Alg1231W12T2 - Tail probabilities and the sign test

Saturday, 20 October 2018

11:54 PM

Machine generated alternative text:
Suppose we toss a weighted coin, for which the probability of getting a head (H) is 80% 
i) If we toss this coin 4 times, then the probability of getting exactly two heads (to two decimal places) is 0.1536 
ii) If we toss this coin 6 times; then the probability of getting exactly four heads (to ftvo decimal places) is 0.245760 
iii) if we toss this coin 9 times, then the probability of getting 7 or more heads (to two decimal places) is 
Note: you can use Maple to evaluate the binomial coefficient 
using the command binomial (n , k) A similar feature should exist on your handheld calculators.

 

Machine generated alternative text:
If we toss a weighted coin, for which the probability of getting a head is 70%, then the probability of getting 7 or more heads from 14 tosses is 
14 
(14)k (0.3)14-k 
(0.7)k (0.3)14-k 
(0.7)k (0.3)14-k 
14

 

Machine generated alternative text:
Recall that to compute the sum 
80 
10-k 
70 
we would use the Maple command: 
sum (binomial (17 ,k) * (O . 6) (O . 1) A (10—k) ,k=70. . 80) 
Now use Maple to calculate the following probabilities to exactly two decimal places. 
If we toss a baised coin which has a 80% of getting a head (H) then: 
i) The probabilty of getting 50 or more heads out of 60 tosses is 0823403275z 
ii) The probability of getting fewer than 18 heads out of 60 tosses is o 
iii) The probability of getting N heads, where 20 < N < 40 out of 60 tosses is 0.010668297t

 

Machine generated alternative text:
A tennis player usually wins 60% of her matches. After getting a new coach, she won 12 of her next 15 matches. Was this just luck or was her new coach a contributing factor? Lets decide 
by calculating a tail probability 
We hypothesise that the coach had no effect on her win rate. Her probability of winning a match is then p 
0.6 
The probability of winning at least 12 of 15 games, with such a probability is the binomial sum (enter your answer as a sum in Maple notation, see the hint at the end of this question) 
c 09050190240 
O To 2 decimal places this evaluates to 0090501902z 
She deduces that this represents roughly a 10% 
O chance Since this is larger than 5% 
O 
she is inclined to think 
O this was a statistically significant event (i.e. the coach really helped) 
@ this was a statistically insignificant event (i e the coach was irrelevant) 
Recall: to compute the sum 
80 
10-k 
k=70 
we would use the Maple command: 
sum (binomial (17 , k) * (O . 6) (O . 1) A (10—k) .80)

 

Machine generated alternative text:
The average yield of an acre of corn is 110 bushels. An experiment with a new variety produced the following average yields on 18 plots: 
112_8, 125_7, 115.0, 91.7, 103.5, 1298, 110.9, 124.0; 128.0; 121 _5; 126.1, 116.1, 1176, 1297, 1049, 125.3, 92.4, 114_7 
Is the new variety better? While we can't be sure: some statistics will guide us 
To investigate, associating a plus sign (+) to each entry larger than 110 , and a minus sign (—) for each entry less than 110 we have U 
14 
4 
O minuses. 
Assuming an equal likelihood of pluses and minuses; so that p 
o 
the tail probability of getting U or more successes from N 
O pluses and D 
18 
samples is (to 3 decimal places) 
This is a probability of about ISO,'O 
1 
O 
S C) (21) c) 
k IV-k 
This means such a yield was 
unlikely to accus by chance 
a 015441894± 
O , so the new variety is 
definitely superior

 

 

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