STAT 381 Summer Semester: Statistical Inference and Hypothesis Testing

This document presents solutions to a STAT 381 homework assignment. The assignment covers several key statistical concepts, including the construction of a 95% confidence interval for a population mean using sample data. It also explores the binomial distribution, calculating probabilities for a scenario involving underweight milk containers. Furthermore, the solutions include hypothesis tests, comparing two population proportions to determine if a new drug is more effective than a standard drug, and a two-sample t-test to compare the sales of sports and action/adventure NES games. Each problem is approached with detailed explanations, formulas, and conclusions drawn in the context of the problems.

Name________________________________
STAT 381 – Summer
Must show work for partial credit. Even if you used technology to calculate something, give a formula or some
indication of what you did to calculate it.
1. The breaking strength of a random sample of 20 bundles of wool fibers have a sample mean of 436.5 and a
sample standard deviation of 11.90. A normal probability plot provides strong support for assuming that
the population distribution of breaking strength is at least approximately normal. Construct a 95%
confidence interval for the average breaking strength. Put your conclusion in the context of the problem.
(10
points)
Answer
Given n=20 , x =436.5 , σ =11.90;
Construct a 95% confidence interval for the average bvreaking strength.
Formula for 95% confidence interval is given by;
P ( X−
Z α
2
∗σ
√ n < μ< X +
Z α
2
∗σ
√ n )
¿ X ± Zα /2
σ
√n
Here we take alpha = 0.05 as the level of significance so the Z(alpha) = Z(0.05) = 1.96
So we have the confidence as follows;
¿ ( X ± Zα /2∗σ
√n )
¿ (436.5 ± 1.96∗11.90
√20 )
¿ (436.5−1.96∗11.90
√20 , 436.5+ 1.96∗11.90
√20 )
¿ ( 436.5−5.2154 , 436.5+5.2154 )
¿ ( 431.2846 , 441.7154 )
From the above calculations, we are 95% confident that the true population mean lies between 431.2846
and 441.7154.
Conclusion: We can conclude that if a sample lies in the interval (431.2846, 441.7154) then we accept the
null hypothesis.
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2. A milk container is supposed to contain 2 Liters of milk, but 10% of the time, it is underweight. The milk
containers are shipped to retail outlets in boxes of 20 containers. (10 points)
A. Let X be the number of underweight containers in a box. Explain why X is a binomial random variable.
Answer
Let x be the no. of underweight container.
The probability that a container is fixed for each container at 10% (p = 0.1).
No. of tails are fixed at 20 →n = 20.
Therefore, X has binomial distribution with parameters n=20 and p=0.1
B. What is the probability that exactly three of the twenty containers are underweight? (On this part,
write out the formula, even if you evaluate with technology).
Answer
P( Exactly three)=P( x=3)
P ( X=x ) = Cx
n ∗px∗(1−p)n− x
P ( X=3 )= Cx
n ∗px∗( 1− p )n−x= C3
20 ∗0.13∗( 1−0.3 )20−3=1140∗0.001∗0.1668
P ( X=3 )=0.1902
C. What is the probability that at most three of the twenty containers are underweight?
Answer
Probability that at most three of the 20 containers are underweight P(X = 3):
P ( Atmost three )=P ( x ≤ 3 ) =P ( X=0 )+ P ( X =1 ) + P ( x =2 )+ P ( X=3 )
P ( x ≤ 3 ) = [ C0
20 ∗0.10∗( 1−0.3 ) 20−0
] + [ C1
20 ∗0.11∗( 1−0.3 ) 20−1
] + [ C2
20 ∗0.12∗( 1−0.3 ) 20−2
] + [ C3
20 ∗0.13∗( 1−0.3 ) 20−3
]
P ( x ≤ 3 ) =0.1216+0.2702+ 0.2852+0.1901=0.8671
P ( x ≤ 3 )=0.8671
D. What is the expected value and variance of the number of containers that will be underweight?
Answer
Expected value E(x) is given as follows;
E ( X ) =np=20∗0.1=2
Variance value V ar ( x ) is given as follows;
Var ( X ) =np ( 1− p ) =20∗0.1∗(1−0.1)=1.8
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Name________________________________
3. A new drug is being compared to a standard drug for treating a particular illness. In the clinical trials, a
group of 200 patients was randomly split into two groups, with one group being given the standard
drug and one given the new drug. Altogether, 83 out of the 100 patients given the new drug improved
their condition, while only 72 out of the 100 patients given the standard drug improved their condition.
Conduct a hypothesis test with a level of significance of .05 to investigate whether the new drug is
better than the standard drug. Comment on appropriateness of the method (based on size) and put
conclusion in the context of the problem. (15 points)
A. Explain why the comparison of two population proportions is the correct method:
Answer
In order to compare two population proportions, the two samples to be tested should be randomly be
drawn and be independent of each other. Each of the sample should have at least 5 cases (sample
size). In this case, we had the two samples having 83 and 72 cases respectively and as such we can
conclude that method is correct.
B. Use the recommended sequence of steps:
1. Identify the parameter of interest and describe it in the context of the problem situation:
Answer
Parameter of interest are as follows;
 Population proportion of patients whose conditions improved , after giving new drug
(denoted by p1)
 Population proportion of patients whose conditions improved , after giving standard drug
(denoted by p2)
2. Determine the null value and state the null hypothesis:
Answer
The null value is zero for the difference between the two samples ( p1− p2=0)
The null hypothesis can be written as follows;
H0 : p1− p2=0
3. State the appropriate alternative hypothesis:
Answer
Alternative hypothesis is given as follows;
H A : p1− p2 >0
4. Give the formula for the computed value of the test statistic (substituting the null value and the
known values of any other parameters, but not those of any sample-based quantities):
Answer
Z= ( ^p1− ^p2 )−( p1− p2 )
√ pq ( 1
n1
+ 1
n2 )
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¿ ( ^p1− ^p2 ) −0
√ pq ( 1
n1
+ 1
n2 )
Where
p= 83+72
100+100 =0.775
q=1− p=1−0.775=0.225
5. Compute any necessary sample quantities, substitute into the formula for the test statistic value
and compute that value:
Answer
n1=100 ,n2=100 ,
^p1= 83
100 =0.83
^p2= 72
100 =0.72
Z= ( ^p1− ^p2 )
√ pq ( 1
n1
+ 1
n2 )
Z= ( 0.83−0.72 )
√0.775∗0.225 ( 1
100 + 1
100 )
¿ 1.86
6. Determine the P-value:
Answer
P-value = P(Z > 1.86) = 0.0314
7. Compare the selected or specified significance level to the P-value to decide whether H0 should be
rejected, and state this conclusion in the problem context:
Answer
α= 0,05
P-value < 0.05
We reject H0
There is sufficient evidence to conclude that new drug is better than the standard drug
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Name________________________________
4. We have a sample of NES games that are either sports games or action/adventures and consider North
American Sales (in millions of dollars). Is there evidence that action/adventure games sold better in
North American than sports games? Assume each sample shows evidence of a normal distribution.
(15
points)
A. Explain why the two-sample t test is the correct method for the following data:
Sports Action
Mean 0.649 1.541667
Std Dev 0.62417 1.082714
size 11 12
Answer
Since the sample sizes presented for the two samples are small and the population standard
deviations are unknown, we use a t-test. So the use of t-test is correct.
B. Use the recommended sequence of steps:
1. Identify the parameter of interest and describe it in the context of the problem situation:
Answer
The parameter of interest is the average sales for either sports games or action/adventures
2. Determine the null value and state the null hypothesis:
Answer
The null value is zero for the difference between the two samples (μ1−μ2=0)
The null hypothesis can be written as follows;
H0 : μ1−μ2=0
3. State the appropriate alternative hypothesis:
Answer
Alternative hypothesis is given as follows;
H A : μ1−μ2> 0
4. Give the formula for the computed value of the test statistic (substituting the null value and the
known values of any other parameters, but not those of any sample-based quantities):
Answer
t= ( μ1 −μ2 )
√ ( s1
2
n1
+ s2
2
n2 )
5. Compute any necessary sample quantities, substitute into the formula for the test statistic
value and compute that value:
Answer
n1=12, n2=11 ,
μ1=1.541667 , μ2=0.649; s1
2 =1.0827142 , s2
2 =0.624172
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t= ( μ1 −μ2 )
√ ( s1
2
n1
+ s2
2
n2 )= ( 1.541667−0.649 )
√ ( 1.0827142
12 + 0.624172
11 )=2.4468
6. Determine the P-value (NOTE: the calculation for degrees of freedom results in 17 df):
Answer
P(t > 2.4468) = 0.0128
7. Compare the selected or specified significance level to the P-value to decide whether H0 should
be rejected, and state this conclusion in the problem context:
Answer
We observe that the p-value , 0.05;
We therefore reject the null hypothesis at 5% level significance and we conclude that there is
sufficient evidence to believe that action games sold in North America are better than the
sports games sold in North America.
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