Question

1.

a. in a study, 68% of 905 randomly selected adults said they believe in the republicans favor the rich. If the margin of error was 4 percentage points, what is the confidence level used for the proportion? a federal report indicated that 26% of children under age 6 live in poverty in alabama, an increase over previous years. how large a sample is needed to estimate the true proportion of children under age 6 living in poverty in alabama within 5% with 99% confidence

n= _____

b. Obesity is defined as a body mass index (BMI) of 30 kg/m^2 or more. A 99% confidence interval for the percentage of U.S. men aged 70 years and over who were obese were found to be 20.0% to 21.0

n= ____ (it is NOT 43, 259)

c. it is believed that 25% of u.s. homes have direct satellite television receiver. 90% confidence within 2 percentage points?

n= _____

d. a recent study indicated that 23% of the 160 women over age 55 in the study were widows. how large a sample must you take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55

n=_____

e. in a recent year, a study found that 78% of adults ages 18-29 had internet access at home. Did research on 182 undergraduates total and found that 166 had access. Estimate the true proportion with 90% confidence.

_____ < p < ____

Answer 1

a)

margin of error= 0.04

Sample Proportion ,    p̂ = x/n =    0.6800
      
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.015506

margin of error , E = Z*SE

Zα/2=E/SE = 0.04/0.0155 = 2.576

α=0.01

confidence level used= 99%

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a)

sample proportion ,   p̂ =    0.26                          
sampling error ,    E =   0.05                          
Confidence Level ,   CL=   0.99                          
                                  
alpha =   1-CL =   0.01                          
Z value =    Zα/2 =    2.576   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   2.576   /   0.05   ) ² *   0.26   * ( 1 -   0.26   ) =    510.6216
                                  
                                  
so,Sample Size required=       511                          
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b)

confidence interval is                  
lower limit =    0.2       upper limit=   0.21  
sample proportion= (lower limit+upper limit)/2= (   0.2   +   0.21   ) / 2 =   0.205
margin of error = (upper limit-lower limit)/2= (   0.2   -   0.21   ) / 2 =   0.005

Confidence Level ,   CL=   0.99                          
                                  
alpha =   1-CL =   0.01                          
Z value =    Zα/2 =    2.576   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   2.576   /   0.005   ) ² *   0.21   * ( 1 -   0.21   ) =    43252.8909
                                  
                                  
so,Sample Size required=       43253                          

c)

sample proportion ,   p̂ =    0.25                          
sampling error ,    E =   0.02                          
Confidence Level ,   CL=   0.90                          
                                  
alpha =   1-CL =   0.1                          
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.645   /   0.02   ) ² *   0.25   * ( 1 -   0.25   ) =    1268.2235
                                  
                                  
so,Sample Size required=       1269                         

d)

sample proportion ,   p̂ =    0.23                          
sampling error ,    E =   0.05                          
Confidence Level ,   CL=   0.90                          
                                  
alpha =   1-CL =   0.1                          
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.645   /   0.05   ) ² *   0.23   * ( 1 -   0.23   ) =    191.6607
                                  
                                  
so,Sample Size required=       192                          

e)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   166          
Sample Size,   n =    182          
                  
Sample Proportion ,    p̂ = x/n =    0.9121          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.020990          
margin of error , E = Z*SE =    1.645   *   0.02099   =   0.0345
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.91209   -   0.03453   =   0.878
Interval Upper Limit = p̂ + E =   0.91209   +   0.03453   =   0.947
                  
90%   confidence interval is (   0.878   < p <    0.947   )

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