Question

According to a? study, 66?% of all males between the ages of 18 and 24 live at home. ? (Unmarried college students living in a dorm are counted as living at? home.) Suppose that a survey is administered and 168 of 227 respondents indicated that they live at home.? (a) Use the normal approximation to the binomial to approximate the probability that at least 168 respondents live at home.? (b) Do the results from part? (a) contradict the? study? ?

(a) ?P(X greater than or equals 168?)=

Answer 1

a&b)

Here we have to use the z test

The formula to calculate the z value is

p=168/227

q=1-p=1-168/227 = 59/227

z= (p-p_o)/sqrt(p*q/n) =(168/227-0.66)/sqrt((168/227*59/227)/(227)) =2.751225

So from the z table for 95% conf level the range is +/-1.96 (look for 2.5%area for let and right tail from z table and check the z vale for that) while the z value calculated 2.75 is not falling in the range provided here 2.75 is not in the range of -/+1.96

So here we can coclude that the results are contradicting the study

P(X>=168) = P(Z>=2.75) From the z table we look for the area under the curve from z=2.75 till z=inf and we get 0.002979763 so we say that 0.002979763 is the probability of X>=168

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