In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 times he missed both, 152 times he made both, 33 times he only made the first, and 37 times he made the second. Is it plausible that the successive free throws are independent?

Answer:Is plausible that the successive throws are independentStep-by-step explanation:1) Table with info given The observed values are given by the following table__________________________________________________First shot          Made          Second shot missed           Total__________________________________________________Made                  152                       33                                185Missed                37                         8                                  45__________________________________________________Total                    189                       41                                2302) Calculations and testWe 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 throwsAlternative hypothesis: No Independence between two successive free throws_____________________________________________________First shot                    Made                                    Second shot missed_____________________________________________________Made                  189(185)/230=152.0217                41(185)/230=32.9783Missed                189(45)/230=36.9783                  41(45)/230=8.0217_____________________________________________________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 statisticThe 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.