Question

4. Comparing two population proportions (independent samples) 

Most major survey research organizations do not include wireless telephone numbers when conducting random-digit-dial telephone surveys. If there are differences between persons with and without landline phones, using a random-digit-dial telephone survey may introduce bias into the survey results. 

Data on a broad range of health topics are collected through personal household interviews of a representative sample of the U.S. civilian noninstitutionalized population in the National Health Interview Survey (NHIS) respondents are also asked about their telephones, the NHIS data allow you to compare the health status of persons with and without landline phones. 

Let pi denote the proportion of adults living in wireless phone-only homes who are obese and p2 denote the proportion of adults living in landline phone homes who are obese. Independent random samples are selected from the two populations. Sample 1, with sample size n - 515, provides a sample proportion p1 0.23. Sample 2, with sample size n2 760, provides a sample proportion p2 0.28. (The sample results are representative of the data collected in the NHIS, but the sample sizes are much smaller.) 

Use the Distributions tool to help you answer the questions that follow. 

The 95% confidence interval for p1-p2 is _______  to _______ . 

Based on the 95% confidence interval for p1-p2, you conclude that obesity is _______ prevalent among adults who live in wireless phone-only homes compared with adults who live in landline phone homes. Thus, a random-digit-dial health survey can be expected to _______ _______  the proportion of U.S adults who are obese.  

Conduct a hypothesis test at o .05 level of significance to determine whether obesity is less prevalent among adults living in wireless phone-only homes compared to adults living in landline phone homes. 

Under the assumption that the null hypothesis is true, the standard error of pi -p2 is _______ 

The test statistic is _______ .The P-value is _______ .Therefore, you  _______ conclude that obesity is less prevalent among adults living in wireless phone-only homes compared to adults living in landline phone homes.

Answer 1

sample #1   ----->  
first sample size,     n1=   515
proportion success of sample 1 , p̂1=   0.230
      
sample #2   ----->  
second sample size,     n2 =    760
proportion success of sample 1 , p̂ 2 =    0.280

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) =     0.0247          
margin of error , E = Z*SE =    1.960   *   0.0247   =   0.0484
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.050   -   0.0484   =   -0.0984
upper limit = (p̂1 - p̂2) + E =    -0.050   +   0.0484   =   -0.0016
                  
so, 95% confidence interval is (   -0.0984   < p1 - p2 <   -0.0016   )  

bases on 95% confidence for p1-p2, you conclude that obesity is less prevalent among adults who live in wireless phone only hones

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sample #1   ----->   wireless
first sample size,     n1=   515
number of successes, sample 1 =     x1=   118.45
proportion success of sample 1 , p̂1=   x1/n1=   0.230
      
sample #2   ----->   landline
second sample size,     n2 =    760
number of successes, sample 2 =     x2 =    212.8
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.280
      
difference in sample proportions, p̂1 - p̂2 =     0.230   -
      
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2598
      
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0250

Z-statistic = (p̂1 - p̂2)/SE = (   -0.050   /   0.0250   ) =   -1.9977
                  
p-value =        0.0229   [Excel function =NORMSDIST(z)      
decision :    p-value<α,Reject null hypothesis               
                  
Conclusion: therefore you can conclude that obsesity is less prevalent...................