Question 7Partially correct answer. Your answer is partially correct. Try again.An article in Knee Surgery, Sports Traumatology,
1 answer:
This question is incomplete, the complete question is;
An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273-279), considered arthroscopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters, 13 of 19 repairs were successful while for shorter tears, 22 of 30 repairs were successful.
a) Is there evidence that the success rate is greater for longer tears? Use alpha=0.05
Answer:
P-value = 0.6443
Now, since the p-value( 0.6443 ) is greater than the significance level (0.05), We fail to reject null hypothesis.
Therefore, there is no significant difference in the successive proportion ( rate ). ₁ =
₂
Step-by-step explanation:
Given the data in the question;
Sample of tears greater than 25;
number of successive repairs X₁ = 13
number of repair n₁ = 19
proportion of successive repairs ₁ = 13/19 = 0.68
q₁ = 1 - ₁ = 1 - 0.68 = 0.32
Sample of tears shorter than 25;
number of successive repairs X₂ = 22
number of repair n₂ = 30
proportion of successive repairs ₂ = 22/30 = 0.73
q₂ = 1 - ₂ = 1 - 0.73 = 0.27
so;
H₀ : There is no significant difference in the successive proportions; ₁ =
₂
Hₐ : The successive rate is greater for longer tears ₁ >
Significance level ∝ = 0.05
Test Statistics
Z = (₁ -
so we substitute
Z = ( 0.68 - 0.73 / √[ (0.68 ×0.32 / 19) + ( 0.73×0.27 / 30 ) ]
Z = -0.05 / √[ 0.01802 ]
Z = -0.05 / 0.1342
Z = -0.37
P-value for the given test, p = P( Z > -0.37 )
= 1 - P( Z ≤ - 0.37 )
= 1 - 0.35569
P-value = 0.6443
Now, since the p-value( 0.6443 ) is greater than the significance level (0.05), We fail to reject null hypothesis.
Therefore, there is no significant difference in the successive proportion ( rate ). ₁ =
₂
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Step-by-step explanation:
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a. 3I 1 3 -34 48
I'm going to do this one in its entirety so you get the idea of how to do it, then you'll be able to do it on your own.
First step is to bring down the first number after the bold line, 1.
3I 1 3 -34 48
_____________
1
then multiply it by the 3 and put it up under the 3. Add those together:
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3
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Now I'm going to multiply the 6 by the 3 after the bold line and add:
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1 6 -16
Same process, I'm going to multiply the -16 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18 -48
___________________
1 6 -16 0
That last zero tells me that x-3 is a factor of that polynomial, AND that the depressed polynomial is one degree lesser and those numbers there under that line represent the leading coefficients of the depressed polynomial:
Factoring that depressed polynomial will give you the remaining zeros. Because this was originally a third degree polynomial, there are 3 zeros as solutions. Factoring that depressed polynomial gives you the remaining zeros of x = -8 and x = 2
I am assuming that since you are doing synthetic division that you have already learned the quadratic formula. You could use that or just "regular" factoring would do the trick on all of them.
Do the remaining problems like that one; all of them come out to a 0 as the last "number" under the line.
You got this!
Answer:
C. Multiply 3 by 18
Step-by-step explanation:
Further explanation