WorksheetsUnit 10 : Data and Statistics
Total questions: 40
Worksheet time: 2hrs 2mins
If waist sizes are normally distributed, determine the z-score of a teenage male with a 33 inch waist.
About what percent of the products last between 12 and 15 days?
A Parks Department employee wants to know if latex paint is more durable than non-latex paint. She has 50 park benches painted with latex paint and has 50 park benches painted with non-latex paint.
Experiment
Survey
Observational study
To test the redesign of it’s website, an online bookseller assembled 96 users of the site and randomly divided them into two groups. One group used the new website to make a purchase and one group used the old website to make the same purchase. Users of the new site were able to complete the purchase 22% faster
Experiment
Survey
Observational study
The teacher samples her class by selecting all students sitting at group 1 and group 5 in her classroom. This sampling technique is called?
sample random
Stratified
Systematic
Cluster
Each student in the school is given a different number. A student is randomly selected from the first 25. That student and every 25th student thereafter are in the sample. This is what type of sampling design?
sample random
Systematic
Multistage
Stratitfied
Determine whether the survey clearly projects a winner. According to the survey of students voting for student council president, 42% of the students said they will vote for Jada and 58% said they would vote for Sean. The survey's margin of error is 5%
Jada wins
Sean wins
It's too close to tell
It's a tie
A survey of 121 subjects found that 84 have suffered the symptoms of depression in the last year.
A 95% confidence interval for p is (0.621, 0.759). Interpret this interval.
Based on this sample, I am 95% confident that the true proportion of people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9%.
95% of the time, the population of people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9%
95% of all of the possible intervals calculated using this method will capture the true proportion of people who have suffered the symptoms of depression in the last year.
The probability that people who have suffered the symptoms of depression in the last year is between 62.1% and 75.9% is 95%.
In a survey of 104 students, it was found that 79 went to the homecoming game this year.
A 95% confidence interval for p is (0.652, 0.868). Interpret this interval.
95% of the time the true proportion of people who went to the homecoming game this year is between 65.2% and 86.6%.
The probability that the population proportion of people who went to the homecoming game this year is between 65.2% and 86.6% is 95%.
Based on this sample, I am 95% confident that the true proportion of people who went to the homecoming game this year is between 65.2% and 86.6%.
95% of all possible intervals calculated this way will capture the true proportion of people who went to the homecoming game this year
For a sample, the 95% confidence interval is (0.202, 0.482).
What is the margin of error of this sample?
0.28
0.14
0.342
0.684
The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 13.2, 5.1, 7.5, 8.0, 12.7, 7.6, 13.8, 14.5, 7.7, and 10.5. Determine a 95 percent confidence interval for the mean time for all players.
12.30 < μ < 7.82
7.82 < μ < 12.30
7.97 < μ < 12.14
12.40 < μ < 7.72
If the 95% Confidence Interval for a population mean is (114.56, 125.54), then is it likely that the true population mean µ is greater than 130?
Yes
No
Not enough information
I have no idea what you're asking of me
A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 95% confidence interval of the true proportion of women in Europe who use the Internet?
0.321 < p < 0.379
0.315< p < 0.384
0.309 < p < 0.391
0.305 < p < 0.395
In a survey of 104 students, it was found that 79 went to the homecoming game this year.
Calculate a 95% confidence interval for p.
(0.750, 0.829)
(0.691, 0.829)
(0.678, 0.842(
(0.685, 0.895)
Previous studies show that the standard deviation for work time is 3 hours, and a survey of 40 students has a mean of 9.8 work hours per week.
The 95% confidence interval is…
(8.84, 10.76)
(8.87, 10.73)
(9.02, 10.58)
(9.00, 10.60)
