Direct mail advertisers send solicitations (that is "junk mail") to thousands of potential customers in the hope that some will buy the company's products. The response rate is usually quite low. Suppose a company wants to test the response rate for a new flyer format. The company randomly selects 1000 from its mailing list of over 200,000 people. Their sample results in orders from 112 people.
Create a 90% confidence interval for the percentage of people the company contacts who may buy something.
The 90% confidence interval is % to %.
What is the margin of error? The margin of error is %. Show your answers as percents rounded to TENTHS of a percent. Do not type in the % sign.
How do I find the confidence interval? Once I get this do I just subtract the intervals to get the margin of error?
n = 1000
x = 112
p = x/n
p = 112/1000 = 0.112
z critical value = 1.645
E = za/2 *sqrt((p (1-p)/)/n))
E = 1.645 *sqrt((.112*.888/1000)
E = 0.016
p ± ('z critical value') * SQRT[p * (1 - p)/n] = 0.112 ± 1.645 * SQRT[0.112 * (1 - 0.112)/1000] = [0.096, 0.128]
The 90% confidence interval is 12.8% to 9.6%
Wow! Thank you for the help! Also, wouldn't the 90% confidence interval be 9.6% to 12.8%? Also, would the margin of error be 3.2 or 1.6?
Thank you!
The 90% confidence interval is 9.6 % to 12.8%.
The margin of error is 1.6.
To create a confidence interval for the percentage of people who may buy something, we need to use the formula:
CI = p̂ ± z * √((p̂(1-p̂))/n)
Where:
- CI represents the confidence interval
- p̂ is the sample proportion (orders/sample size)
- z represents the z-score corresponding to the desired confidence level (90% confidence corresponds to a z-score of 1.645)
- n is the sample size
In this case, we have a sample size of 1000 and 112 people who placed orders. Therefore, the sample proportion is 112/1000 = 0.112.
Substituting the values into the formula, we get:
CI = 0.112 ± 1.645 * √((0.112(1-0.112))/1000)
Calculating this gives us:
CI = 0.112 ± 1.645 * √((0.112 * 0.888)/1000)
= 0.112 ± 1.645 * √(0.099456/1000)
= 0.112 ± 1.645 * √(0.000099456)
= 0.112 ± 1.645 * 0.009972
Therefore, the 90% confidence interval for the percentage of people who may buy something is:
CI = 0.112 ± 0.016389
To determine the actual values for the confidence interval, we calculate the upper and lower limits:
Upper limit = 0.112 + 0.016389 = 0.128389
Lower limit = 0.112 - 0.016389 = 0.095611
Thus, the 90% confidence interval is 0.0956 to 0.1284 (rounded to four decimal places).
To find the margin of error, we subtract the lower limit from the upper limit:
Margin of Error = Upper limit - Lower limit
= 0.1284 - 0.0956
= 0.0328
Therefore, the margin of error is 0.0328 or 3.28% (rounded to two decimal places).
In summary:
- The 90% confidence interval for the percentage of people who may buy something is 0.0956 to 0.1284.
- The margin of error is 3.28%.