The 2006 Statistical Abstract of the United States reports on a survey that asked a national sample of 80,000 American households about pet ownership. Suppose that one-third of all American households own a pet cat. The survey discovered that 31.6% of all the households sampled owned a pet cat. What is the z-score of this?From the standard deviation formula for a sample proportion, I found that standard deviation is 0.0016. From there, I plugged that into the z-score formula, and got (0.316-0.333)/0.0016 = -10.625. However, a z-score that high baffles me and I cannot imagine getting a z-score that high. Where did I go wrong?

Question

Keely Fernandez · Accepted Answer

Your calculations are correct. The extreme z score you get indicates that if the percentage of households was really 33%, then this sample is highly unrepresentative. The conclusion you would therefore draw is that the actual percentage must in reality be lower than 33%, assuming that the sample is random and representative of the population as a whole.