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d) Using this portion of the standard normal chart and part (c), find the probability P(x 2 60). .00 .01 .02 .03 .04 .05 1.8 .9641 .9713 9772 9656 .9726 9783 .9671 1.9 2.0 .9649 9719 9778 .9664 .9732 .9788 .9738 9793 .9678 .9744 9798

d) Using this portion of the standard normal chart and part (c), find the probability P(x 2 60). .00 .01 .02 .03 .04 .05 1.8 .9641 .9713 9772 9656 .9726 9783 .9671 1.9 2.0 .9649 9719 9778 .9664 .9732 .9788 .9738 9793 .9678 .9744 9798

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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can you answer 27a, b, c
d) Using this portion of the standard normal chart and part (c), find the probability P(x 2 60). .00 .01 .02 .03 .04 .05 .9649 .9656 .9664 .9732 .9788 .9671 .9738 9793 1.8 .9678 .9641 .9713 9772 .9719 .9778 1.9 .9726 .9783 .9744 2.0 .9798 e) You should have found that the answer to (d) is much smaller than the answer to (b). This means that the probability of getting 60% or more tails depends on the sample size n. That is: Probability of 6 or more tails out of 10 > Probability of 60 or more tails out of 100 even though the proportion, p = 60%, is the same in both cases. Why does this happen? f) Without doing any new calculations, resolve these two hypothesis tests to a 3% level of significance. The null hypothesis is that the coin is fair. The alternate hypothesis is that the coin is biased toward tails. Ho:p = 0.5 H1:p > 0.5 I. If the coin is flipped 10 times and it comes up tails 6 times, what is your conclusion? Compare your result from part (b) to the cutoff of 3%. II. If the coin is flipped 100 times and it comes up tails 60 times, what is your conclusion? Compare your result from part (d) to the cutoff of 3%.
Transcribed Image Text:d) Using this portion of the standard normal chart and part (c), find the probability P(x 2 60). .00 .01 .02 .03 .04 .05 .9649 .9656 .9664 .9732 .9788 .9671 .9738 9793 1.8 .9678 .9641 .9713 9772 .9719 .9778 1.9 .9726 .9783 .9744 2.0 .9798 e) You should have found that the answer to (d) is much smaller than the answer to (b). This means that the probability of getting 60% or more tails depends on the sample size n. That is: Probability of 6 or more tails out of 10 > Probability of 60 or more tails out of 100 even though the proportion, p = 60%, is the same in both cases. Why does this happen? f) Without doing any new calculations, resolve these two hypothesis tests to a 3% level of significance. The null hypothesis is that the coin is fair. The alternate hypothesis is that the coin is biased toward tails. Ho:p = 0.5 H1:p > 0.5 I. If the coin is flipped 10 times and it comes up tails 6 times, what is your conclusion? Compare your result from part (b) to the cutoff of 3%. II. If the coin is flipped 100 times and it comes up tails 60 times, what is your conclusion? Compare your result from part (d) to the cutoff of 3%.
27. This problem is about probabilities and coin flipping. The goal is to compare the probability of getting 6 tails or greater when flipping a coin 10 times, versus the probability of getting 60 tails or greater when flipping a coin 100 times. In each case, the proportion of tails is the same, 2 60%, but the probabilities are different. Assume that we have a fair coin, so the odds of getting tails on any one flip are 0.5. a) First, let's calculate the odds of getting exactly 6 tails out of 10 total coin flips. Use the binomial probability formula P(x) = „Cx (p)*(1 – p)"-* to find P(6). %3D b) You don't have to find the other probabilities by hand; they can be found from this piece .50 of the binomial chart. Using this information, determine the probability of getting 6 tails or more. That is, find P(x > 6). Hint: it could be 6 or 7 or 8 or 9 or 10. Note that the answer to part (a) is right there next to x = 6, so you can check to see if you got part (a) correct. n x 10 0 .001 .010 1 2 .044 3 .117 .205 4 .246 5 6 .205 .117 7 8 .044 9.010 10 .001 Now we would like to consider n = 100 coin flips and determine P(x > 60). It is too difficult to add up each of the possibilities by hand from the binomial chart, so instead we will approximate the binomial distribution as a normal distribution. This is valid because n is large enough. 60 c) Using the appropriate z-score formula, determine the z-score for ôp = 0.60, p = .5, and n = 100. 100
Transcribed Image Text:27. This problem is about probabilities and coin flipping. The goal is to compare the probability of getting 6 tails or greater when flipping a coin 10 times, versus the probability of getting 60 tails or greater when flipping a coin 100 times. In each case, the proportion of tails is the same, 2 60%, but the probabilities are different. Assume that we have a fair coin, so the odds of getting tails on any one flip are 0.5. a) First, let's calculate the odds of getting exactly 6 tails out of 10 total coin flips. Use the binomial probability formula P(x) = „Cx (p)*(1 – p)"-* to find P(6). %3D b) You don't have to find the other probabilities by hand; they can be found from this piece .50 of the binomial chart. Using this information, determine the probability of getting 6 tails or more. That is, find P(x > 6). Hint: it could be 6 or 7 or 8 or 9 or 10. Note that the answer to part (a) is right there next to x = 6, so you can check to see if you got part (a) correct. n x 10 0 .001 .010 1 2 .044 3 .117 .205 4 .246 5 6 .205 .117 7 8 .044 9.010 10 .001 Now we would like to consider n = 100 coin flips and determine P(x > 60). It is too difficult to add up each of the possibilities by hand from the binomial chart, so instead we will approximate the binomial distribution as a normal distribution. This is valid because n is large enough. 60 c) Using the appropriate z-score formula, determine the z-score for ôp = 0.60, p = .5, and n = 100. 100
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