Question

1. A random sample of size n = 10 is taken from a large population. Let μ be the unknown population mean. A test is planned of H0: μ=12vs. HA: μ̸=12usingα=0.1. A QQ plot indicates it is reasonable to assume a normal population. From the sample, x̄ = 14.2 and s = 4.88.

(I suggest doing this problem with a calculator and table as practice for exams. You may check your answers with R if you wish.)

(a) Since the data leave it plausible that the population is normal, and the population standard deviationσ is unknown, a t-test is appropriate. Compute the p-value of the test. Do you reject or not reject H0?

(b) Based on the test (and without calculating the interval), say whether you expect a 90% confidence interval to include 12.

(c) Using s = 4.88 as our best guess of σ (that is, pretending we know σ = 4.88), compute the power of a future test of H0: μ=12vs. HA: μ̸=12, if the true population mean, is μA =15.

(d) Using s = 4.88 as our best guess of σ (that is, pretending we know σ = 4.88), approximately what sample size would be required to achieve a power of 0.8 if the true population mean is μA = 15? Give your answer as the smallest whole number that meets the criterion.

Answer 1

1)

A)

Ho :   µ =   12
Ha :   µ ╪   12
      
Level of Significance ,    α =    0.1
sample std dev ,    s =    4.8800
Sample Size ,   n =    10
Sample Mean,    x̅ =   14.200
      
degree of freedom=   DF=n-1=   9
      
Standard Error , SE =   s/√n =   1.5432
      
t-test statistic=   (x̅ - µ )/SE =    1.4256
      
  
p-Value   =   0.1877
Conclusion:     p-value>α=0.10, Do not reject null hypothesis   

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b)

since, we do not reject null hypothesis at α=0.10, so it will contain the null hypothesis ,µ=12

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c)

true mean ,    µ =    15                          
                                  
hypothesis mean,   µo =    12                          
significance level,   α =    0.1                          
sample size,   n =   10                          
std dev,   σ =    4.88                          
                                  
δ=   µ - µo =    3                          
                                  
std error of mean,   σx = σ/√n =    1.5432                          
                                  
Zα/2   = ±   1.645   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.645   and   1.645
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-1.645   ≤(x̄ - µo)/σx≤   1.645                          
9.462   ≤ x̄ ≤   14.538                          
                                  
now, type II error is ,ß =        P (   9.462   ≤ x̄ ≤   14.538   )          
       Z =    (x̄-true mean)/σx                      
       Z1 =   -3.589                      
       Z2 =    -0.299                      
                                  
   so, P(   -3.589   ≤ Z ≤   -0.299   ) = P ( Z ≤   -0.299   ) - P ( Z ≤   -3.589   )
                                  
       =   0.382   -   0.000   =   0.3822      
                                  
power =    1 - ß =   0.6178                          
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d)

True mean,   µ =    15
hypothesis mean,   µo =    12
      
Level of Significance ,    α =    0.1
std dev =    σ =    4.88
power =    0.8

ß=1-power = 0.2
      
δ=   µ - µo =    3

      
Z (α/2)=       1.6449
      
Z (ß) =        0.8416
      
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    16.3593
      
so, sample size =        17