Comparing two proportions Please answer as soon as possible No need to show your

Question

Comparing two proportions

Please answer as soon as possible

No need to show your steps

1. Suppose that I am interested in determining if the proportion of students in set A who live in New York proper (as opposed to a suburb) is different from the proportion of students in set B who live in New York proper.

If I ask 18 students in set A and find that 5 of them live in New York, and 16 students in set B and find that 7 of them live in New York, construct an 80% confidence interval for the difference in proportions of New Yorkites between set B and set A. Take set B as population 1 and set A as population 2.

Enter the upper end of the confidence interval, as a decimal value. Answer to 4 decimal places of precision.

Explanation / Answer

p1 = 7/16 = 0.44

p2 = 5/18 = 0.28

The pooled sample proportion (P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                                       = (0.44 * 16 + 0.28 * 18)/(16 + 18)

                                                       = 0.355

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

      = sqrt(0.355 * (1 - 0.355) * (1/16 + 1/18))

      = 0.1644

At 80% confidence interval the critical value is z0.1 = 1.28

The 80% confidence interval is

(p1 - p2) +/- z0.1 * SE

= (0.44 - 0.28) +/- 1.28 * 0.1644

= 0.16 +/- 0.2104

= -0.0504, 0.3704

As the confidence interval cointains 0, so we can conclude that there is no difference in the proportion of student in set A from the proportion of student in set B.

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