Question 888626
note that 262 / 359 = 0.729805014 or 73%
now 0.76 of 359 = 272.84 or 273 (expected value) vs 262 (observed value)
now 0.24 0f 359 =  86.16 or  86 (expected value vs 97 (observed value)
A) The proportion of college students who use only a cellular telephone (no land line) is 76%.
B) Level of significance is 10% or alpha = 0.10
C) Degree of freedom is 2 -1 = 1
x = ((262 - 273)^2 / 273)  + ((86 - 97)^2 / 97)
x = (-11)^2 / 273  + (-11)^2 / 97 =  0.443223443 + 1.24742268 = 1.690646123 = 1.69
our chi-square value is 1.69, now we consult the chi-square table to determine our p-value, using our degree of freedom of 1, we scan that row from left to right looking for a chi-square value greater than 1.69 and we find that value of 2.07 with a p-value of 15%
thus our p-value lies between 15% - 20%
D) This means  there's a 15-20% chance that the results we observed weren't the result of a change in location (instructor's students, as opposed to the entire nation), but instead just happened by chance. Since we were looking for a chance of less than 10%, we can't say that we're sure our instructor's students are less biased towards using cell phones only - there's a small but statistically significant chance they aren't.