Chapter 11.1, Problem 16E

a)

To determine

To perform chi-square test.

Answer to Problem 16E

There is no convincing evidence that heads and tails are not equally likely when an euro coin is spin.

Explanation of Solution

Given:

Sample size = n = 25-

Level of significance = $\alpha$ = 0.05

Formula:

Test statistic:

  ${\chi }^2={\displaystyle \sum \frac{{(O-E)}^2}{E}}$

Calculation:

The null and alternative hypotheses:

  $H_0:p_1=p_2=\frac{1}{2}=0.5$

  $H_a:Atleastoneofthe{p}_i’sisincorrect.$

Expected values can be found as,

  $\begin{array}{l}{E}_1=n{p}_1=250\times 0.5=125\\ E_2=n{p}_2=250\times 0.5=125\end{array}$

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, fail to reject H0.

Conclusion: There is no convincing evidence that heads and tails are not equally likely when an euro coin is spin.

b)

To determine

To perform z-test for proportion.

Answer to Problem 16E

There is no convincing evidence that heads and tails are not equally likely when an euro coin is spin.

Explanation of Solution

Given:

Sample size = n = 25-

Level of significance = $\alpha$ = 0.05

Number of successes = $x$ = 140

Calculation:

The null and alternative hypotheses:

  $H_0:p=0.5$

  $H_a:p\ne 0.5$

Using excel,

Observed	Hypothesized
0.56	0.5	p (as decimal)
140/250	125/250	p (as fraction)
140.	125.	X
250	250	n
	0.0316	std. error
	1.90	z
	.0578	p-value (two-tailed)

Decision: P-value > 0.05, fail to reject H0.

Conclusion: There is no convincing evidence that heads and tails are not equally likely when an euro coin is spin.

Here, z²= (1.90)2=3.6= ${\chi }^2$

Also, P-value is same for both tests.