Question

Smoke detectors are highly recommended safety devices for early fire detection in homes. It is very important that they emit a loud warning signal when needed, and if they don't, they are considered defective. Officials at the companies that make Brand A and Brand B smoke detectors each took random samples of their products and tested the warning signals. Twenty of the 200 Brand A detectors were faulty, while only twelve of the 200 Brand B detectors were defective. Assume the samples are representative of their corresponding populations and that the normal criteria are met.

Part A) Find and interpret a 95% confidence interval for the true difference in the proportions of faulty detectors manufactured by the two companies.

Part B) Based on the confidence interval, is there statistical evidence that one brand is better than the other at producing a lower rate of faulty detectors? Why or why not?

Answer 1

Answer)

N1 = 200, P1 = 20/200

N2 = 200, P2 = 12/200

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 20

N1*(1-p1) = 180

N2*p2 = 12

N2*(1-p2) = 188

All the conditions are met so we can use standard normal z table to ontain the interval

Critical value z from z table for 95% confidence level is 1.96

Margin of error (MOE) = Z*Standard error

Standard error = √{((p1*(1-p1)/n1) + (p2*(1-p2))/n2}

MOE = 0.05302877709

Interval is given by

(P1-P2) - MOE < (P1-P2) < (P1-P2) + MOE

−0.0130287770 < P1-P2 < 0.09302877709

B)

Null hypothesis Ho : P1 - P2 = 0

As the interval contains the null hypothesised value 0 in it we fail to reject the null hypothesis Ho

So, we do not have enough evidence to conclude that if one is better than the other