91Ó°ÊÓ

$P=0.0384$

Given claim: Roulette wheel is fair and therefore the ball lends in a read slot $18$ times out of$38$on average. }

The claim is either the null hypothesis or the alternative hypothesis. The null hypotheses statement is that the population mean is equal to the value given in the claim. If the null hypothesis is the claim then the alternative hypothesis statement is the opposite of null hypothesis.

$H_0:p=\frac{18}{38}=0.4737$

$H_1:p≠0.4737$

The P-value is the probability of getting the value of the test statistic or a value more extreme, when the null hypothesis is true.

There is a $0.0384$ probability that get$31$successes among $50$ trials or more extreme, when the roulette wheel is fair.

$\mathrm{Î\pm }=0.05$

$n=50$

$x=31$

$\text{Given claim is that the proportion is}18\text{out of every}38$

The claim is either the null hypothesis or the alternative hypothesis. The null hypotheses statement is that the population mean is equal to the value given in the claim. If the null hypothesis is the claim then the alternative hypothesis statement is the opposite of null hypothesis.

$H_0:p=\frac{18}{38}=0.4737$

$H_1:p≠0.4737$

Conditions

The three conditions are: Random, independent, Normal (large counts)

Random: Satisfied, because it is safe to assume that the different spins of the wheel are random.

Independent: satisfied, the reason is that the sample of $50$spins is less than $10\%$of the population of all spins (assuming that there are more than $500$spins with the roulette wheel).

Normal: Satisfied, because

$n{p}_0=50(0.4737)=23.685\text{and}n\left(1-p_0\right)=50(1-0.4737)=50(0.5263)=26.315\text{are both at least}10$

Since all condition are satisfied, it is suitable to use a hypothesis test for the population proportion $P$

Hypothesis test

$\hat{p}=\frac{x}{n}$

$=\frac{31}{50}$

$=0.62$

The test- statistic is

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

$=\frac{0.62-0.4737}{\sqrt{\frac{0.4737(1-0.4737)}{50}}}=2.07$

The P-value is the probability of getting the value of the test statistic, or a value more extreme, when the null hypothesis is true. Find the P-value using the normal probability table

$P=P(Z<-2.07\text{or}Z>2.07)$

$=2P(Z<-2.07)$

$=2(0.0192)$

$=0.0384$

If the P-value is lesser than the significance level $\mathrm{Î\pm }$then reject the null hypothesis:

$P<0.05⇒\mathrm{R}\text{eject}{H}_0$

There is enough convincing proof that the American roulette wheel is not fair.

$99\%\text{confidence level:}(0.44,0.80)$

$p=\frac{18}{38}$

$=0.4737$

The Researcher is correct.

A $99\%$confidence interval associates with a significance test at the $\mathrm{Î\pm }=0.01$ level.

A $95\%$confidence interval associates with a significance test at the $\mathrm{Î\pm }=0.05$level.

The significance test at the $\mathrm{Î\pm }=0.05$level and therefore the associating $95\%$confidence interval, would lead to the opposite conclusion. Thus the there is not enough convincing evidence.