusted index is larger than 0.75.

Ans: The number of the adjusted index (z-score) values which are more than 0.75 is 18. Hence, the probability that the adjusted index is larger than 0.75 is 18/77=0.2338.

5. Compute the first and third quartiles using the normal approximation and compare them with the sample first and third quartiles (25th & 75th percentile) computed by SPSS. Write the answers in the Word document. You should also include the graphs with your answers. Just copy and paste the graphs in your word document.

Ans:

(-01336))F0-25 o. 420 OL+0.1 336 , 420 (- > K4 0,1336 Shu K-(-0.1336 ) o 4 21 o e( 0.75 0-42 70 ラ > 2( t 0.1336 0.42700.6145The first and third quartiles using the normal approximation and them with the sample first and third quartiles (25th & 75th percentile) computed by SPSS are approximately equal.

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