Chapter 9, Problem 5CRE

Solution

(a)

To perform: an appropriate test to at $\alpha =0.05$ significance level to find if the wheel is unfair.

Given Information:

The number of slots in the American roulette wheel = $38$

The number of red slots = $18$

The sample size $n=50$

In a random sample of $50$ , the number of times the ball lands in red slot = $31$ .

Concept used: Describe the null hypothesis and alternate hypothesis. On the basis of test statistics reject or accept the null hypothesis.

Solution: Let $p$ is the proportion of red slots in the wheel.

The value of $p$ is calculated as $p=\frac{18}{38}$ . Simplify to get the value $p=0.4737$

The sample size $n=50$

In a random sample of $50$ , the number of times the ball lands in red slot = $31$ .

Forming the hypothesis:

Describe the null hypothesis and the alternate hypothesis as follows

The null hypothesis $H_0=p$

Rewrite the null hypothesis as $H_0=p=0.4737$

The alternate hypothesis $H_1\ne 0.4737$

Find the sample proportion $\widehat{p}$ as follows

Formula used:

  $\begin{array}{c}\text{Sample Proportion <mover accent="true"> <mo>p </mo><mo>^</mo></mover> =}\frac{\text{number of successes}}{\text{Sample Size}}\\ =\frac{31}{50}\\ =0.62\end{array}$

Formula used:

Describe the test statistics $z$ as follows

  $z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Calculations:

  $\begin{array}{c}z=\frac{0.62-0.4737}{\sqrt{\frac{0.4737\left(1-0.4737\right)}{50}}}\\ =\frac{0.1463}{\sqrt{\frac{0.4737\left(0.5263\right)}{50}}}\end{array}$

  $\begin{array}{c}=\frac{0.1463}{\sqrt{\frac{0.2493}{50}}}\\ =\frac{0.1463}{\sqrt{0.004986}}\\ =\frac{0.1463}{0.07061}\\ =2.071\end{array}$

The value of test statistics $z=2.071$

Interpretation:

Use the test statistics to find the P- value. The P- value gives the evidence whether to accept or reject the null hypothesis.

The alternate hypothesis is $H_1\ne 0.4737$ , therefore, use two tail tests.

  $P\left(z<-2.071\text{or }z>-2.071\right)$ , since the distribution is normally distributed, therefore, rewrite

  $P\left(z<-2.071\text{or }z>-2.071\right)=2P\left(z<-2.071\right)$

Using tables, write the value at $\alpha =0.05$ level of significance.

  $2P\left(z<-2.071\right)=0.0384$

Result:

Reject null hypothesis if the level of significance is more than P- value.

Here level of significance $\alpha =0.05$ and $0.0384<0.05$ .

Reject null hypothesis $H_0=p=0.4737$ .

Therefore, alternate hypothesis is true $H_1\ne 0.4737$ .

This implies that there is convincing evidence to prove that the wheel is unfair.

(b)

To check: if the data at $99\%$ confidence interval proves the wheel is fair.

Given Information:

The casino manager uses the data to produce $99\%$ confidence interval for $p$ which is proportion for the red slot .

The manager gets confidence interval $\left(0.44,0.80\right)$ .

Proof:

The number of slots in the American roulette wheel = $38$

The number of red slots = $18$

Let $p$ is the proportion of red slots in the wheel.

The value of $p$ is calculated as $p=\frac{18}{38}$ . Simplify to get the value $p=0.4737$

The $99\%$ confidence interval gives the level of significance $\alpha =0.01$ .

The level of significance is $\alpha =0.05$ ,

Find the value of test statistics as

Formula used:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$ given by $z=2.071$ and the P − value is $\alpha =0.0384$

Interpretation:

Reject null hypothesis if the level of significance is less than P- value.

Here level of significance $\alpha =0.05$ and $0.0384<0.05$ .

Reject null hypothesis $H_0=p=0.4737$ .

Therefore, alternate hypothesis is true $H_1\ne 0.4737$ .

Result:

This implies that there is convincing evidence that wheel is unfair at $95\%$ confidence interval.

The same evidence would lead to contradiction at $99\%$ confidence interval.

Therefore, the casino manager is right in saying that $99\%$ confidence interval is convincing to prove that wheel is fair.