Probability Math question help

Question

 A random sample of 500 adults in Flagstaff found 15 adults who had lost a job in the past year. Find a 95% confidence interval for the true percent of adults in Flagstaff who have lost a job in the past year. Express your results to the nearest hundredth of a percent

Answer 1

For those able to use a TI-83 or TI-84:

STAT --> TESTS --> A: 1-PropZInt

x:15 (since we are interested in the 15 adults in Flagstaff who lost a job in the past year)

n: 500 (since we sampled 500 adults in Flagstaff)

C-Level: .95 (since we are interested in a 95% confidence interval)

Hit "Calculate."

(.01505, .04495).

Round to the nearest hundredth of a percent. (1.51%, 4.50%).

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For those who are to do it by hand, use the formula for a one-proportion confidence interval:

In this case, p-hat should be equal to 15/500, or 0.03, and n is 500, since we sampled 500 Flagstaff adults, 15 of which lost a job in the past year. Missing from this formula is a subscript attached to the z, so it should read as z_(a/2), where a is the significance level suggested by the interval (.95 confidence, so a=0.05). a/2 is equal to 0.05/2, or 0.025. The notation z_0.025 means that we are looking for the z quartile such that the area to the right of it is 0.025. This happens at z=1.96. (This can be found in a z-table or in a table of commonly used confidence and significance levels based on the z distribution.) Plugging in these values, one arrives at the statement 0.03 ± 0.0149526292. This suggests that the confidence interval is (0.01505, 0.04495). Re-writing as rounded percentages, we get (1.51%, 4.50%).

In either case, we are 95% confident that the true proportion of Flagstaff adults who have lost a job in the past year is in between 1.51% and 4.50%.