Question

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 11 of the plates have blistered (a) Does this provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of 0.05 Ho: p = 0.10 H: p0.10 Ho: p = 0.10 Ha p 0.10 Ho: p > 0.10 Ha: p-0.10 Ho: p = 0.10 Ha: p0.10 Calculate the test statistic and determine the p-value. (Round your test statistic to two decimal places and your p. value to four decimal places.) p-value State the conclusion in the problem context. | ⓔ Do not reject the null hypothesis. There is not sufficient evidence that more than 10% of plates blister under the experimental conditions. Reject the null hypothesis. There is sufficient evidence that more than 10% of plates blister under the experimental conditions. Reject the null hypothesis. There is not sufficient evidence that more than 10% of plates blister under the experimental conditions. Do not reject the null hypothesis. There is sufficient evidence that more than 10% of plates blister under the experimental conditions. In reaching your conclusion, what type of error might you have committed? type type II b If it is really the case that 16% of all plates blister under these circumstances and a sample size 100 is used how likely is it that the null hypothesis of part a will not be ejected by the 0.05 test? Round yo answer to four decimal places.) If it is really the case that 16% of all plate5 blister under these circumstances and a sample size 20 answer to four decimal places.) 5 used how like hypothess of part not be ree s t that the n wil he a ound your (c) How many plates would have to be tested to have (0.16) 0.10 for the test of part (a)? (Round your answer up to the next whole number.) plates

Answer 1

a)

Ho :   p =    0.1                  
H1 :   p >   0.1       (Right tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   11                  
Sample Size,   n =    100                  
                          
Sample Proportion ,    p̂ = x/n =    0.110                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0300                  
Z Test Statistic = ( p̂-p)/SE = (   0.1100   -   0.1   ) /   0.0300   =   0.33
                          
  
p-Value   =   0.3694 [Excel function =NORMSDIST(-z)              
Decision:   p value>α ,do not reject null hypothesis

b)

true mean ,    p=   0.16                          
                                  
hypothesis mean,   po=    0.1                          
significance level,   α =    0.05                          
sample size,   n =   100                          
                                  
                                  
                                  
                                  
std error of sampling distribution,   σx = po*(1-po)/√n =    0.1000   *   0.9   /√   100   =       0.0300

Zα =       1.6449   (right tailed test)                  
                              
We will fail to reject the null (commit a Type II error) if we get a Z statistic <               1.645              
this Z-critical value corresponds to X critical value( X critical), such that                              
                              
(x̄ - po)/σx ≤ Zα                              
x̄ ≤ Zα*σx + po                              
x̄ ≤    1.645   *   0.030   +   0.1   =   0.149   (acceptance region)
                              
now, type II error is ,ß =    P( x̄ ≤    0.149   given that µ =   0.16              
                              
   = P ( Z < (x̄ - p)/σx )                          
   = P ( Z < (   -0.355   )                  
   =   0.3612

------------------------

when n=200

std error of sampling distribution,   σx = po*(1-po)/√n =    0.1000   *   0.9   /√   200   =       0.0212

Zα =       1.6449   (right tailed test)                  
                              
We will fail to reject the null (commit a Type II error) if we get a Z statistic <               1.645              
this Z-critical value corresponds to X critical value( X critical), such that                              
                              
(x̄ - po)/σx ≤ Zα                              
x̄ ≤ Zα*σx + po                              
x̄ ≤    1.645   *   0.021   +   0.1   =   0.135   (acceptance region)
                              
now, type II error is ,ß =    P( x̄ ≤    0.135   given that µ =   0.16              
                              
   = P ( Z < (x̄ - p)/σx )                          
   = P ( Z < (   -1.184   )                  
   =   0.1183

c)

True mean,   p =    0.11  
hypothesis mean,   po =    0.10   
          
Level of Significance ,    α =    0.05  
ß=       0.16  
          
          
Z ( α ) =       1.645  
           [excel function: =normsinv(α) ]

Z (ß) =        0.994  
           [excel function: =normsinv(ß)

sample size needed =    n = p*(1-p)[Z(α) + Z(ß) ]² / (p-po)² = 0.11*(1-0.11)[1.645+0.994]²/(0.11-0.10)² = 6819.6800  
          
so, sample size =        6820