2a) X~Po(20/3), P(X>4) = 0.794 2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219 2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106
3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828
4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks 4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation) 4c) s=17.67, t=9.62 (or the other way around maybe) 4d) v=4, X2 = 8.749 < 9.488, Accept H0 4e) Wild not cultivated, because poisson
2a) X~Po(20/3), P(X>4) = 0.794 2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219 2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106
3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828
4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks 4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation) 4c) s=17.67, t=9.62 (or the other way around maybe) 4d) v=4, X2 = 8.749 < 9.488, Accept H0 4e) Wild not cultivated, because poisson
but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize. and 5(c) i remember it was at least 2 out of 8 months, shouldnt it be P(X>=2) ?
2a) X~Po(20/3), P(X>4) = 0.794 2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219 2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106
3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828
4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks 4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation) 4c) s=17.67, t=9.62 (or the other way around maybe) 4d) v=4, X2 = 8.749 < 9.488, Accept H0 4e) Wild not cultivated, because poisson
2a) X~Po(20/3), P(X>4) = 0.794 2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219 2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106
3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828
4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks 4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation) 4c) s=17.67, t=9.62 (or the other way around maybe) 4d) v=4, X2 = 8.749 < 9.488, Accept H0 4e) Wild not cultivated, because poisson
but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize. and 5(c) i remember it was at least 2 out of 8 months, shouldnt it be P(X>=2) ?
Why is the degrees of freedom for 4b 3? Isn’t 4 (5-1)? And for 1a) shouldnt the probability be X=3 as it the probability of winning a prize??
Degrees of freedom is 3 because there are two constraints (one for estimating p, one for expected frequencies adding up to 80). Degree of freedom = 5 - 2 = 3
omg finally a paper i actually got similar answers to others for... maybe i'll get the A i need in FM after all. i did so bad on core pure 2 so im glad this one can pull me up
6 columns to begin with Combining columns 5 and 6 so that E>5 makes 5 columns. -1 because the total is constrained to 80 makes 4 DOF. -1 because p has been estimated from the data makes 3 DOF.
In the poisson one, it's 4 DOF because you don't need to combine columns 5 and 6.
I calculated the parameter in part a just in case and got p=0.55. then the expected values we were given in part b matched p=0.55 exactly, so I reckon it was calculated.
Degrees of freedom is 3 because there are two constraints (one for estimating p, one for expected frequencies adding up to 80). Degree of freedom = 5 - 2 = 3
But it didnt say the p probability was estimated. It just said that there was a constant value that the person believed. It was only the second part with poisson that the mean was estimated where you had to take away 1.
I calculated the parameter in part a just in case and got p=0.55. then the expected values we were given in part b matched p=0.55 exactly, so I reckon it was calculated.
Perhaps, I felt it wasn't clear in the exam so it could be either way really. Maybe they'll accept both.
I decided to go with it because that meant the first test was rejected and then the second was accepted, which makes more sense. Otherwise they'd both be accepted (contradictory)
But it didnt say the p probability was estimated. It just said that there was a constant value that the person believed. It was only the second part with poisson that the mean was estimated where you had to take away 1.