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Answers without the blur. Sign up and see all textbooks for free! Q 16.

Expert-verified Found in: Page 718 ### The Practice of Statistics for AP Examination

Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339 # Spinning heads? When a fair coin is flipped, we all know that the probability the coin lands on heads is $0.50$ However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin $250$ times. It landed heads $140$ times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met.a. Carry out a chi-square test for goodness of fit to test if heads and tails are equally likely when a euro coin is spun.b. In Chapter $9$ Exercise $50$ you analyzed these data with a one-sample $z$ test for a proportion. The hypotheses were ${H}_{0}:p=0.5$ and ${H}_{a}:p\ne 0.5$where $p=$the true proportion of heads. Calculate the $z$ statistic and P-value for this test. How do these values compare to the values from part (a)?

Part (a) When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

Part (b) When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

See the step by step solution

## Part (a) Step 1: Given information

Sample size $=n=25$

Level of significance $=\alpha =0.05$

## Part (a) Step 2: Concept

Test statistic: ${\chi }^{2}=\Sigma \frac{{\left(O-E\right)}^{2}}{E}$

## Part (a) Step 3: Calculation

The null and alternative hypotheses:

${H}_{0}:p1=p2=\frac{1}{2}=0.5\phantom{\rule{0ex}{0ex}}{H}_{a}:Atleastoneofthe{p}_{i}’sisincorrect.$

Expected values can be found as,

${E}_{1}={np}_{1}=250×0.5=125\phantom{\rule{0ex}{0ex}}{E}_{2}={np}_{2}=250×0.5=125$

Therefore, test statistic is,

${\chi }^{2}=\frac{{\left(140-125\right)}^{2}}{125}+\frac{{\left(110-125\right)}^{2}}{125}=3.6$

P-value using excel formula, $=CHIDIST\left(3.6,1\right)$

$P-$value $=0.0578$

Decision: $P-value>0.05,failtoreject{H}_{0}$

When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

## Part (b) Step 1: Calculation

The null and alternative hypotheses:

${H}_{0}:p=0.5\phantom{\rule{0ex}{0ex}}{H}_{a}:p\ne 0.5$

Using excel, Decision: $P-value>0.05,failtoreject{H}_{0}$

When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

Here, ${z}^{2}=\left(1.90\right)2=3.6={\chi }^{2}$

Also, $P-$value is same for both tests. ### Want to see more solutions like these? 