Question

QUESTION ONE:

A company has set a goal of developing a rechargeable battery that lasts over 5 hours (300 minutes) in continuous use,
on average. A random sample of 10 of these batteries measured the following lifespans (in minutes): 319, 294, 336, 353,
341, 329, 315, 329, 302, and 289.

(a) Display the sample data in a stemplot and describe the distribution.

(b) Is there convincing evidence that the company has met its goal? Provide statistical evidence to support your answer.

QUESTION TWO:

Researchers conducted a study to test a potential side effect of a new allergy medication. 200 subjects with allergies
were used for the study. The new “improved” Brand A medication was randomly assigned to 100 subjects, and the
current Brand B medication was randomly assigned to the other 100 subjects. 13 of the 100 patients with Brand A
reported drowsiness, and 23 of the 100 patients with Brand B reported drowsiness.
(a) Construct and interpret a 90% confidence interval for the difference in proportions of subjects reporting drowsiness.

(b) Describe in detail a method that could have been used to randomly assign the Brand A and Brand B medications to
the subjects.

Answer 1

A)

Frequency	Stem	Leaf
1	28	9
1	29	4
1	30	2
2	31	5 9
2	32	9 9
1	33	6
1	34	1
1	35	3
10

Distribution seems to be normal

b)

Ho :   µ =   300                  
Ha :   µ >   300       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   20.8968                  
Sample Size ,   n =    10                  
Sample Mean,    x̅ = ΣX/n =    320.7000                  
                          
degree of freedom=   DF=n-1=   9                  
                          
Standard Error , SE = s/√n =   20.8968   / √    10   =   6.6082      
t-test statistic= (x̅ - µ )/SE = (   320.700   -   300   ) /    6.6082   =   3.13
                          
  
                          
p-Value   =   0.0060   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence that battery lasts over 5 hour.

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a)

level of significance, α =   0.10              
Z critical value =   Z α/2 =    1.645   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0539          
margin of error , E = Z*SE =    1.645   *   0.0539   =   0.0886
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.100   -   0.0886   =   -0.1886
upper limit = (p̂1 - p̂2) + E =    -0.100   +   0.0886   =   -0.0114
                  
so, confidence interval is (   -0.1886   < p1 - p2 <   -0.0114   )  

b)

We can use SRM simple random sampling

THANKS

revert back for doubt

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