Exercise 71

Question 7.101

71. Rasmussen Report conducted a national telephone survey of a random sample of 1000 U.S. adults from June 19 to 20, 2013. Results indicated that 63% of adults nationwide would agree with the statement “Most Americans want the government to have less power and money.”

Use the information from the report to calculate a 95% confidence interval for the proportion of Americans who would agree with the statement above. Restate your confidence interval in terms of percentages. What is the margin of error?
The report concluded with the following statement: “The margin of error is ±3% with a 95% level of confidence.” Compare this statement with the margin of error you calculated in part (a).
Was a sample of size 1000 sufficiently large to guarantee that the margin of error was less than 3% even if the sample percentage had been as low as 50% or as high as 80%? Explain.
How large a sample size was needed to guarantee that the margin of error was below 3% regardless of the sample proportion?

Solving for One Variable in Terms of Another

71.

(a) $0.63 \pm 2 \sqrt{\frac{(0.63) (1 - 0.63)}{1000} \approx 0.63 \pm 0.0305}$ ; from approximately 0.60 to 0.66, or from 60% to 66%. The margin of error, to two decimals, is 0.03 or ±3%. (It was 3.05%, which we rounded to 3%.)

(b) They match if rounded to the nearest whole percent.

(c) Corresponding to 50%: margin of error is $2 \sqrt{\frac{(0.5) (0.5)}{1000}} \approx 0.0316; 3.1 %$ . So, the margin of error would be 3% only if we round to the nearest whole percent.

Corresponding to 80%, margin of error is $2 \sqrt{\frac{(0.8) (0.2)}{1000}} \approx 0.0253$ , or approximately 2.5%.

(d) Solve $2 \sqrt{\frac{(0.5) (0.5)}{n}} = 0.03$ for $n$ ; $n = {(\frac{0.5}{0.015})}^{2} \approx 1111.11$ . In order to guarantee that the margin of error was less than 3%, a sample size of at least 1112 should be used.