Question

A)In our sample of 204 students, 27 have September birthdays. Calculate a 90% confidence interval for the proportion of 332 students that have a September birthday.

B) Interpret the interval from part b.

C) If the birth months are equally likely (ignoring the difference in the number of days), what proportion of births would be in September?

D) For various reasons, September is considered to be the most common birth month. Using your confidence interval from part A and response from part B, explain if there is evidence to support that the proportion of September birthdays is greater than we would expect by chance.

Answer 1

(A)

n = 204    

p = 0.132352941    

% = 90    

Standard Error, SE = √{p(1 - p)/n} =    √(0.132352941176471(1 - 0.132352941176471))/204 = 0.023725931

z- score = 1.644853627    

Width of the confidence interval = z * SE =     1.64485362695147 * 0.0237259310586548 = 0.03902568

Lower Limit of the confidence interval = P - width =     0.132352941176471 - 0.039025683754629 = 0.09332726

Upper Limit of the confidence interval = P + width =     0.132352941176471 + 0.039025683754629 = 0.17137862

The confidence interval is [0.093, 0.171]

(B)

We can say with 95% confidence that the true proportion of the number of students having September birthdays falls in the above range

(C)

204/12 = 17

17/204 = 0.0833 (8.33%)

(D)

The entire confidence interval computed in part (A) is above 0.0833. So, there is sufficient evidence that the proportion of September birthdays is greater than we would expect by chance.