Hypothesis Testing in Statistics

STATISTICS
[Document subtitle]
[DATE]

Question 1
The requisite hypotheses are as stated below.
Null Hypothesis: μ = 22o C
Alternative Hypothesis: μ ≠ 22o C
Level of significance = 0.05
As the population standard deviation is known and also the sample size is greater than 30,
hence in accordance with the Central Limit Theorem, the relevant test statistics would be Z
and not T.
The formula for computation of t stat is shown below.
Z=(x−μ ¿/ (σ /n0.5)
Putting the respective input values, we get
Z = (20-22)/(1.5/400.5) = -8.43
The requisite two tail p value for the above z value comes out to be zero.
Since the computed p value (0.00) is lower than the assumed level of significance (0.05),
hence the available evidence would lead to rejection of null hypothesis and acceptance of
alternative hypothesis. Hence, it can be concluded that the population mean temperature is
different from 22oC. As a result, the original claim is false.
Question 2
The requisite hypotheses are as stated below.
Null Hypothesis: sf2 ≤ sM2 i.e. the variation in blood pressure of females is lower than or equal
to variation in blood pressure of males
Alternative Hypothesis: sf2 > sM2 i.e. the variation in blood pressure of females is greater than
variation in blood pressure of males
Level of significance = 0.05

The relevant test statistic for the given case would be F value
Test statistic i.e. F value = sf2 / sM2 = 22.72/20.12 = 1.275
The critical value for F with df1 = 15, df2 = 16 and level of significance = 0.05 comes out as
2.35
Since computed test statistic is lower than F critical value, hence the available evidence
would not lead to rejection of null hypothesis and acceptance of alternative hypothesis. Thus,
it can be concluded that the variation in blood pressure of females is lower than or equal to
variation in blood pressure of males.
Question 3
 Step 1
Null hypothesis H0 : p ≤ 0.03Production process is not out of control when defect does not
exceed 3%.
Alternative hypothesis Ha : p >0.03Production process is out of control when defect does
exceed 3%.
 Step 2
Significance level = 0.01
 Step 3
The z stat would be computed as shown below.
Sample proportion P= 5.9%
Sample size = 85 items
z= P− p
√ pq
n
= 0.059−0.03
√ 0.03∗(1−0.03)
85
=1.5 7
 Step 4
The p value for z = 1.57 and right tailed hypothesis testing