Question

7 .53 Surfing the Net Do you use the Internet to gather information for a project? A survey reports that the percentage of students who used the Internet as their major resource for a school project in a recent year was 66o/o. 16 Suppose that you take a sample of n = I 000 sn1dents, and record the number of students who used the Internet as their major resource for their school project during the past year. Let p be the proportion of students swveyed who used the Internet as a major resource in the past year. . How can you approximate the distribution of p? b. What is the probability that the sample proportion p exceeds 68%? c. What is the probability that the sample proportion lies between 64% and 68%? d. Would a sample proportion of 70% contradict the reported value of 66%?

Answer 1

a)

exact distribution of phat is binomial with n=1000,p=0.66

since, n=1000, is very large, sampling distribution can be aprroximated by a normal distribution with

mean of sampling distribution=   0.66
n=   1000
  
std dev of sampling distribution = √( p(1-p)/n ) =    0.0150

b)

population proportion ,p=   0.66      
n=   1000      
          
std error , SE = √( p(1-p)/n ) =    0.0150      
          
sample proportion , p̂ =   0.68      
Z=( p̂ - p )/SE=   1.335      
P ( p̂ >    0.68   ) =P(Z > ( p̂ - p )/SE) =  
          
=P(Z >   1.335   ) =    0.0909
c)

population proportion ,p=   0.66  
n=   1000  
      
std error , SE = √( p(1-p)/n ) =    0.0150  
      
we need to compute probability for       
0.64   < p̂ <   0.68
                       Z1 =( p̂1 - p )/SE=   -1.335          
                       Z2 =( p̂2 - p )/SE=   1.335          
P(   0.64   < p̂ <   0.68   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.335   < Z <   1.335   )
                                      
= P ( Z <   1.335   ) - P (    -1.335   ) =    0.9091   -   0.091   =   0.8182  

d)

population proportion ,p=   0.66      
n=   1000      
          
std error , SE = √( p(1-p)/n ) =    0.0150      
          
sample proportion , p̂ =   0.7      
Z=( p̂ - p )/SE=   2.670      
P ( p̂ >    0.7   ) =P(Z > ( p̂ - p )/SE) =  
          
=P(Z >   2.670   ) =    0.0038
the probability than sample proportion exceess 0.70 is 0.0038

so, a sample proportion of 70% contradict the reported value of 66%