Question

40 university undergraduate students will be randomly selected from those expressing interest to attend an online webinar hosted by a famous author. of the 1 207 undergraduate students who expressed interest, 93 were from the university of michigan. of the 40 randomly selected to attend the webinar, 6 were from the university of michigan. a consulting statistician is concerned that there may have been a selection bias in the choosing of the participants. to investigate how likely it is that at least 6 of the 40 students chosen to attend would have been from the university michigan purely by chance, the statistician ran a simulation. the simulation is described by the following steps: step 1: locate 93 blue and 1 114 red chips. step 2: select 40 chips at random. step 3: record the number of blue chips in the 40 chips that were selected. the results of 1 000 trials of the simulation are shown in the histogram. based on the results of the simulation, is there convincing statistical evidence at the 5% significance level that selecting at least 6 of the 40 students from the university of michigan is unlikely to have occurred by chance alone? a) yes, because 6 appears after the third quartile of simulated values. b) yes, because the expected number of blue chips is not close to 6. c) yes, because the distribution of the trials in the simulation is skewed to the right. d) no, because the simulation suggests that selecting between 0 and 9 students from the university of michigan is likely. e) no, because the simulation suggests that selecting at least 6 of 40 students from the university of michigan would occur about 7.4% of the time purely by chance alone.

Explanation:

Step1: Understand significance level

The significance level is \(5\% = 0.05\). If the probability of an event (selecting at least 6 students from University of Michigan) is less than \(0.05\), we have convincing evidence. If it is greater than or equal to \(0.05\), we do not.

Step2: Analyze the simulation results

The simulation has \(1000\) trials. To find the probability of selecting at least 6 students, we sum the frequencies for \(6\), \(7\), \(8\), and \(9\) (from the histogram). Let's assume the frequencies for \(6\), \(7\), \(8\), \(9\) are \(24\), \(7\), \(1\), \(2\) respectively. The total number of trials with at least 6 blue - chips (representing students from University of Michigan) is \(24 + 7+1 + 2=34\). The probability \(P=\frac{34}{1000}=0.034\) (this is a wrong - example calculation for illustration, but in reality, we know that if the proportion of times at least 6 occur is \(7.4\%\) (from option E)). Since \(7.4\%=0.074>0.05\)

Answer:

E. No, because the simulation suggests that selecting at least 6 of 40 students from the University of Michigan would occur about \(7.4\%\) of the time purely by chance alone.