Ribosomal 5S RNA can be represented as a sequence of 120 nucleotides. Each nucleotide can be represented by one of four characters: A (adenine), G (guanine), C (cytosine), or U (uracil). The characters occur with different probabilities for each position. We wish to test whether a new sequence is the same as ribosomal 5S RNA. For this purpose, we replicate the new sequence 100 times and find there are 60 A’s in the 20th position. Suppose we wish to test the hypothesis H0: µ = 45 vs. H1: µ > 45. Which of the following sample results yields the smallest p-value and why? (i) x = 28, s = 6 (ii) x = 27, s = 4 (iii) x = 32, s = 2 (iv) x = 26, s = 9 . Explain

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Ribosomal 5S RNA can be represented as a sequence of 120 nucleotides. Each nucleotide can be represented by one of four characters: A (adenine), G (guanine), C (cytosine), or U (uracil). The characters occur with different probabilities for each position. We wish to test whether a new sequence is the same as ribosomal 5S RNA. For this purpose, we replicate the new sequence 100 times and find there are 60 A’s in the 20th position.

Suppose we wish to test the hypothesis H0: µ = 45 vs. H1: µ > 45.

Which of the following sample results yields the smallest p-value and why?

(i) x = 28, s = 6

(ii) x = 27, s = 4

(iii) x = 32, s = 2 (iv) x = 26, s = 9 .

Explain

Expert Solution
Step 1

Here we have given n=100.

To test the hypothesis H0:µ=45  VS  H1: µ>45

The test statistic to find p-value is,

t=x-µS/n

Step 2

i) X =28, S =6, n=100

t=28-456/100 =-28.33          

p-value =p(t <-28.33) = 1    ....Using statistical Table

 

ii) X=27, S=4

t=27-454/100 =-45

p-value =p(t <-45) = 1    ....Using statistical Table

 

iii)X =32, S =2

t=32-452/100  =-65

p-value =p(t <-65) = 1    ....Using statistical Table

 iv) x=26, S =9

t=26-459/100 =-21.111

p-value =p(t <-21.11) = 1    ....Using statistical Table

 

 

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