Respuesta :
Answer:
Is plausible that the successive throws are independent
Step-by-step explanation:
1) Table with info given
The observed values are given by the following table
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First shot Made Second shot missed Total
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Made 152 33 185
Missed 37 8 45
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Total 189 41 230
2) Calculations and test
We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:
Null hypothesis: Independence between two successive free throws
Alternative hypothesis: No Independence between two successive free throws
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First shot Made Second shot missed
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Made 189(185)/230=152.0217 41(185)/230=32.9783
Missed 189(45)/230=36.9783 41(45)/230=8.0217
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On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.
3) Calculate the chi square statistic
The statistic for this case is given by:
[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]
Where O represent the observed values and E the expected values. Replacing the values that we got we have this
[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]
Now with the calculated value we can find the degrees of freedom
[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.
Now we can calculate the p value
[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]
On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.
