INTRODUCTION TO BIOSTATISTICS Page 1 of 11 Introduction to Biostatistics 2020 Assignment 3 [5 questions in total] This assignment is due to be submitted by 4pm Monday 1st June XXXXXXXXXXAll...

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Answer To: INTRODUCTION TO BIOSTATISTICS Page 1 of 11 Introduction to Biostatistics 2020 Assignment 3 [5...

Aimy answered on May 21 2021
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Question 1 a)
The probability for P (t > 2.31) is 0.0189.
B)
The probability for P (t > -1.8727) is 0.94.
c)
The probability for P (t < -1.7247) is 0.95.
d)
The probability for P (Χ2 > 6.68) is 0.01.
e)
The probability for P (Χ2 > 9.50) is 0.237
2) Solution (A):
Given
,
Before treatment
=24
= 30.07
=6.10
After treatment
=24
=27.26
=5.20
The null and alternate hypothesis are as follows:
Here, the s1 > s2, but s1 < 2s2, we will use the case when population standard deviation of 1 and 2 are same.
The test statistic is calculated as:
t =
=
=
=1.899
Now, the test statistics is 1.899.
The t critical value can be found with alpha = 0.05 for two tail and (n1+n2-2) = (24+24-2) = 46 degree of freedom. Therefore, we use critical value is 2.014. As this is the 2-tail test, the critical value will be -2.014 for the left tail and 2.014 for the right tail.
We can see that; the test statistic lies between the 2 critical values i.e. -2.014 < 1.899 < 2.014.
Therefore, we will not reject the null hypothesis.
Solution (B):
We can see from part (A) that we do not reject the null hypothesis, i.e. the two means before and after the treatment are equal. There is insignificant difference in the mean score of before and after treatment.
3) Solution (A):
Given,
Intervention = 415
=0.56
= 0.87
Control
=414
=0.43
=0.73
The null and alternate hypothesis are as follows:
Here, the s1 > s2, but s1 < 2s2, we will use the case when population standard deviation of 1 and 2 are same.
The test statistic is calculated as:
Now, the test statistics is 0.0557.
The t critical value can be found with alpha = 0.05 for two tail and (n1+n2-2) = (415+414-2) = 827 degree of freedom. Therefore, we use critical value is 1.96. As this is the 2-tail test, the critical value will be -1.96 for the left tail and 1.96 for the right tail.
We can see that; the test statistic lies between the 2 critical values i.e. -1.96 < 0.0557 < 1.96.
Therefore, we will not reject the null hypothesis.
Solution (B):
We can see from part (A) that we do not reject the null hypothesis, i.e. the two means intervention and control are equal. There is insignificant difference in the mean score of intervention and control.
Solution (C):
If we use the one tail test, we now have to calculate the t critical value for 1 tail. The t critical value for 1 tail (0.05) and 827 degree of freedom is 1.646. Now, t statistics < t critical value. We do not reject the null hypothesis.
Here, both the results, either with one tail or, two tail are same. In both the cases, we do not reject the null hypothesis.
4) Solution (A):
H0 = There is no relationship between defendant’s ethnicity and sentencing to the death penalty.
H1 = There is a relationship between defendant’s...
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