Statistics Assignment: Hypothesis Testing and Statistical Analysis

This statistics assignment presents solutions to six hypothesis testing problems. The first problem involves testing a claim about a population mean using a z-test. The second problem uses an F-test to compare the variance of blood pressure between men and women. The third problem tests a claim about a production process using a z-test for proportions. The fourth problem compares the mean response times of two companies using a z-test. The fifth problem uses a t-test to compare the scores of gymnasts before and after training, and the final problem uses a t-test to test a claim about the mean lifetime of car engines. Each solution includes the null and alternative hypotheses, the test statistic, the p-value, and a final conclusion.

STATISTICS
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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

p value=0.0582
 Step 5
The null hypothesis would only be rejected when the p value is lower than the significance
level.
In present case, the p value comes out to be higher than the significance level and hence, null
hypothesis would not be rejected. Therefore, alternative hypothesis would not be accepted.
Therefore, it can be concluded that the production process is not out of control when defect
does not exceed 3%.
Question 4
Sample size = 50
Response time for two companies A and B
 Step 1
Null hypothesis H0 : μA =μBMean response time of company A is same as company B.
Alternative hypothesis Ha : μA ≠ μBMean response time of company A is not same as company
B.
 Step 2
Significance level = 0.01
 Step 3
The z stat would be computed as shown below.
z= 7.6−6.9
√ ( 1.42 )
50 + ( 1.72 )
50
=2.2 5

 Step 4
The p value for z = 2.25 and two tailed hypothesis testing
p value=0.02444 9
 Step 5
The null hypothesis would only be rejected when the p value is lower than the significance
level.
In present case, the p value comes out to be higher than the significance level and hence, null
hypothesis would not be rejected. Therefore, alternative hypothesis would not be accepted.
Therefore, it can be concluded that the mean response time of company A is same as
company B.
Question 5
Competitors scores before and after training
Sample size = 7 gymnasts
 Step 1
Null hypothesis H0 : μBefore=μAfterMean scores of gymnasts before and after training is same.
Alternative hypothesis Ha : μBefore < μAfterMean scores of gymnasts before is lower than after
training.
 Step 2
Significance level = 0.01
 Step 3
The t stat would be computed as shown below.

d =1
n ∑ d=−0.4
7 =−0.057
Standard deviation= √ 1
n−1 ∑ ¿ ¿ ¿ ¿
t=−0.057−0
0.1718
√7
=−0.8799
 Step 4
Critical value approach
Degree of freedom = 7-1 = 6
Significance level = 0.01
The given critical value = -3.143 (left tailed hypothesis test)
 Step 5
The null hypothesis would only be rejected when the t calculated is lower than the t critical
value.
In present case, the t calculated comes out to be higher than the t critical value and hence, null
hypothesis would not be rejected. Therefore, alternative hypothesis would not be accepted.
Therefore, it can be concluded that mean scores of gymnasts before and after training is
same. Thus, it is essential to improve the training program so as to make the gymnasts
competition score effective.
Question 6

The requisite hypotheses are as stated below.
Null Hypothesis: μ = 2,20,000 miles
Alternative Hypothesis: μ > 2,20,000 miles
Significance level = 0.01
Since the population standard deviation is unknown and also the sample size is less than 30,
hence the appropriate test statistic is T.
The formula for computation of t stat is shown below.
T =( x−μ ¿/ (s/n0.5)
Putting the respective input values, we get
Test statistics (T) = (226450-220000)/(11500/230.5) = 2.690
The t critical value for one tail with df = 23-1 = 22 and 0.01 significance level comes out as
2.508.
Since computed test statistic exceeds the t critical value, hence the available evidence would
lead to rejection of null hypothesis and acceptance of alternative hypothesis. Hence, it can be
concluded that the mean lifetime of car engines is greater than 2,20,000 miles.