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Yakvenalex [24]
3 years ago
11

Jane is playing a game in which he spins a spinner with 6 equal-sized slices numbered 1 through 6. The spinner stops on a number

ed slice at random This game is this: Jane spins the spinner once. She wins $1 if the spinner stops on the number 1, $4 if the spinner stops on the number 2, $7 if the spinner stops on the number 3, and $10 if the spinner stops on the number 4. She loses $8.75 if the spinner stops on 5 or 6.
Find the expected value of playing the game.
Mathematics
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

The expected value of playing the game is $0.75.

Step-by-step explanation:

The expected value of a random variable is the weighted average of the random variable.

The formula to compute the expected value of a random variable <em>X</em> is:

E(X)=\sum x\cdot P(X=x)

The random variable <em>X</em> in this case can be defined as the amount won in playing the game.

The probability distribution of <em>X</em> is as follows:

Number on spinner:   1           2           3          4           5              6

Amount earned (<em>X</em>):   $1        $4         $7        $10     -$8.75     -$8.75

Probability:                 1/6       1/6         1/6        1/6         1/6           1/6

Compute the expected value of <em>X</em> as follows:

E(X)=\sum x\cdot P(X=x)

         =(1\times \frac{1}{6})+(4\times \frac{1}{6})+(7\times \frac{1}{6})+(10\times \frac{1}{6})+(-8.75\times \frac{1}{6})+(-8.75\times \frac{1}{6})

         =\frac{1}{6}+\frac{4}{6}+\frac{7}{6}+\frac{10}{6}-\frac{8.75}{6}-\frac{8.75}{6}

         =\frac{1+4+7+10-8.75-8.75}{6}

         =0.75

Thus, the expected value of playing the game is $0.75.

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Answer:

z=\frac{0.224 -0.2}{\sqrt{\frac{0.2(1-0.2)}{624}}}=1.499  

p_v =P(Z>1.499)=0.0669  

If we compare the p value obtained and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly higher than 0.2 or 20%.  

Step-by-step explanation:

How would you make the decision if you were Callaway management? Would you use hypothesis testing?

The best way to test the claim if with a proportion test. The procedure is explained below.

1) Data given and notation

n=624 represent the random sample taken

X=140 represent the golf ball purchasers will buy a Callaway golf ball

\hat p=\frac{140}{624}=0.224 estimated proportion of golf ball purchasers will buy a Callaway golf ball

p_o=0.2 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is more than 0,2 or 20%:  

Null hypothesis:p\leq 0.2  

Alternative hypothesis:p > 0.2  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.224 -0.2}{\sqrt{\frac{0.2(1-0.2)}{624}}}=1.499  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level is not provided but let's assume \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(Z>1.499)=0.0669  

If we compare the p value obtained and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly higher than 0.2 or 20%.  

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