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)?

Step by step solution

01

 Part (a) Step 1: Given information

Sample size $=n=25$

Level of significance $=\alpha =0.05$

02

Part (a) Step 2: Concept

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

03

Part (a) Step 3: Calculation

The null and alternative hypotheses:

$$\begin{gathered} H_0:p1=p2=\frac{1}{2}=0.5 \\ {H}_a:Atleastoneofthe{p}_i’sisincorrect. \end{gathered}$$

Expected values can be found as,

$$\begin{gathered} E_1={np}_1=250\times 0.5=125 \\ {E}_2={np}_2=250\times 0.5=125 \end{gathered}$$

Therefore, test statistic is,

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

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

$P-$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.

04

Part (b) Step 1: Calculation

The null and alternative hypotheses:

$$\begin{gathered} H_0:p=0.5 \\ {H}_a:p\ne 0.5 \end{gathered}$$

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

Also, $P-$value is same for both tests.

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