Question

A cell phone battery manufacturer claims that one of their batteries for a particular cell phone will outperform a competitor's equivalent brand. To establish this claim, a researcher selected samples of the two brands of batteries and perform accelerated tests on them in the lab under identical conditions. A random sample of 55 of the manufacturer's battery was selected and placed on test. A corresponding random sample of 55 of the competitor's battery was also put on test. The number of batteries lasting beyond 2000 hours (successes) and sample sizes are given in the following table. Manufacturer Competitor X2 = 44 n2= 55 = 41 n1 = 55 Step 1 of 2: Construct a 95 percent confidence interval for the difference in the proportions of batteries which lasted beyond 2000 hours for the manufacturer's brand relative to the competitor's

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Answer 1

1)

step 1)

sample #1   ----->   manufacture
first sample size,     n1=   55
number of successes, sample 1 =     x1=   41
proportion success of sample 1 , p̂1=   x1/n1=   0.7455
      
sample #2   ----->   competitor
second sample size,     n2 =    55
number of successes, sample 2 =     x2 =    44
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.8000

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0797          
margin of error , E = Z*SE =    1.960   *   0.0797   =   0.1563
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.055   -   0.1563   =   -0.2108
upper limit = (p̂1 - p̂2) + E =    -0.055   +   0.1563   =   0.1018
                  
so, confidence interval is (   -0.2108   < p1 - p2 <   0.1018   )  

step 2)

we are 95% confident that true difference in proportions of batteries which lasted beyond 2000 hours for manufacture's brand relative to competitor's lies within confidence interval

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