Q.5 - The Practice of Statistics for AP

Question

Roulette an American roulette wheel has $18$ red slots among its $38$ slots.In a random sample of $50$pins,the ball lands in a red slot $31$ times.

(a) Do the data give convincing evidence that the wheel is unfair? Carry out an appropriate test at the$\alpha =0.05$ significance level to help answer this question.

(b) The casino manager uses your data to produce a $99\%$ confidence interval for $p$ and gets $\left(0.44,0.80\right)$. He says that this interval provides convincing evidence that the wheel is fair. How do you respond?

- If conditions are met, conduct a one-sample \(t\) test about a population mean \(\mu\).

Step-by-Step Solution

Answer

(a) There is convincing evidence to prove that the wheel is unfair.

(b)The casino manager is right in saying that $99\%$ confidence interval is convincing to prove that wheel is fair.

Part (a) Step 1; Given Information

Given In the question that,

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 $=31$. we have to find Do the data give convincing evidence that the wheel is unfair.

Part (a) Step 2: Explanation

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

 $p=\frac{18}{38}=0.4737$

The sample size $n=50$

calculate the null hypothesis and alternative hypothesis as follows:

The null hypotheis $H_0=p$

                                $H_0=p=0.4737$

The Alternative hypothesis $H_1\ne 0.4737$

Find the sample proportion $\hat{p}$ as follows:      $$\begin{gathered} \hat{p}=\frac{numuber of success}{sample size} \\ =\frac{31}{50} \\ =0.62 \end{gathered}$$

compute the test statistic $z$ as follows:

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

  $$\begin{gathered} =\frac{0.62-0.4737}{\sqrt{{\displaystyle \frac{0.4737\left(1-0.4737\right)}{50}}}} \\ =\frac{0.1463}{\sqrt{{\displaystyle \frac{0.2493}{50}}}} \\ =\frac{0.1463}{\sqrt{0.004986}} \\ =\frac{0.1463}{0.07061} \\ =2.071 \end{gathered}$$

the value of test statistics is $2.071$

use the test statistics to find the  $p$ value gives the evidence whether to accept or reject the null hypothesis.

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

$P\left(z<-2.071 or z>-2.071\right)$ it can be written as $2P\left(z<-2.071\right)$.

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

$2P\left(z<-2.071\right)=0.0384$

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 $H_1:P\ne 0.4737$.

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

  

Part (b) Step 1: Given Information

Given in the question that 

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 $\left(0.44,0.80\right)$ we have to find that He says that this interval provides convincing evidence that the wheel is fair. How do you respond.

Part (b) Step 2: Explanation

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 $p=\frac{18}{38}=0.4737$. S

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

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

compute the value of test statistics as:

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

given by $z=2.071$ and $p$ value is$\alpha =0.0384$

Reject the 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 the null hypothesis $H_0:P=0.4737$.

Therefore, the alternate hypothesis is true $H_1:P\ne 0.4737$

At a $95\%$ confidence interval, this means there is compelling evidence that the wheel is unfair.

At a $99\%$ confidence interval, the same evidence would lead to the contradiction.

As a result, the casino management is correct in claiming that the $99\%$ confidence interval is sufficient to establish that the wheel is fair.