Statistics Review - Two Sample Hypothesis Tests for Means and Probability

Today was the first of two days of review. The quiz on one-sample hypothesis tests for means indicated that students were still struggling with this idea. A larger segment than I would care to see had found the sample mean and simply compared it to the hypothesized value. There was no referencing the sampling model nor the probability that they would draw a sample with the characteristics they saw.

To start things off, I reiterated the ideas behind statistical analysis and inference. I drew a large cloud on the board and said this was the population we were studying. Inside the cloud I drew a small circle and said this represented the sample we drew. The idea is to draw conclusions about the entire population from the small snapshot that we took via our sample.

I drew more circles throughout the population cloud, some of which overlapped. Each sample we draw provides a different snapshot of the population. We need to account for every possible sample we draw. This is where the sampling distribution model comes into play. The sampling distribution model describes what we should expect given the sample size we have drawn from the population.

We cannot simply compare our sample mean to the hypothesized population mean. Sure, this time it may be greater than our hypothesized value, the next drawn sample could show the mean less or more. Every sample could and probably will be different. We need to take the one sample we drew and use that to draw conclusions about all the possible samples that could be drawn and from this draw a conclusion about the population we are studying.

With that said, we went through the quiz questions, working through results. As students did this, I passed out two sets of die; one of the dice was colored green and the other blue. I asked students to roll both die six times and to count the number of times each colored die won. As soon as we finished going through the quiz problems I collected the die and we started working with the data that was generated.

I described the situation to the entire class, since not every student had a die set passed to them. I described the data collected and asked them what they expected to happen. Several students said they expected the number of wins for each die to be the same. From here I asked them to state null and alternative hypotheses.

H₀: The mean number of blue dice wins equals the mean number of green dice wins (μ_b = μ_g)
H_a: The mean number of blue dice wins does not equal the mean number of green dice wins (μ_b ≠ μ_g)

I then asked the students if the samples we had were independent or not. Here there was some disagreement. One student said that if you knew the number of blue wins then you would also know the number of green wins. Another said that wouldn't necessarily be true since you could have ties. I then brought up that you might not know the exact number of wins but you certainly would know the  maximum number of wins. This indicates that the two samples are not independent of each other. Students need to carefully consider the samples they draw as to whether or not there is any direct connection between the two samples.

I asked students to analyze the results of these samples and draw a conclusion about the data. There was a lot of confusion about how to analyze the data. With a matched pair test, you take the difference in values between the matched numbers and then analyze the differences as a single sample.

Their analysis resulted in a p-value of 0.04. Students were somewhat shocked by this result. The conclusion was to reject the null hypothesis and conclude that the number of wins for the blue die and the green die were different. I told them not to be too shocked as the green die had two number fives on its face and no number two.

Next, I asked several students to roll an individual die 3-4 times and record the values rolled. I made two columns on the board, one for blue die rolls and one for green die rolls. Students recorded their results on the board. I then asked if these two samples were independent. The response this time was a resounding "Yes!"

I asked what an appropriate null hypothesis would be and what should the alternative hypothesis be. Because students now knew that the green die was "unfair" they concluded that the alternative should be the green die rolls would exceed the blue die rolls. We had

H₀: The mean roll of the blue die equals the mean roll of the green die  (μ_b = μ_g)
H_a: The mean roll of the blue die is less than the mean roll of the green die  (μ_b < μ_g)

I asked students to analyze the samples and draw a conclusion. In this situation, students had questions about how to calculate their degrees of freedom. While the text does provide a formula, it is long and complicated and a calculator or computer can easily do the computation for you. I told students this and for the few students who did not have a calculator that would figure the degrees of freedom for them, I told them to simply add the degrees of freedom for the two individual samples, which basically provides an upper-bound to the number of degrees of freedom. For most reasonably large samples the results will not be affected.

For this analysis, students calculated that the p-value was 0.056. Using a 5% significance level we would fail to reject the null hypothesis. A student pointed out that this was the wrong decision. For our sample, we committed a Type II error. This error was discussed versus a Type I error.

To confirm that we, in fact, committed a Type II error, I asked students to calculate the expected value (mean) of rolling the green die. This entails constructing a probability model and determining the expected value. Many students struggled with this but were able to complete the task with some assistance. I then asked the class to compute the expected value of the blue die. The values were 4 and 3.5 respectively. So, we did indeed commit a Type II error.

I then asked students to determine the probability of the green die roll exceeding the blue die roll. Students seemed baffled as to how to proceed. I told them they needed to consider all the possible outcomes and which of those met the desired criteria. Listing out the 36 possible outcomes, it becomes readily apparent that there are 18 outcomes when the green roll exceeds the blue roll, resulting in a P(green > blue) = 0.5. Proceeding further I asked what P(green = blue) equals? Students used their outcomes and came to a result of 1/6. This baffled them for some moments as they seemed to expect it would be different.

I asked students to pick 2-3 questions from past quizzes and exams or chapter review problems that they did not know how to complete. We will use these as a basis for further review next class.

I concluded by working with students who had questions about the project reports. Most of the reports that I was shown looked to be on the right track. A few needed to focus more on comparing the two data sets rather than simply viewing them as two distinct, non-related entities. I am looking forward to seeing what they produce for their final versions.