Previous research indicates that 20% of homeowners do not carry sufficient homeowner's insurance to adequately protect their investment.An insurance agency conducts a major advertising campaign in an attempt to rectify this situation.After the campaign, the agency randomly surveys 150 homeowners and find that 20 of them do not carry sufficient insurance.Does it appear that the proportion of homeowners without sufficient insurance was less after the ad campaign? State the critical value (CV), test statistic (TS), and decision from the test.(Use α = 0.05.)
A) CV = -1.2816; TS = -2.40; fail to reject H₀
B) CV = -2.3263; TS = -2.04; reject H₀
C) CV = 2.3263; TS = -2.04; fail to reject H₀
D) CV = 1.2816; TS = 2.04; reject H₀

Question

Quizplus · Accepted Answer

We will use a one-sample proportion test.

Hypotheses: 
H₀: p = 0.20 (proportion of homeowners without sufficient insurance before the ad campaign)
H₁: p < 0.20 (proportion of homeowners without sufficient insurance after the ad campaign)

Level of significance: α = 0.05

The critical value for a left-tailed test with α = 0.05 is -1.645 (using a z-table or calculator).

The test statistic can be calculated as:
TS = (p̂ - p) / √[p(1-p) / n]
where p̂ = 20/150 = 0.1333 (proportion of homeowners without sufficient insurance in the sample)
n = 150 (sample size)

TS = (0.1333 - 0.20) / √[0.20(0.80) / 150] = -2.04

Since TS (-2.04) is less than the critical value (-1.645), we reject the null hypothesis. This means that there is evidence to suggest that the proportion of homeowners without sufficient insurance decreased after the ad campaign.

Previous Research Indicates That 20% of Homeowners Do Not Carry