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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