For this question, you will use a fair coin to take some samples and analyze the

Question

For this question, you will use a fair coin to take some samples and analyze them. First, take any fair coin and flip it 12 times. Count the number of heads out of the 12 flips. This is your first sample. Do this 4 more times and count the number of heads out of the 12 flips in each sample. Thus, you should have 5 samples of 12 flips each. The important number is the number of heads in each sample (this can be any whole number between 0 and 12). Then answer the following questions: First, list the number of heads in each of the 5 samples. Combining all 5 samples, calculate the mean number of heads. Show your work. Combining all 5 samples, calculate the median number of heads. Show your work. If you had a very large number of samples, what value(s) should the mean and median have? Why? For each of your 5 samples of 12 flips, use the binomial distribution to test the null hypothesis that the coin is fair, using alpha = .05 (two-tailed). (The alternative hypothesis is that the coin is not fair.) For each sample, calculate the p value (the probability of getting the number of heads you did or a more extreme number of heads ("two-tailed") purely by chance). Show your work and organize the data into a small table. For each sample indicate the number of heads out of 12, the p value, your decision about the null hypothesis, and the explicit reason for your decision, in terms of the decision rule. If each sample involved flipping the coin 6 times (rather than 12 times), and you did this for a very large number of samples, what would be the expected number of heads, on average, per set of 6 flips? On a series of 17 flips of a fair coin, what is the probability of getting 3 or fewer heads OR 14 or more heads? If I give you a coin to flip and I tell you it is not a fair coin, but is weighted to come up heads on average only 25% of the time (instead of 50%), what is the most likely number of heads you would observe in a series of 20 flips? If the coin was truly weighted to come up heads on average only 25% of the time, what is the probability that you would observe 9 or more heads in a series of 20 flips? What is the probability that you would observe 3 or fewer heads in a series of 20 flips? In general, what proportion of samples of fair coin flips (when N is sufficiently large that a rejection region of the binomial distribution exists) should result in rejecting the null hypothesis (alpha .05, two-tailed)?

Explanation / Answer

I have tossed a fair coin for 12 times and did this procedure 5 times.

(a) I got 5,5,5, 6,7 Heads in these different samples out of 12 flips

(b) mean of the sample = (5 + 5 + 5 + 6 + 7)/5 = 5.6

(c) median number of heads = 5

(d) for infinite or large number of samples

mean = 6 and median = 6

for large number of sample mean = number of trials * its probability = 12 * 0.5 = 6; similarly this distribution will be normal for large number of samples so median will be the mean value = 6

(e) For each of my samples, i have to test the null hypothesis. As my first three results are 5 heads out of 12 flips.

Null Hypothessis : H0 : Coin is fair

H1: Coin is biased.

Probability of getting 5 tosses out of 12 flips.

Probability of getting 5 heads = P(x= 5; 12; 0.50)= 0.1933

Probability of getting atleast 5 heads p (x >=5; 12; 0.5) = 0.8061

so this is unbiased towards any tail as this is not under 0.05 level significance. Null hypothesis is correct and coin is fair

Similarly, for getting 6 heads out of 12 flips

Probability of getting 6 heads = P(x= 6; 12; 0.50)= 0.2255

Probability of getting atleast 6 heads p (x >=6; 12; 0.5) = 0.6127

so this is unbiased towards any tail as this is not under 0.05 level significance. Null hypothesis is correct and coin is fair.

Similarly, for getting 7 heads out of 12 flips

Probability of getting 7 heads = P(x= 7; 12; 0.50)= 0.1933

Probability of getting atleast 7 heads p (x >= 7; 12; 0.5) = 0.3872

so this is unbiased towards any tail as this is not under 0.05 level significance. Null hypothesis is correct and coin is fair

(f) Expected number of Heads for 6 flips = 6 * 0.5 = 3 Heads

(g) P ( X<=3; 17; 0.5) = (0.5)17 [17C0 + 17C1 + 17C2 + 17C3] = 0.0063 = P ( X>=14; 17; 0.5)

(h) Most likely number of heads in 20 flips = number of trials * its probability = 20 * 0.25 = 5

P( X>=9. 20, 0.25) = we can calculate it by either bionomial or poisson distribution

P( X>=9. 20, 0.25) = 0.04090

P( X<=3, 20, 0.25) = 20Cn * (0.25)n(0.75)20-n

by bionomial distribution calculator

P( X<=3, 20, 0.25) = 0.2251

(i) in general , around 5% of the samples of fair coin flips will result in rejecting the nulll hypothesis because significance level is 0.05 or say 5 %.

Number of heads p value null hypothesis Decision rule 5 0.8061 Valid greater than significance level 5 0.8061 Valid greater than significance level 5 0.8061 Valid greater than significance level 6 0.6127 Valid greater than significance level 7 0.3872 Valid greater than significance level

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