STAT101‐19S1 ASSIGNMENT 3.

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STAT101‐19S1 Assignment 3
Name
Institution

STAT101‐19S1 ASSIGNMENT 3 2
STAT101‐19S1 ASSIGNMENT 3
Question 1
Suppose that past experience suggests the following properties of these assessment items
(each out of 100%): E(X) = 61, sd(X) = 20, E(Y) = 72, sd(Y) = 22 and E(Z) =65, sd(Z) = 24.
a) Find the distribution parameters, E(T) and Var(T), for the total mark, T, where: T
= 0.47X+0.34Y+0.19Z.
Considering a random sample of variables X and Y; E(X+Y) =E(X) +E(Y).
and E(aX+b) =aE(X)+b.(in case there are constants)
Therefore; E(T) =E(0.47X+0.34Y+0.19Z)
E(T) =0.47E(X) +0.34E(Y)+0.19E(Z) But E(X) = 61, E(Y) = 72 and E(Z) =65
Therefore E(T) =0.47*61+0.34*72+0.19*65=65.5
For a random variable X and Y, Var (aX+b) =a2Var(X) and ThenVar(X+Y) =Var(X)
+Var(Y).
Therefore VAR(T) = VAR(0.47X+0.34Y+0.19Z) =0.47^2 VAR(X) + 0.34^2
VAR(Y) +0.19^2 VAR (Z)
VAR(T) =88.36+55.95+20.79=165.10
b) Assume the on‐line assessment, exam and assignment scores are normally
distributed. If the pass mark is 50%, calculate the probability that a randomly
selected student will pass.
P(X>50) =Z= X−μ
σ = (50-65.5)/√165.1=-1.206
P(Z<.1206)=0.1139
P(Z>.1206)=0.8861 Probability of getting 50 and above
c) Find the expected number of A+ grades (90% and above) to be awarded in July if
there are 840 students on the course this semester.
P(X>90) =Z= X−μ
σ = (90-65.5)/√165.1=1.906
P(Z<.1.906)=0.9717

STAT101‐19S1 ASSIGNMENT 3 3
P(Z>1.906)=0.02827 Probability of getting 90 and above
Expected number is 0.02827*840=23.75 approximately 24
d) If a random sample of 16 STAT101 students is selected, what is the probability that
their average grade is at least a B (that is, on average they get 70% or more in total)?
P(X>70) =Z= X−μ
σ = (70-65.5)/√165.1=0.3502
P(Z<.0.3502)=0.6369
P(Z>0.3502)=0.3631 Probability of getting 70 and above.
Question 2
e) Use the data to construct a 90% confidence interval for the population proportion
of first-year UC students who were working while studying at the start of Semester 1
2019.
Proportion is the ratio of the total number of successful events (m) out of the total
number of outcomes. (n) i,e proportion (p) = m / n
The Proportion of UC students working while studying at the start of Semester 1 2019
=355/627= 0.5662 (4.d.p).Newcombe (2012) outlines the confidence interval of the proportion
formula to be;
For a 90% confidence level the value of z score is 1.645
Considering that p=0.5662,the value of confidence interval is =0.5662±1.645
√ 0.5662(1−0.5662)
627
C.I =0.5662 ±1.645(0.01979)
Therefore Confidence Interval =0.5662±0.03256
Lower boundary =0.5336 and Upper boundary is 0.5987
f) Use the data to construct a 90% confidence interval for the population proportion
of first-year UC students who were working while studying at the start Semester 2
2018.

STAT101‐19S1 ASSIGNMENT 3 4
The Proportion of UC students working while studying at the start of Semester 2 2018
=250/510=0.4902
Considering that p=0.4902,the value of confidence interval is =0.4902±1.645
√ 0.4902(1−0.4902)
510
C.I =0.4902 ±1.645(0.02214)
Therefore Confidence Interval =0.4902±0.03641
Lower boundary =0.4538 and Upper boundary are 0.5266 . ( Altman, Machin, and
Gardner,2013).
g) Use your confidence intervals to comment on the suggestion that the proportion of
first-year UC students who were working while studying was higher at the start of
Semester 1 2019 than at the start of Semester 2 2018. Give your response in one or two
brief sentences.
The population proportion of first-year UC students who were working at the start of
Semester 1 2019 had an interval of between 53.4% to 59.9% compared to a proportion of 45.4%
to 52.7%. This shows that a higher proportion was working while studying in semester 1 than
semester 2.
Question 3
(i) Use the sample data and a significance level of 0.01 to carry out a hypothesis test to
test the administrator’s claim. Assume the number of beds filled per day is normally
distributed in the population.
# beds occupied per day
Mean 175
Standard Error 3.57
Standard Deviation 14.28
Sample Variance 204
Count 16
Hypothesis; Null Hypothesis is H0: At least 185 beds are occupied on any given day
and Alternative Hypothesis is H1: Less than 185 beds are occupied on any given day

STAT101‐19S1 ASSIGNMENT 3 5
Since the sample size n=16 is less than 30, the t-test was used to compute and test the
hypothesis.
The claim is that on average, at least 185 beds are occupied daily. (x=185)
Sample mean = X = ∑ X
N = 173+189+…+187
16 = 2800
16 =175
Standard deviation=√ ¿ ¿ =14.28
T-statistic formula is given by; t= x−μ
σ / √n = 185−175
14.28 /4 =2.801
The calculated t value is 3.571. Since n=16, the degrees of freedom, in this case, is n-
1=15
From the t table at alpha=.01 is 2.602. ( King'oriah, 2004).
Since the calculated t-test value is larger than the tabled value at alpha = .01, we, therefore, reject
the null hypothesis and accept the alternative hypothesis.
This, therefore, suggest that the administrator claim that at least 185 beds are occupied daily is
not true. Probably this is inflated
(ii) Explain, in words, the meaning of type I and a type II error in this situation.
Type 1 error is the probability of rejecting the null hypothesis when true (Rejecting a
correct hypothesis)
While type II is the probability of accepting a false hypothesis
For instance, in this case, it would be the likelihood of not adopting the administrator claim when
it is true (when actually the occupied beds are more than 185). Type II will occur when one
accepts that the occupied beds are more than 185 when they were less than the number. These
errors are caused by the setting of a narrow and wide confidence interval respectively.
(iii) Write down, in a few sentences, the consequences for the hospital for making
each of these two statistical errors.
Making a type I error will result in the management under planning. They will end up
budgeting for fewer patients while the actual number is higher hence causing congestion and
pressure on available staff. Making a type II error will result in the management over planning.
This is by allocating of more resources than the actual number required. This will lead to
wastage of resources.

STAT101‐19S1 ASSIGNMENT 3 6
(b) The Excel file Ass3_Q3b.xlsx contains data on the carpace lengths, in mm, of two
samples of Midland Painted Turtles (Chrysemys picta marginata). One sample is for female
turtles and the other sample is for male turtles.
(i) Use the sample data to carry out a hypothesis test to test the claim that the
variability of the carpace lengths is different for female Midland Painted Turtles
compared to male Midland Painted Turtles. Use a significance level of 0.05.
Your answer should include:
A statement of any assumptions required for the test.
 The data is involved is normally distributed.
 Data values from the turtles were randomly selected.
 The predictor and predicted variables are stochastic
A comment on the strength of the evidence on which the final decision and
conclusion are based.
Turtles N Mean Std. Deviation Variance
Females carpace
length (mm)
N1=24 X1=136.04mm (S1)=21.25 (S12)=451.52
Males carpace length
(mm)
N2=24 X2=113.38mm (S2)=11.78 (S22)=138.77
The hypothesis to test this is; Null Hypothesis H0: μ1- μ2 = 0 and Alternative Hypothesis is H1:
μ1- μ2 ≠ 0
The two sample t-test statistic is:
t=( X1− X2 )−( μ1−μ2 )
√ s p
2 (1
n1
+1
n2
)
d . f .=n1+ n2−2 since n < 30
Pooled Variance:
s p
2 =( n1−1) s1
2+(n2−1)s2
2
n1 +n2−2
= (24-1)*451.52 + (24-1)*138.77/ (24+24-2) = 295.14
t=22.67/4.96= 4.571; t computed value is 4.5717

STAT101‐19S1 ASSIGNMENT 3 7
Checking from t tables at 46 degrees of freedom and alpha=.05 t=1.645
At α=5%, t = 4.571>1.645, therefore we reject null hypothesis.
Using the t-test tables; At α=0.05, t = 2.028>1.645, therefore we reject the null hypothesis (Gupta
and Kapoor,2019).
Since the computed value is greater than the tabulated value, the null hypothesis is rejected
and the alternative hypothesis accepted. Therefore, the variability of the carpace lengths is
different for female Midland Painted Turtles compared to male Midland Painted Turtles
(ii) Based on your decision and conclusion from the test in part (i), explain in one or two
sentences how you would proceed if you wanted to carry out a hypothesis test to test the
claim that the mean of the carpace lengths is different for female Midland Painted Turtles
compared to male Midland Painted Turtles.(You are not required to complete this test but
you must make it clear which test you would use).
This can be done using an independent sample t-test or analysis of variance test.Statistically,
these both tests will be able to measure if there significant between the two means

STAT101‐19S1 ASSIGNMENT 3 8
References
Altman, D., Machin, D., Bryant, T. and Gardner, M. eds. (2013). Statistics with confidence:
confidence intervals and statistical guidelines. John Wiley & Sons.
Field, Andy. (2009), Discovering statistics using SPSS. 3rd ed. London: Sage Publications Ltd.
Gupta, S.C. and Kapoor, V.K. (2019), Fundamentals of applied statistics. Sulthan Chand &
Sons.
King'oriah, G. K.,(2004). Fundamentals of applied statistics. Nairobi: The Jomo Kenyatta
Foundation.
Newcombe, R.G.(2012). Confidence intervals for proportions and related measures of effect
size. CRC press.