Answer to Question #260604 in Statistics and Probability for boo

Question

A pinball machine has five holes of varying sizes in which the ball may land. The payoff for the

ball landing in a given hole indicates that the probabilities of success for various holes should be

0.28, 0.24, 0.19, 0.15 and 0.14.

A contestant plays the game 42 times and obtains the following frequencies: 15, 10, 7, 6 and 4. Test

at a 5% level of significance whether the probabilities corresponding to the payoffs are correct.

Answer 1

This question requires us to perform a goodness of fit test. To perform this test, we shall use the Chi-square distribution for goodness of fit.

The hypotheses tested are,

$H_0:$  probabilities corresponding to the payoffs are correct.

$Against$

$H_1:$  probabilities corresponding to the payoffs are not correct.

The expected frequencies $(E_i)$ are first determined using the formula,

$E_i=n*p_i$ where $n=15+10+7+6+4=42$ and $p_i$ are the probabilities of success for each observed count $O_i$ .

The expected frequencies are given as,

$E_1=n*p_1=42*0.28=11.76$

$E_2=n*p_2=42*0.24=10.08$

$E_3=n*p_3=42*0.19=7.98$

$E_4=n*p_4=42*0.15=6.3$

$E_5=n*p_5=42*0.14=5.88$

A summary of the above information is given in the table below

$O_i$ $p_i$ $E_i$

15 0.28 11.76

10 0.24 10.08

7 0.19 7.98

6 0.15 6.3

4 0.14 5.88

Where $i=1,2,3,4,5$ is the number of holes.

The test statistic is given as,

$\chi^2=\displaystyle\sum^5_{i=1}(O_i-E_i)^2/E_i$

$\chi^2=(15-11.76)^2/11.76+(10-10.08)^2/10.08+(7-7.98)^2/7.98+(6-6.3)^2/6.3+(4-5.88)^2/5.88= 1.629013$

$\chi^2$ is compared with the table value at $\alpha=5\%$ with $(i-1)=5-1=4$. The table value is $\chi^2_{\alpha,4}=\chi^2_{0.05,4}=9.48773$ and the null hypothesis is rejected if $\chi^2\gt \chi^2_{0.05,4}$

Since $\chi^2=1.629013\lt \chi^2_{0.05,4}=9.48773$, we fail to reject the null hypothesis and we conclude that sufficient evidence exist to support the claim that probabilities corresponding to the payoffs are correct at 5% level of significance.