Stat Analysis Case Study Assignment

STAT ANALYSIS CASESTUDY
ASSIGNMENT 3
Student id
[Pick the date]

Case 1
(a) i) For the given hypothesis testing, it would be appropriate to use a z test statistic instead
of a t test statistic. This is because in the given case, the given binomial distribution can
be approximated as normal distribution owing to the satisfaction of the underlying
assumptions.
Sample size (n)=200
Observed proportion ( p )= 165
200 =0.825
Hence, np>10∧also np (1− p)> 10
Considering the above, it would be appropriate to assume the given sample as normal and
hence z test would be the appropriate choice.
ii) This would be a test for population proportion as the given data relates to the proportion
of the students passed and not the average students that pass. Also, the mean would be
subject to change due to count of students graduating and hence proportion would be more
suitable as it expresses pass performance in relative terms and not absolute terms.
iii) The appropriate formula for computation of z test statistic is given below.
z sample statistics= ^p− p0
√ p0 ( 1− p0 )
n
Here,
^p=sample proportion=0.82 5
p0=hypothesize population proportion=0.77
n=sample ¿ 200
z= 0.825−0.77
√ 0.77 ( 1−0.77 )
200
z=1.848
Hence, the value of sample test statistics comes out to be 1.848.
(b) Hypothesis testing
Step 1: Hypotheses

Null hypothesis H0 : Percentage of graduates is equal to 0.77. p=0.77
Alternative hypothesis H1 :Percentage of graduates is not equal to 0.77. p ≠ 0.77
Step 3: Test statistic and significance level
Level of significance = 0.01
Test statistic
z= 0.825−0.77
√ 0.77 ( 1−0.77 )
200
→ z =1.848
Step 3: Decision rule
According to the decision rule, null hypothesis would be rejected when the p value is lower
than the level of significance. Similarly, as a result alternative hypothesis would be accepted.
Step 4: p value and decision
The p value for z statistic 1.848 comes out to be 0.9677.
It can be seen that the p value is 0.9677 which is higher than the level of significance (0.01).
Hence, insufficient evidence presents to reject the null hypothesis. As a result, alternative
hypothesis would not be accepted.
Step 5: interpretation of the decision
It would be fair to conclude that null hypothesis would not be rejected and thus, percentage of
graduates is equal to 0.77.
(c) 95% confidence interval for population proportion
The z value for 95% confidence interval = 1.96
Confidence interval ¿ ^p ± z √ ^p ( 1− ^p )
n
Upper limit of 95% confidence interval ¿ ^p+z √ ^p (1− ^p )
n
¿ 0.825+1.96 √ 0.825 ( 1−0.825 )
200
¿ 0.877

Lower limit of 95% confidence interval ¿ ^p−z √ ^p ( 1− ^p )
n
¿ 0.825−1.96 √ 0.825 ( 1−0.825 )
200
¿ 0.772
Therefore, the 95% confidence interval is [0.7720.877 ].
Case 2
(a) The relevant data is shown below:
(b)

(i) A t test would be more appropriate in the given case as the population standard deviation
is not known and also the sample size chosen is less than 30. A z test would have been
preferable if the sample standard deviation was known and the size of the sample would
have been at least 30.
(ii) The given test would be on mean and not the population proportion as there exists a
particular value of hypothesised mean from which the comparison of the sample data needs to
be drawn. Further, in the given case, the outcome is not dependent on the proportion of
students who study for desired hours but rather their average hours of study.
(iii) The appropriate formula for computation of z test statistic is given below.
t sample statistics= x−μ
s
√n
x=sample mean=8
μ= population mean=5
s=Sample standard deviation=2.723
Now,
t= 8−5
2.723
√ 25
t=5.50 7
Hence, the value of sample test statistics comes out to be 5.50 7.
(iv) Hypothesis testing
Step 1: Hypotheses
Null hypothesis H0 : μ ≤5 hours
Alternative hypothesis H1 : μ> 5 hours
Step 3: Test statistic and significance level
Level of significance = 0.01 (assumption)
Degree of freedom¿ n−1=25−1=24
Test statistic

t= 8−5
2.723
√ 25
→ t =5.507
Step 3: Decision rule
According to the decision rule, null hypothesis would be rejected when the p value is lower
than the level of significance. Similarly, as a result alternative hypothesis would be accepted.
Step 4: p value and decision
The p value for t statistic 5.507 and 24 degree of freedom and one tailed test comes out to be
0.00001.
It can be seen that the p value is 0.00001 which is lower than the level of significance (0.01).
Hence, sufficient evidence presents to reject the null hypothesis. As a result, alternative
hypothesis would be accepted.
Step 5: interpretation of the decision
It would be fair to conclude that null hypothesis would be rejected and thus
, μ>5 hours . Therefore , the claim that students who study more than 5 hours per week would
achieve an above average grade on the subject.
(j) 95% confidence interval for population proportion
The t value = 5.50 7
Confidence interval ¿ x ± t ( s
√n )
Upper limit of 95% confidence interval ¿ x +t ( s
√ n )
¿ x +t ( s
√n )=8+ (5.507∗2.723
√25 )=11
Lower limit of 95% confidence interval ¿ x−t ( s
√n )
¿ x−t ( s
√n )=8+
( 5.507∗2.723
√25 )=5
Therefore, the 95% confidence interval is [511].