power the versus the alternative H1 3.6 at a 0.05, what sample size would you re

Question

power the versus the alternative H1 3.6 at a 0.05, what sample size would you recommended? 19. DNA Random Walks. DNA random walks are numerical transcriptions of a of nucleotides. The imaginary walker starts at 0 and goes one step up (s +1) if a purine nucleotide (A, is encountered, and one step down (s if pyramidine nucleotide (CTis encountered. Peng et proposed identifying coding/noncoding regions by measuring the al. (1992) irregularity of associated DNA random walks. A standard irregularity mea- sure is the Hurst exponent H, an index that ranges from 0 to 1. Numerical sequences with H close to 0 are irregular, while the sequences with H close to 1 appear more smooth. Figure 9.3 shows a DNA random walk in the DNA of a spider monkey CAte- m a noncoding region and has a Hurst exponent of H 0.61. Fig. 9.3 A DNA random walk formed by a noncoding region from the DNA of a spider monkey. The Hurst exponent is 0.61. A researcher wishes to design an experiment in which n nonoverlapping DNA random walks of a fixed length will be constructed, with the goal of testing to see if the Hurst exponent for noncoding regions is 0.6. The researcher would like to develop a test so that an effect luo pilo will be detected with a probability of 1-B 0.9. The test should be two- sided with a significance level of 0.05. Previous analyses of noncoding regions in the DNA of various species suggest that exponent H is approx-

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu = 0.6 Vs H1 : mu not= 0.6

where mu is population mean.

Assume alpha = level of significance = 0.05

Here sample size = 24

Here we use one sample t-test to the data.

The test statistic is,

t = (Xbar - mu) / (s/sqrt(n))

where Xbar is sample mean

mu is population mean

s is sample standard deviation

n is sample size.

Rejection region approach is nothing but critical value approach.

This we can done in MINITAB.

Steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 1-Sample t --> Samples in columns : select data column --> Perform hypothesis test --> Hypothesized mean : 0.6 --> Options --> Confidence level :95.0 --> Alternative : not equal --> ok --> ok

One-Sample T: observation

Test of mu = 0.6 vs not = 0.6

Variable N Mean StDev SE Mean 95% CI T
observation 24 0.591667 0.029291 0.005979 (0.579298, 0.604035) -1.39

Variable P
observation 0.177

Test statistic = -1.39

P-value = 0.177

Also critical value we can find in EXCEL.

syntax :

=TINV(probability, deg_freedom)

where probability = 0.05

deg_freedom = n-1 = 24-1 = 23

Critical value = 2.069

P-value > alpha

Also |test statistic| < critical value

Accept H0 at 5% l.o.s.

Conclusion : The population mean is differ than 0.6

Now we have to find power of the test.

Power we can find by using MINITAB.

steps :

STAT --> Power and sample size --> 1-Sample t --> sample sizes : 24 --> differences : 0.02 --> standard deviation :0.0293 --> Options --> Alternative hypothesis : not equal --> ok --> ok

Power and Sample Size

1-Sample t Test

Testing mean = null (versus not = null)
Calculating power for mean = null + difference
Alpha = 0.05 Assumed standard deviation = 0.0293

Sample
Difference Size Power
0.02 24 0.892792

Power = 0.892792

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