401077 Introduction to Biostatistics, Autumn 2019
Assignment 2
Due Sunday September 15, 2019
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software for the purpose of detecting possible plagiarism. This software may retain
a copy of this assignment on its database for future plagiarism detection.
I understand that failure to uphold this declaration may result in academic
proceedings in line with the UWS Student Academic Misconduct Policy.”
Your name: Zaiba Dabhoiwala
Your student number: 19773155
Question 1
a) As per the data set x=32.12, s=4.14 , n=233. The sample size here is large enough
(n>30) to assume that it meets the Central Limit Theorem (CLT).
s
From Sullivan Table 6.3 the formula is x ± z
√n
From Sullivan Appendix Table 1B setting z=1.96 gives a 95% confidence interval (as it meets
the CLT)
s 4.14
Therefore the 95% confidence interval is x ± z = 32.12 ±1.96 =32.12± 0.27
√n √233
Hence the Upper Limit (UL) = 32.39 and Lower Limit (LL) = 31.85 of 95% Confidence Interval.
We can say that we are 95% sure that the mean grip strength of the grandparent carers in
Parramatta is 32.12 kgs with more or less 0.27 kgs.
, b) Yes. The difference in means is statistically significant because the 95% confidence
interval does not include the possibility of equality (zero difference) between group
means.
c) As the sample sizes are both groups are >30 we do not need the data to be Normally
distributed for the Central Limit Theorem to apply. Further there is less than a two-
fold difference in the standard deviations. Therefore we can use the formula in
Sullivan Section 7.5.
Let us denote that the group of male is Group 1 and that of female is group 2. We know
from the data that x 1=31.641, s1=4.151 and n1=128 and that x 2=32.709, s2=4.069
and n2 =105.
Step 1: Set up hypotheses and determine the level of significance
Let’s take H 0=mean grip strength does not differ between males∧females .
H 0 : μ1=μ2
H 1=mean grip strengthdiffers between males∧females .
H 1 : μ 1 ≠ μ2
Or
H 0 : μ1−μ2=0 and H 1 : μ 1−μ2 ≠ 0
And
α =0.05
Step 2: Select the appropriate test statistic
From Sullivan Section 7.5 the appropriate formula given when both samples are larger than
30, is:
x 1−x 2
z=
S p √1/n1 +1/n2
Where,
S p=
√ ( n1 −1 ) s12+ ( n21−1 ) s 22
n1 +n2−2
Step 3: Set up the decision rule
This is a two-tailed test, using a z-statistics at the 5% significance level. The appropriate
critical values are found in Sullivan Appendix Table 1C (or using R Commander). This gives a
decision rule of:
Assignment 2
Due Sunday September 15, 2019
“When submitting your assignment to Turnitin you are implicitly ticking these statements:
I retain a backup file of this assignment in case the original file is lost or damaged.
I hereby certify that no part of this assignment or product has been copied from any
other student’s workor from any other source except where due acknowledgement
is made in the assignment.
I hereby certify that no part of this assignment or product has been submitted by me
in another (previous or current) assessment.
I hereby certify that no part of the assignment has been written or produced by any
person.
I hereby certify that no part of this assignment has been made available to any other
student.
I am aware that this work will be reproduced and submitted to plagiarism detection
software for the purpose of detecting possible plagiarism. This software may retain
a copy of this assignment on its database for future plagiarism detection.
I understand that failure to uphold this declaration may result in academic
proceedings in line with the UWS Student Academic Misconduct Policy.”
Your name: Zaiba Dabhoiwala
Your student number: 19773155
Question 1
a) As per the data set x=32.12, s=4.14 , n=233. The sample size here is large enough
(n>30) to assume that it meets the Central Limit Theorem (CLT).
s
From Sullivan Table 6.3 the formula is x ± z
√n
From Sullivan Appendix Table 1B setting z=1.96 gives a 95% confidence interval (as it meets
the CLT)
s 4.14
Therefore the 95% confidence interval is x ± z = 32.12 ±1.96 =32.12± 0.27
√n √233
Hence the Upper Limit (UL) = 32.39 and Lower Limit (LL) = 31.85 of 95% Confidence Interval.
We can say that we are 95% sure that the mean grip strength of the grandparent carers in
Parramatta is 32.12 kgs with more or less 0.27 kgs.
, b) Yes. The difference in means is statistically significant because the 95% confidence
interval does not include the possibility of equality (zero difference) between group
means.
c) As the sample sizes are both groups are >30 we do not need the data to be Normally
distributed for the Central Limit Theorem to apply. Further there is less than a two-
fold difference in the standard deviations. Therefore we can use the formula in
Sullivan Section 7.5.
Let us denote that the group of male is Group 1 and that of female is group 2. We know
from the data that x 1=31.641, s1=4.151 and n1=128 and that x 2=32.709, s2=4.069
and n2 =105.
Step 1: Set up hypotheses and determine the level of significance
Let’s take H 0=mean grip strength does not differ between males∧females .
H 0 : μ1=μ2
H 1=mean grip strengthdiffers between males∧females .
H 1 : μ 1 ≠ μ2
Or
H 0 : μ1−μ2=0 and H 1 : μ 1−μ2 ≠ 0
And
α =0.05
Step 2: Select the appropriate test statistic
From Sullivan Section 7.5 the appropriate formula given when both samples are larger than
30, is:
x 1−x 2
z=
S p √1/n1 +1/n2
Where,
S p=
√ ( n1 −1 ) s12+ ( n21−1 ) s 22
n1 +n2−2
Step 3: Set up the decision rule
This is a two-tailed test, using a z-statistics at the 5% significance level. The appropriate
critical values are found in Sullivan Appendix Table 1C (or using R Commander). This gives a
decision rule of:
