The Pew Research Center Internet Project, conducted on the 25th anniversary of the Internet, involved a survey of 857 Internet users (Pew Research Center website, April 1, 2014). It provided a variety of statistics on Internet users. For instance, in 2014, 87% of American adults were Internet users. In 1995 only 14% of American adults used the Internet.
a. The sample survey showed that 90% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally (to 4 decimals).
95% Confidence Interval: ___ to ___.
I know that the formula is TS: z = p bar - p / sqrt p(-p)/n
n = 857
I don't know how to solve for p bar and p.
0.8976 to 0.9024
To solve for p-bar and p in order to calculate the confidence interval, we need to gather some additional information about the sample. Specifically, we need to know the number of respondents who said the Internet has been a good thing, as well as the total number of respondents in the sample.
Given that the sample size is 857 and 90% of respondents said the Internet has been a good thing, we can calculate p-bar (the sample proportion) and p (the population proportion) as follows:
p-bar = number of respondents who said the Internet has been a good thing / total number of respondents
p = estimated population proportion (since we don't know the exact value)
Using these values, we can then calculate the margin of error and construct the confidence interval.
Let's assume that the number of respondents who said the Internet has been a good thing is x.
p-bar = x / 857
To find p, we use the proportion of American adults in 2014 who were Internet users (87%):
p = 0.87
Now, we can calculate the standard error (SE) using the formula:
SE = sqrt((p * (1 - p)) / n)
where n is the sample size (857).
Once we have the SE, we can calculate the margin of error (ME) using the z-value for a 95% confidence level. Typically, for a 95% confidence level, the z-value is approximately 1.96.
ME = z-value * SE
Finally, we can construct the confidence interval by subtracting the margin of error from p-bar (lower bound) and adding the margin of error to p-bar (upper bound):
Lower bound = p-bar - ME
Upper bound = p-bar + ME
These values will give us the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally.
To find the values for p̄ (p bar) and p, you will need to use the information given in the problem.
The problem states that 90% of the respondents said the Internet has been a good thing for them personally. This means that the proportion of respondents who said the Internet has been a good thing (p) is 0.90.
The sample survey was conducted with 857 Internet users, so the sample size (n) is 857.
Using these values, you can calculate p̄ (p bar) by dividing the number of respondents who said the Internet has been a good thing (90% of 857) by the sample size (857).
p̄ = (0.90 * 857) / 857 = 0.90
With p̄ and p known, you can proceed to calculate the confidence interval using the formula:
z = (p̄ - p) / sqrt(p * (1 - p) / n)
First, you need to find the value of z for the 95% confidence level. This can be obtained from a standard normal table or using a calculator. For a 95% confidence level, the z-value is approximately 1.96.
Plugging in the known values:
1.96 = (p̄ - p) / sqrt(p * (1 - p) / n)
Rearranging the formula to solve for the interval:
p̄ - p = 1.96 * sqrt(p * (1 - p) / n)
Now you can substitute the known values:
0.90 - p = 1.96 * sqrt(0.90 * (1 - 0.90) / 857)
Simplifying the square root:
0.90 - p = 1.96 * sqrt(0.09 / 857)
Calculating the square root:
0.90 - p = 1.96 * 0.032
Finally, subtracting 1.96 * 0.032 from 0.90:
0.90 - 0.06272 = 0.83728
The lower bound for the confidence interval is 0.83728.
Similarly, adding 1.96 * 0.032 to 0.90:
0.90 + 0.06272 = 0.96272
The upper bound for the confidence interval is 0.96272.
Therefore, the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally is 0.8373 to 0.9627 (rounded to four decimals).