Ok. . . Now to expand from CI's to hypothesis testing. . .
I understand that for small n, and data approx. Normal, you use the formula
T= (x-bar - mu initial)/(s/ sqr n)
Now, what If i have no mu initial given. . .
Example-
Sample mean= 0.8
St. D= 0.1789
n=6
I was under the impression that 95% C.I requires that the critical value in the error term comes from a t-distribution with 25 degrees of freedom.
I was taught in class today that a 25% CI requires that the critical value in the error term comes from a t-distribution with 23 degrees of...
I have ran thru a couple of other problems that are a lot easier to see if I have improved at all. . . Thought maybe you would enjoy checking over them :-)?
1. A game of Yahtzee, you get 3 rolls of the dice per turn, and you get to choose which ones you re-roll each time. My brother calls me...
Here I was, thinking I was on a roll. . . Ugh.
Well, would this involve using the same formula? The:
(n/x)p^x(1-p)^n-x
the p would still equal 0.05 since its 5%?
Yes, I see how that works. I can NOT believe I was so easily confused by some basic algebra! AH! I feel quite ridiculous having stumbled through that the way I had. Thank you so much for your support!
For the question about the friends coming down to attend a football game, I do believe I have to use the equation. . .
(n/x)p^x(1-p)^n-x
Here, I think, you would use 125 for n (since there are 125 students asked), 4 for x (since there is a total of four tickets needed), and the p would be...
0 = E(x)= (-5)(12/13)+(W)(1/13),
so, W = 13W?
Meaning -4.6153 should be divided by 13W?
And no! I am really trying to learn this. . . I greatly appreciate your help! I am a psychology major with an emphasis with behavior analysis- I never work with stats on my day to day life! But I...
So for a fair game, E(x)=0
E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited
Alright. . .
~amount lost= $5, SO -5
~prob. of loss= (number of cards that are not aces/full deck)= (48/52) OR (12/13)
~amount gained (Win)= x(win)...
E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited
Alright. . .
~amount lost= $5
~prob. of loss= (number of cards that are not aces/full deck)= 48/52
~amount gained is what we are trying to figure out so we can determine the...
I do believe that I have an idea as to how to do number 5.
E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited?
3. Total amount of defective springs = 1+4+2= 7%
If there is 7 percent of the total productive of springs that are defective, shouldn't the chances that a randomly defective spring also be 7%?
4. Dice= numbers 1-6.
Here is where i get confused. You could roll a two which could be higher if...
2. Both of us have a dice. (1/6 probability of rolling any number on the die)
Probability of pulling a red = 5/8
Probability of pulling blue= 3/8
Probability of having the same number OR same color?
Same number = (since 1-3 are the only similar numbers) 3/8
Same color= pulling...
I am really awful at statistics, so I literally have to talk myself thru every step, but that will help everyone identify just what I am doing wrong!
1. So to find the probability that four will go to the game:
1-P(of not getting 4 tickets)
~keeping in mind the 5% chance of an extra...
Hey guys. I am having some issues understanding how to calculate these. If anyone could help, it would be greatly appreciated.
1. My friends came down to visit and wanted to go to a football game. Tickets were sold out, but I planned on asking my student's in lab throughout the day to see if...