Question

2. A home inspector claims that less than half of all homes have carbon monoxide detectors. In a survey of 240 randomly selected homes, they found that 102 have carbon monoxide detectors. Test the home inspector’s claim at a 5% significance level.

   1. a) Define the parameter and random variable of interest.

   2. b) State the null and alternative hypotheses, and identify the claim.

   3. c) Determine the distribution of the test statistic. (Check the relevant criteria.)

   4. d) Calculate the test statistic.

   5. e) Find the p-value.

   6. f) State your decision.

   7. g) State your conclusion.

Answer 1

Solution:

Part a

The parameter for this test is population proportion of all homes that have carbon monoxide detectors. The random variable of interest is the proportion of homes that have carbon monoxide detectors.

Part b

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

H0: p = 0.5 versus Ha: p < 0.5

This is a lower tailed test.

We are given

Level of significance = α = 0.05

Part c

For this distribution we have to use z or normal distribution for the test statistic because sample size is adequate and np and nq > 5.

Part d

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 102

n = sample size = 240

p̂ = x/n = 102/240 = 0.425

p = 0.5

q = 1 - p = 0.5

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.425 – 0.5)/sqrt(0.5*0.5/240)

Z = -2.3238

Test statistic = -2.3238

Part e

P-value = 0.0101

(by using z-table)

Part f

P-value < α = 0.05

So, we reject the null hypothesis

Part g

There is sufficient evidence to conclude that less than half of all homes have carbon monoxide detectors.