Question

A random sample of 5009 adults in a country includes 751 who do not use the Internet. Construct a 95% confidence interval estimate of the percentage of adults in the country who do not use the Internet. Based on the result, does it appear that the percentage of adults in the country who do not use the Internet is different from 49%, which was the percentage in the year 2000? Construct a 95% confidence interval estimate of the percentage of adults in the country who do not use the Internet. %

Answer 1

A 95% confidence interval for the percentage of adults not using the Internet is constructed using the sample proportion formula. The calculated interval is approximately (13.87%, 16.11%), indicating that the true percentage of non-users is likely within this range and significantly different from the 2000 percentage of 49%.

Step-by-step explanation:

To construct a 95% confidence interval estimate for the percentage of adults in a country who do not use the Internet, we'll use the provided data: out of a random sample of 5009 adults, 751 do not use the Internet. The point estimate of the proportion is the sample proportion, which is p = 751 / 5009. From this, we can use the standard formula for a confidence interval of a proportion:

Confidence Interval = p ± Z*(√(p(1-p)/n))

In this formula:

p is the sample proportion

n is the sample size

Z* is the Z-score corresponding to the desired confidence level, which is 1.96 for 95% confidence

Plugging in the numbers:

p = 751 / 5009 = 0.1499 (approximately)

Z* = 1.96

√(p(1-p)/n) = √(0.1499(1-0.1499)/5009) = 0.0057 (approximately)

The error margin is Z* × SE = 1.96 × 0.0057 ≈ 0.0112

The 95% confidence interval is therefore (0.1499 - 0.0112, 0.1499 + 0.0112) or approximately (0.1387, 0.1611). This means we are 95% confident that the true proportion of adults who do not use the Internet is between 13.87% and 16.11%.

Since 49% is not within this confidence interval, it appears that the percentage of adults in the country who do not use the Internet has changed since the year 2000.

Answer 2

The 95% confidence interval for the percentage of adults in the country who do not use the Internet is estimated to be between 13.93% and 16.85%. This interval is significantly lower than the year 2000 benchmark of 49%, indicating a decrease in the percentage of adults who do not use the Internet since then.

To construct a 95% confidence interval for the percentage of adults in the country who do not use the Internet, we can use the formula for a confidence interval for a population proportion:

CI = p ± z*(√(p(1-p)/n))

where:

- CI is the confidence interval

- p is the sample proportion (number who do not use the Internet / total number in the sample)

- z is the z-score associated with a 95% confidence level (approximately 1.96)

- n is the sample size

In this case, the sample proportion (p) is 751/5009. The sample size (n) is 5009. Plugging these into the formula, we get:

CI = (751/5009) ± 1.96*(√((751/5009)(1 - 751/5009)/5009))

Next, calculate the margin of error (ME) which is the second part of the formula after the ± symbol and then plug into the formula to get the interval:

CI = (0.1499) ± 1.96*(√(0.1499*0.8501/5009))

CI = (0.1499) ± 1.96*(0.0054)

CI = (0.1499) ± 0.0106

So, we estimate with 95% confidence that between 13.93% and 16.85% of all adults in the country do not use the Internet. Comparing this to the year 2000 benchmark of 49%, we can see that the current percentage is significantly different (and lower) than the percentage in the year 2000, since 49% is not within our confidence interval.