A college professor would like to estimate the proportion of students who pull an "all-nighter,” meaning they study all night for an upcoming exam. She selects a random sample of 100 students from her large college and finds that the 99% confidence interval for the true proportion of students who have pulled an all-nighter to be 0.48 to 0.62. If the professor had randomly selected 50 students rather than 100 students, what effect would this have had on the width of the interval? A) It would have doubled.
B) It would have been cut in half.
C) It would have remained the same.
D) It would have been larger, but it would not have doubled.

On a certain portion of an experiment, a statistical test result yielded a p-value of 0.21. what can you conclude? 2(0.21) = 0.42 < 0.5; the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 21% of the time, so the test is not statistically significant. 0.21 > 0.05; the test is statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 79% of the time, so the test is not statistically significant. p = 1 - 0.21 = 0.79 > 0.05; the test is statistically significant.