Randomness Refer to the following ages at inauguration of the elected presidents of the United States (from Data Set 15 “Presidents” in Appendix B). Test for randomness above and below the mean. Do the results suggest an upward trend or a downward trend?

57

61

57

57

58

57

61

54

68

49

64

48

65

52

46

54

49

47

55

54

42

51

56

55

51

54

51

60

62

43

55

56

52

69

64

46

54

47

















Short Answer

Expert verified

There is insufficient evidence to conclude that the age distribution is random.

Since the null hypothesis was not rejected, it can be concluded that the given values do not indicate any trend.

Step by step solution

01

Given information

The ages of the inauguration of the elected presidents are provided.

The size of the sample (n) is 38.

02

Identify the hypothesis of the test

The researcher wants to test for randomness above and below the mean.

The null hypothesis is as follows:

The given sample of ages is random.

The alternative hypothesis is as follows:

The given sample of ages is not random.

03

Calculate the mean

The mean value of the sample is computed below:

\(\begin{array}{c}\bar x = \frac{{\sum {{x_i}} }}{n}\\ = \frac{{57 + 61 + .... + 47}}{{38}}\\ = 54.76\\ \approx 54.8\end{array}\)

04

Data Transformation

Suppose the values above the mean are denoted by A.

Suppose the values below the mean are denoted by B.

The following table shows the values along with their symbols:

Age

Symbol

Age

Symbol

Age

Symbol

57

A

54

B

55

A

61

A

49

B

56

A

57

A

47

B

52

B

57

A

55

A

69

A

58

A

54

B

64

A

57

A

42

B

46

B

61

A

51

B

54

B

54

B

56

A

47

B

68

A

55

A



49

B

51

B



64

A

54

B



48

B

51

B



65

A

60

A



52

B

62

A



46

B

43

B



05

Calculate the test statistic

The sequence is as follows:

A

A

A

A

A

A

A

B

A

B

A

B

A

B

B

B

B

B

A

B

B

B

A

A

B

B

B

A

A

B

A

A

B

A

A

B

B

B








Let the number of times A occurs be denoted by\({n_1}\).

Let the number of times B occurs be denoted by\({n_2}\).

Here,

\(\begin{array}{l}{n_1} = 19\\{n_2} = 19\end{array}\)

The runs are computed below:

\(\underbrace {AAAAAAA}_{{1^{st}}run}\underbrace B_{{2^{nd}}run}\underbrace A_{{3^{rd}}run}\underbrace B_{{4^{th}}run}\underbrace A_{{5^{th}}run}\underbrace B_{{6^{th}}run}\underbrace A_{{7^{th}}run}\underbrace {BBBBB}_{{8^{th}}run}\underbrace A_{{9^{th}}run}\underbrace {BBB}_{{{10}^{th}}run}\underbrace {AA}_{{{11}^{th}}run}\underbrace {BBB}_{{{12}^{th}}run}\underbrace {AA}_{{{13}^{th}}run}\underbrace B_{{{14}^{th}}run}\underbrace {AA}_{{{15}^{th}}run}\underbrace B_{{{16}^{th}}run}\underbrace {AA}_{{{17}^{th}}run}\underbrace {BBB}_{{{18}^{th}}run}\)

The number of runs (G) is equal to 18.

Since\({n_1} \le 20\)and\({n_2} \le 20\), the value of the test statistic is denoted by G and has a value of 18.

The critical values of G corresponding to\({n_1} = 19\)and \({n_2} = 19\) are 13 and 27.

Since the value of the test statistic is neither less than the smaller critical value of 13 and nor greater than the larger critical value, the null hypothesis fails to reject.

06

Conclusion of the test

There is not enough evidence to conclude that the given sample of ages is random.

Since the null hypothesisfails to reject, no trend is depicted by the given values.

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Arkansas

4.8

4.9

5

5.4

5.4

5.4

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California

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5.4

5.4

5.5

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5.6

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5.6

5.8

6.6

6.9

6.9

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7.1

7.3

7.5

7.6

7.7

7.7

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