Due to some gray area in a TA's grading, I am going to review my statistics final with my professor on Friday. There's some possibility of a grade change. One of the questions I'm most anxious about, the final question, went essentially like this:

"Tommy finds a deck of cards, but he is not sure whether it is his normal deck of cards, or his little brother Billy's trick deck of cards. Like a normal deck, Billy's trick deck has 52 cards, but it has no deuces (no "twos"); instead it has eight aces. Tommy decides to do an experiment: he will draw a single card from the deck. He will begin with the null case hypothesis that the deck is normal, but he will reject the null case and conclude that it is the trick deck if he draws an ace. What is alpha for Tommy's experiment? What is beta?"

I constructed a tree diagram with two root branches: null case hypothesis is true and null case hypothesis is false (i.e., alternative hypothesis is true). Then I drew two further branches from each of these two root branches. Thus, if the null was true, and the deck was the normal deck, there should be 4 chances in 52 of drawing an ace, and 48 chances in 52 of drawing some other card. And, if the null was false, and the deck was the trick deck, there should be 8 chances in 52 of drawing an ace, and 44 chances in 52 of drawing some other card. So the four possibilities should be:

Drawing an ace when the null is true;

Drawing some other card when the null is true;

Drawing an ace when the null is false;

Drawing some other card when the null is false.

It was my further understanding that these probabilities should sum to one. One problem I had, however, is that I didn't know what the probabilities were of the null or alternative hypotheses in the first place. So I decided to assign them equal probabilities. (Right or wrong, that is what I thunk with four minutes left on the exam.) So I concluded that my overall probabilities were 2/52 chances (4/52 x 1/2) of drawing an ace from the normal deck; 24/52 chances (48/52 x 1/2) of drawing some other card from the normal deck; 4/52 chances (8/52 x 1/2) of drawing an ace from the trick deck, and 22/52 chances (44/52 x 1/2) of drawing some other card from the trick deck. These probabilities sum to one (52/52). I therefore said that alpha was 2/52--the chances of drawing an ace from the normal deck--and beta was 22/52--the chances of not drawing an ace from the trick deck.

Was I right or wrong? Why or why not?

Many Thanks,

Antikythera (Google it or YouTube it!)

"Tommy finds a deck of cards, but he is not sure whether it is his normal deck of cards, or his little brother Billy's trick deck of cards. Like a normal deck, Billy's trick deck has 52 cards, but it has no deuces (no "twos"); instead it has eight aces. Tommy decides to do an experiment: he will draw a single card from the deck. He will begin with the null case hypothesis that the deck is normal, but he will reject the null case and conclude that it is the trick deck if he draws an ace. What is alpha for Tommy's experiment? What is beta?"

I constructed a tree diagram with two root branches: null case hypothesis is true and null case hypothesis is false (i.e., alternative hypothesis is true). Then I drew two further branches from each of these two root branches. Thus, if the null was true, and the deck was the normal deck, there should be 4 chances in 52 of drawing an ace, and 48 chances in 52 of drawing some other card. And, if the null was false, and the deck was the trick deck, there should be 8 chances in 52 of drawing an ace, and 44 chances in 52 of drawing some other card. So the four possibilities should be:

Drawing an ace when the null is true;

Drawing some other card when the null is true;

Drawing an ace when the null is false;

Drawing some other card when the null is false.

It was my further understanding that these probabilities should sum to one. One problem I had, however, is that I didn't know what the probabilities were of the null or alternative hypotheses in the first place. So I decided to assign them equal probabilities. (Right or wrong, that is what I thunk with four minutes left on the exam.) So I concluded that my overall probabilities were 2/52 chances (4/52 x 1/2) of drawing an ace from the normal deck; 24/52 chances (48/52 x 1/2) of drawing some other card from the normal deck; 4/52 chances (8/52 x 1/2) of drawing an ace from the trick deck, and 22/52 chances (44/52 x 1/2) of drawing some other card from the trick deck. These probabilities sum to one (52/52). I therefore said that alpha was 2/52--the chances of drawing an ace from the normal deck--and beta was 22/52--the chances of not drawing an ace from the trick deck.

Was I right or wrong? Why or why not?

Many Thanks,

Antikythera (Google it or YouTube it!)

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