Point Estimate/Error
Critical Values
Sample Size
Proportions
Means
100
Find the point estimate based off of the following information:
99% Confidence level, n= 827, x=174
Round to 4 decimal places
0.2104
100
Determine the critical values for a z-distribution with a 90%, 95%, and 99% confidence level.
z = +/-1.645, +/-1.96, +/-2.575
100
Use the given data to find the minimum sample size required to estimate the population proportion. Margin of Error: 0.024; confidence level: 99%; 
n=2874
100
Express the confidence interval, (0.291, 0.749), in inequality form.
0.291 < p < 0.749
100
Use these summary statistics given: n=200, “x-bar”= 32.6 hg, s= 8.1. Construct a 95% confidence interval.
31.47 < μ < 33.73
200
Find the point estimate of the proportion of people who enjoy reading, in a random sample of 749 people, 61 enjoyed reading.
“p-hat”=0.08
200
Find the critical value Zα/2 that corresponds to the given confidence level: 92%. Round your answer to 2 decimal places.
z = +/-1.75
200
Use the given data to find the minimum sample size required to estimate the population proportion. Margin of Error: 0.055; confidence level: 90%;
n=224
200
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones.. Of the 500 people surveyed, 392 responded yes – they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones. Round to 3 decimal places, and put in inequality form.
0.748 < p < 0.820
200
Use the given degree of confidence and sample data to construct a confidence interval for the population mean. Assume the population has a normal distribution.
n=82, “x-bar”=5.3, s= 2.4, 96% confidence
4.75 < μ < 5.85
300
A confidence interval was constructed to estimate the proportion of smokers in the U.S. Use the confidence interval 0.167 < p < 0.458 to find the margin of error, E. Round answer to 3 decimal places.
E=0.146
300
Determine the critical values for a t-distribution with degrees of freedom equal to 19 and a confidence level of 99%.
t = +/-2.86
300
Suppose you want to construct a 95% confidence interval for the average speed that cars travel on the highway. You want a margin of error of no more than plus or minus 0.5 mph and know that the population standard deviation is 7 mph. At least how many cars must you clock?
n=753
300
In a study of the accuracy of fast food drive-through orders, Burger King had 375 accurate orders and 65 that were not accurate. Construct a 95% confidence interval estimate of the percentage of orders that are accurate.
0.819 < p < 0.885
300
A group of 10 foot surgery patients had a mean weight of 240 pounds. The sample standard deviation was 25 pounds. Find a confidence interval for a sample for the true mean weight of all foot surgery patients. Use a 95% confidence level. Round to the nearest pound.
222 < μ < 258
400
Find the point estimate based on the following confidence interval
(0.629, 0.936).
0.7825
400
Do one of the following as appropriate: (a) find the critical value 𝑡𝛼/2 OR (b) find the critical value 𝑧𝛼/2. Round your answer to 2 decimal places.
Confidence Level = 95%; n = 32; s = 13; population appears to be normally distributed
t = +/-2.04
400
Suppose you want to construct a 90% confidence interval for the proportion of people who will stop to pick up a nickel if they see one on the ground. You want a margin of error of no more than plus or minus 4 percentage points. How many people must you survey?
n=423
400
The drug OxyContin is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 278 subjects were treated with OxyContin and 76 of them developed nausea. What is the best point estimate? Construct a 90% confidence interval estimate of the percentage of OxyContin users who develop nausea. Write the answer as an open interval rounded to 3 decimal places.
Point estimate: p = 0.273, (0.229, 0.317)
400
Construct a 92% confidence interval of an experiment that found the mean temperature for the city in August was 101.82, with a population standard deviation of 1.2. There were 6 samples in this experiment.
100.96 < μ < 102.68
500
Use the confidence level, 23827 < μ < 39262, to find the Margin of error, E.
7717.5
500
Do one of the following as appropriate: (a) find the critical value tα/2, OR (b) find the critical value 𝑧𝛼/2. Round your answer to 2 decimal places.
Confidence Level = 92%; n = 19; σ = 18; population appears to be normally distributed.
z = +/-1.75
500
Last year, you did a study and found out that 67% of the students who eat in the cafeteria have a salad. This year you want to construct a 95% confidence interval for the proportion of students who have a salad in the cafeteria. You want a margin of error of no more than plus or minus 3 percentage points. How many students must you observe?
n=944
500
In a 1997 survey done by the Marist College Institute for Public Opinion, 36% of a randomly selected sample of 935 American adults said that they do not get enough sleep each night. Create a 95% confidence interval and determine the margin of error for the percentage that feels they don’t get enough sleep each night. Express the confidence interval as p^±E, as an open interval, and in the inequality form.
.36 ± .031
500
11) A randomly selected sample of 12 students at a university are asked, “How much did you spend on textbooks this semester?”. The responses, in dollars, are:
200, 175, 450, 300, 350, 250, 150, 200, 320, 370, 404, 250
Create a 99% confidence level and determine the margin of error. Express the confidence interval as 
284.92±86.125, (198.79, 371.04), 198.79 <