A political scientist thinks that 50% of a population would favor a ban on smoki

Question

A political scientist thinks that 50% of a population would favor a ban on smoking in all public places. A survey was sent out to 200 randomly selected people in this population, and only 90 (45%) said they would favor the ban. Which of the following statements best characterizes this result?

a. The difference between the expected 50% and the observed 45% is probably the result of the random variability of the survey.

b. Getting only 45% in a sample of 200 who favor the ban is a good indication that less than 50% of the population favors the ban.

Explanation / Answer

We first get the z score for the critical value:          
          
x = critical value =    90.5      
u = mean = np =    100      
          
s = standard deviation = sqrt(np(1-p)) =    7.071067812      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.343502884      
          
Thus, the left tailed area is          
          
P(z <   -1.343502884   ) =    0.089554596

Hence, the probability of getting 90 or less is not rare (0.08955 > 0.05.)

Hence,

OPTION A: a. The difference between the expected 50% and the observed 45% is probably the result of the random variability of the survey. [ANSWER]

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