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Using Nonparametric Tests. In Exercises 1–10, use a 0.05 significance level with the indicated test. If no particular test is specified, use the appropriate nonparametric test from this chapter.

Old Faithful Listed below are time intervals (min) between eruptions of the Old Faithful geyser. The “recent” times are within the past few years, and the “past” times are from 1995. Test the claim that the two samples are from populations with the same median. Does the conclusion change with a 0.01 significance level?

Recent

78

91

89

79

57

100

62

87

70

88

82

83

56

81

74

102

61

Past(1995)

89

88

97

98

64

85

85

96

87

95

90

95






Short Answer

Expert verified

At 0.05 level of significance, there is enough evidence to warrant rejection of the claim that two samples come from populations with equal medians.

At 0.01 level of significance, there is not enough evidence to warrant rejection of the claim that two samples come from populations with equal medians.

Yes, the conclusion of the test changes with a 0.01 level of significance.

Step by step solution

01

Given information

Two samples are given showing the time intervals between eruptions of the Old Faithful geyser.

02

Appropriate test

Since the 2 samples are independent, the Wilcoxon rank-sum test is used to claim that the two samples come from populations with the same median.

03

Widentify the hypothesis

The null hypothesis is as follows:

The two samples come from populations with equal medians.

The alternative hypothesis is as follows:

The median of the first population is different from the second population.

04

Assign ranks

Combine the two samples and write the sample number against each value.

Compute the ranks of all the values by assigning a rank of 1 to the lowest value, 2 to the next lowest value and so on until the greatest value.

If some values are equal, assign the mean of the ranks to all those values.

The following table shows the signed-ranks:

Time Interval

Sample Number

Ranks

78

Sample 1

8

91

Sample 1

22

89

Sample 1

19.5

79

Sample 1

9

57

Sample 1

2

100

Sample 1

28

62

Sample 1

4

87

Sample 1

15.5

70

Sample 1

6

88

Sample 1

17.5

82

Sample 1

11

83

Sample 1

12

56

Sample 1

1

81

Sample 1

10

74

Sample 1

7

102

Sample 1

29

61

Sample 1

3

89

Sample 2

19.5

88

Sample 2

17.5

97

Sample 2

26

98

Sample 2

27

64

Sample 2

5

85

Sample 2

13.5

85

Sample 2

13.5

96

Sample 2

25

87

Sample 2

15.5

95

Sample 2

23.5

90

Sample 2

21

95

Sample 2

23.5

05

Calculate the test statistic

The sum of the ranks corresponding to sample 1 is computed below:

\(\begin{array}{c}R = 8 + 22 + .... + 3\\ = 204.5\end{array}\)

The mean value of R is computed using the given formula:

\(\begin{array}{c}{\mu _R} = \frac{{{n_1}\left( {{n_1} + {n_2} + 1} \right)}}{2}\\ = \frac{{17\left( {17 + 12 + 1} \right)}}{2}\\ = 255\end{array}\)

The standard deviation of R is computed using the given formula:

\(\begin{array}{c}{\sigma _R} = \sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} \\ = \sqrt {\frac{{17\left( {12} \right)\left( {17 + 12 + 1} \right)}}{{12}}} \\ = 22.58\end{array}\)

The test statistic is computed as shown below:

\(\begin{array}{c}z = \frac{{R - {\mu _R}}}{{{\sigma _R}}}\\ = \frac{{204.5 - 255}}{{22.58}}\\ = - 2.24\end{array}\)

06

Determine the critical value at the significancelevel 0.05 and the conclusion of the test

The critical value of Z at\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Since the absolute value of the test statistic is greater than the critical value, the null hypothesis is rejected.

There is enough evidence to warrant rejection of the claim that two samples comes from populations with equal medians.

07

Determine the critical value at the significancelevel 0.01 and the conclusion of the test

The critical value of Z at\(\alpha = 0.01\)for a two-tailed test is equal to 2.575.

Since the absolute value of the test statistic is less than the critical value, the null hypothesis is failed to reject.

Thus, there is not enough evidence to warrant rejection of the claim that two samples comes from populations with equal medians at 1% level of significance.

It can be said that the conclusion of the test changes if the level of significancechanges from 0.05 to 0.01.

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Odd and Even Digits in Pi A New York Times article about the calculation of decimal places of\(\pi \)noted that “mathematicians are pretty sure that the digits of\(\pi \)are indistinguishable from any random sequence.” Given below are the first 25 decimal places of\(\pi \). Test for randomness in the way that odd (O) and even (E) digits occur in the sequence. Based on the result, does the statement from the New York Times appear to be accurate?

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3

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Foot Length , Height Listed below are foot lengths (cm) and heights (cm) of males from Data Set 2 “Foot and Height” in Appendix B. Which method of nonparametric statistics should be used? What characteristic of the data is investigated with this test?

Foot Length 27.8 25.7 26.7 25.9 26.4 29.2 26.8 28.1 25.4 27.9

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Using Nonparametric Tests. In Exercises 1–10, use a 0.05 significance level with the indicated test. If no particular test is specified, use the appropriate nonparametric test from this chapter.

California Lottery Listed below are consecutive first-digits drawn in the California Daily 4 lottery. Test for randomness of even and odd integers. Does the lottery appear to be working as it should?

5

2

2

8

4

8

8

7

1

0

6

4

1

5

1

5

5

3

1

4

1

5

0

0

3

9

6

6

3

7





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