Question

30 randomly selected college students were asked whether they prefer to watch football or basketball. 8 students said they prefer to watch basketball, while the remaining 22 said they prefer to watch football. Go to www.lock5stat.com/StatKey/Links to an external site., but click 'CI for Single Proportion' instead of ‘CI for Single Mean, Median, St.Dev.’ Click on the button near the top of the page that says ‘Edit Data.’ Enter the count and sample size giving in this problem. Then, click Ok. Generate a bootstrap distribution with 10,000 samples for the proportion of college students who prefer to watch basketball. [Click ‘Generate 1000 Samples’ ten times (there might be some delay between each click).] What is the mean of the bootstrap distribution? What is the standard deviation of the bootstrap distribution? Construct a 95% confidence interval for the true proportion of college students who prefer to watch basketball instead of football. Interpret the confidence interval in context.

Answer 1

1. Follow the link www.lock5stat.com/StatKey (Links appended is not working i guess)

2. Click on CI for Single Proportion'

3. Click on ‘Edit Data.’ on top left

4. Enter count =8 and Sample =30. We are interested in basketball preferring count (8) out of our total sample which is 30

5. Enhance the number of samples using bootstrapping i.e. Click ‘Generate 1000 Samples’ ten times. This will generate a histogram chart with 10000 samples.

6. The chart (assuming normal distribution) also shows Mean=.266

7. Standard Deviation of the bootstrapped sample = Standard Error of the graph.= .081 (shown just below Mean

8. 2-tailed 95% Confidence Interval will be Mean +/- 1.96SD which means .266 - 1.96*.081 to .266+1.96*.081

from .107 to .425 is the desired 95% confidence interval which means that we can say with 95% certainty that the basketball watching people will be between 10.7% to 42.5%