Smartphone adoption among American teens has increased substantially, and mobile

Question

Smartphone adoption among American teens has increased substantially, and mobile access to the Internet is pervasive. One in four teenagers are "cell mostly" Internet users-that is, they mostly go online using their phone and not using some other device such as a desktop or laptop computer. (Source: Teens and Technology 2013, Pew Research Center, bitly/1O1ciF1.) If a sample of 10 American teens is selected, what is the probability that 4 are "cell mostly" Internet users? at least 4 are "cell mostly" Internet users? at most 8 are "cell mostly" Internet users? If you selected the sample in a particular geographical area and found that none of the 10 respondents are "cell mostly" Internet users, what conclusions might you reach about whether the percentage of "cell mostly" Internet users in this area was 25%?

Explanation / Answer

a)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.25      
x = the number of successes =    4      
          
Thus, the probability is          
          
P (    4   ) =    0.145998001 [ANSWER]

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b)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.25      
x = our critical value of successes =    4      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   3   ) =    0.775875092
          
Thus, the probability of at least   4   successes is  
          
P(at least   4   ) =    0.224124908 [ANSWER]

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c)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.25      
x = the maximum number of successes =    8      
          
Then the cumulative probability is          
          
P(at most   8   ) =    0.999970436 [ANSWER]

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d)

Getting the probability of 0 out of 10 are cell mostly users,

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.25      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.056313515

We see that this only happens 5.6% of the time. If at 10% significance, we might conclude that the proportion of "cell mostly" internet users in that area is not 0.25.

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