Question

1) You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=20.5σ=20.5. You would like to be 90% confident that your esimate is within 10 of the true population mean. How large of a sample size is required?

n =

Use a critical value accurate to three decimal places, and do not round mid-calculation — this is important for the system to be able to give hints for incorrect answers.

2)Given ˆpp^ = 0.2286 and N = 35 for the high income group,

Test the claim that the proportion of children in the high income group that drew the nickel too large is smaller than 50%. Test at the 0.1 significance level.

a) Identify the correct alternative hypothesis:

• μ=.50μ=.50
• μ>.50μ>.50
• p<.50p<.50
• μ<.50μ<.50
• p=.50p=.50
• p>.50p>.50

Give all answers correct to 3 decimal places.

b) The test statistic value is:   

c) Using the P-value method, the P-value is:

d) Based on this, we

• Reject H0H0
• Fail to reject H0H0

e) Which means

• There is not sufficient evidence to support the claim
• The sample data supports the claim
• There is not sufficient evidence to warrant rejection of the claim
• There is sufficient evidence to warrant rejection of the claim

3) Links claims that at least half the bars of Ivory soap they produce are 99.44% pure (or more pure) as advertised. Unilever, one of Link's competitors, wishes to put this claim to the test. They sample the purity of 102 bars of Ivory soap. They find that 41 of them meet the 99.44% purity advertised.

Calculate the p-value.

Answer 1

1)

Standard Deviation ,   σ =    20.5                  
sampling error ,    E =   10                  
Confidence Level ,   CL=   90%                  
                          
alpha =   1-CL =   10%                  
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.645   *   20.5   /   10   ) ² =   11.370
                          
                          
So,Sample Size needed=       12          

..........

2)

Ho :   p =    0.5  
H1 :   p <   0.5   (Left tail test)
          
Level of Significance,   α =    0.10  
Number of Items of Interest,   x =   8.001  
Sample Size,   n =    35  
          
Sample Proportion ,    p̂ = x/n =    0.2286  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0845  
Z Test Statistic = ( p̂-p)/SE =    (0.2286-0.5)/0.0845=   -3.21  
          
critical z value =        -1.282   [excel function =NORMSINV(α)]
          
p-Value   =   0.0007   [excel function =NORMSDIST(z)]
Decision:   p-value<α ,

reject null hypothesis

• The sample data supports the claim

....

3)

  
        Ho :   p =    0.5  
H1 :   p <   0.5   (Left tail test)
             
Number of Items of Interest,   x =   41  
Sample Size,   n =    102  
          
Sample Proportion ,    p̂ = x/n =    0.4020  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0495  
Z Test Statistic = ( p̂-p)/SE =    (0.402-0.5)/0.0495=   -1.98  
          
          
p-Value   =   0.0238   [excel function =NORMSDIST(z)]

                          

............

Please let me know in case of any doubt.

Thanks in advance!

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