Question #260604

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.

1
Expert's answer
2021-11-10T15:38:18-0500

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,

H0:H_0:  probabilities corresponding to the payoffs are correct.

AgainstAgainst

H1:H_1:  probabilities corresponding to the payoffs are not correct.

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

Ei=npiE_i=n*p_i where n=15+10+7+6+4=42n=15+10+7+6+4=42 and pip_i are the probabilities of success for each observed count OiO_i .

The expected frequencies are given as,

E1=np1=420.28=11.76E_1=n*p_1=42*0.28=11.76

E2=np2=420.24=10.08E_2=n*p_2=42*0.24=10.08

E3=np3=420.19=7.98E_3=n*p_3=42*0.19=7.98

E4=np4=420.15=6.3E_4=n*p_4=42*0.15=6.3

E5=np5=420.14=5.88E_5=n*p_5=42*0.14=5.88

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

OiO_i pip_i EiE_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=1,2,3,4,5i=1,2,3,4,5 is the number of holes.

The test statistic is given as,

χ2=i=15(OiEi)2/Ei\chi^2=\displaystyle\sum^5_{i=1}(O_i-E_i)^2/E_i

χ2=(1511.76)2/11.76+(1010.08)2/10.08+(77.98)2/7.98+(66.3)2/6.3+(45.88)2/5.88=1.629013\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

χ2\chi^2 is compared with the table value at α=5%\alpha=5\% with (i1)=51=4(i-1)=5-1=4. The table value is χα,42=χ0.05,42=9.48773\chi^2_{\alpha,4}=\chi^2_{0.05,4}=9.48773 and the null hypothesis is rejected if χ2>χ0.05,42\chi^2\gt \chi^2_{0.05,4}

Since χ2=1.629013<χ0.05,42=9.48773\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.


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